Sa ch10 se2 compressed

‫اﻟﻜﻴﻤﻴﺎء‬

‫ﻟﻠﺼﻒ ا ول اﻟﺜﺎﻧﻮي‬



´ÉÑj ’h Ék fÉ› ´Rƒj

ájOƒ©°ùdG á«Hô©dG áµ∏ªªdÉH º«∏©àdGh á«HôàdG IQGRh äQôb É¡à≤Øf ≈∏Y ¬©ÑWh ÜÉàµdG Gòg ¢ùjQóJ

á«ÑjôéàdG á©Ñ£dG Ω 2009 - `g 1430

Original Title:

Chemistry Matter and Change By: Thandi Buthelezi Cheryl Wistrom Nicholas Hainen Laurel Dingrando Dinah Zike

‫اﻟﻜﻴﻤﻴﺎء‬ ôjƒ£àdGh çÉëHCÓd ¿Éµ«Ñ©dG ácô°T :á«Hô©dG áî°ùædG sóYCG áeAGƒªdGh á©LGôªdGh ôjôëàdG

¬fhGô£dG ˆG É£Y ≈°Sƒe …ô°ShódG ºLôW øH óªëe øH ô°UÉf ¢SÉÑYO º«∏°S ôªY …ƒ¨∏dG ôjôëàdGh Öjô©àdG

ø«°ü°üîàªdG øe áÑîf Qƒ°üdG OGóYEG

êGôØdG õjõ©dGóÑY øH Oƒ©°S .O ±Gô°TE’G

™«aQ óªëe óªMCG .O

www.macmillanmh.com

English Edition Copyright © 2008 the McGraw-Hill Companies, Inc. All rights reserved.

Arabic Edition is published by Obeikan under agreement with The McGraw-Hill Companies, Inc. © 2008.

.Ω2008 ,©πghôLÉe ácô°ûd áXƒØëe ájõ«∏éfE’G á©Ñ£dG ¥ƒ≤M Qɪãà°SÓd ¿Éµ«Ñ©dG áYƒªée :á«Hô©dG á©Ñ£dG .`g1429 /Ω2008 ©πghôLÉe ácô°T ™e É¡à«bÉØJ’ Ék≤ah

øjõîàdG hCG ,π«é°ùàdG hCG ,z»HƒcƒJƒa{ ï°ùædÉH ôjƒ°üàdG ∂dP »a ɪH ,᫵«fɵ«e hCG á«fhôàµdEG âfÉcCG Ak Gƒ°S ,ᣰSGh hCG πµ°T …CG »a ¬∏≤f hCG ÜÉàµdG Gòg QGó°UEG IOÉYEÉH íª°ùj ’ .ô°TÉædG øe »£N ¿PEG ¿hO ,´ÉLôà°S’G h

‫َا ْﻟ ُﻤ َﻘ ﱢﺪ َﻣ ُﺔ‬

‫ﹺ‬ ‫ﺍﻟﺮ ﹾﺣ ﹺ‬ ‫ﺍﻟﺮﺣﻴﻢﹺ‬ ‫ﻤﻦ ﱠ‬ ‫ﺑﹺ ﹾﺴﻢﹺ ﺍﻟﻠﻪ ﱠ‬

‫ﻳﺄﺗﻲ ﺍﻫﺘﻤﺎﻡ ﺍﻟﻤﻤﻠﻜﺔ ﺍﻟﻌﺮﺑﻴﺔ ﺍﻟﺴﻌﻮﺩﻳﺔ ﺑﺘﻄﻮﻳﺮ ﻣﻨﺎﻫﺞ ﺍﻟﺘﻌﻠﻴﻢ ﻭﺗﺤﺪﻳﺜﻬﺎ ﻓﻲ ﺇﻃﺎﺭ ﺍﻟﺨﻄﺔ ﺍﻟﻌﺎﻣﺔ‬

‫ﻟﻠﻤﻤﻠﻜﺔ‪ ،‬ﻭﺳﻌﻴﻬﺎ ﺇﻟﻰ ﻣﻮﺍﻛﺒﺔ ﺍﻟﺘﻄﻮﺭﺍﺕ ﺍﻟﻌﺎﻟﻤﻴﺔ ﻋﻠﻰ ﻣﺨﺘﻠﻒ ﺍﻷﺻﻌﺪﺓ‪.‬‬

‫ﻭﻳﺄﺗﻲ ﻛﺘﺎﺏ ﺍﻟﻜﻴﻤﻴﺎﺀ ﻟﻠﺼﻒ ﺍﻷﻭﻝ ﺍﻟﺜﺎﻧﻮﻱ ﻓﻲ ﺇﻃﺎﺭ ﻣﺸﺮﻭﻉ ﺗﻄﻮﻳﺮ ﻣﻨﺎﻫﺞ ﺍﻟﻜﻴﻤﻴﺎﺀ‪ ،‬ﺍﻟﺬﻱ‬

‫ﻳﻬﺪﻑ ﺇﻟﻰ ﺇﺣﺪﺍﺙ ﺗﻄﻮﺭ ﻧﻮﻋﻲ ﻓﻲ ﺗﻌﻠﻴﻢ ﻭﺗﻌﻠﻢ ﺍﻟﻜﻴﻤﻴﺎﺀ‪ ،‬ﻳﻜﻮﻥ ﻟﻠﻄﺎﻟﺐ ﻓﻴﻬﺎ ﺍﻟﺪﻭﺭ ﺍﻟﺮﺋﻴﺴﻲ‬

‫ﻭﺍﻟﻤﺤﻮﺭﻱ ﻓﻲ ﻋﻤﻠﻴﺘﻲ ﺍﻟﺘﻌﻠﻴﻢ ﻭﺍﻟﺘﻌﻠﻢ‪ .‬ﻭﻗﺪ ﺟﺎﺀ ﻫﺬﺍ ﺍﻟﻜﺘﺎﺏ ﻓﻲ ﺟﺰﺃﻳﻦ‪ ،‬ﻳﻀﻢ ﺍﻟﻔﺼﻞ ﺍﻟﺪﺭﺍﺳﻲ‬ ‫ﺍﻟﺜﺎﻧﻲ ﻓﺼﻠﻴﻦ ﺍﺷﺘﻤﻠﺖ ﻋﻠﻰ‪ :‬ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﻟﻤﻮﻝ‪.‬‬

‫ﻭﻗﺪ ﺟﺎﺀ ﻋﺮﺽ ﻣﺤﺘﻮ￯ ﺍﻟﻜﺘﺎﺏ ﺑﺄﺳﻠﻮﺏ ﻣﺸﻮﻕ‪ ،‬ﻭﺗﻨﻈﻴﻢ ﺗﺮﺑﻮﻱ ﻓﺎﻋﻞ‪ ،‬ﻳﻌﻜﺲ ﺗﻮﺟﻬﺎﺕ‬

‫ﺍﻟﻤﻨﻬﺞ ﻭﻓﻠﺴﻔﺘﻪ‪ ،‬ﻭﻳﺘﻤﺜﻞ ﻓﻲ ﺩﻭﺭﺓ ﺍﻟﺘﻌﻠﻢ‪ .‬ﻭﻗﺪ ﻛﺘﺐ ﺑﺄﺳﻠﻮﺏ ﻳﺴﺎﻋﺪ ﺍﻟﻄﺎﻟﺐ ﻋﻠﻰ ﺗﻨﻤﻴﺔ ﻣﻬﺎﺭﺍﺕ‬ ‫ﺍﻟﺘﺤﻠﻴﻞ ﻭﺍﻟﺘﻔﺴﻴﺮ ﻭﺍﻻﺳﺘﻨﺘﺎﺝ ﻭﺍﻟﺘﻌﺒﻴﺮ‪ ،‬ﻭﺫﻟﻚ ﻣﻦ ﺧﻼﻝ ﺍﻫﺘﻤﺎﻣﻪ ﺑﺎﻟﺠﺎﻧﺐ ﺍﻟﺘﺠﺮﻳﺒﻲ‪ .‬ﻛﺬﻟﻚ ﺍﺷﺘﻤﻞ‬

‫ﺍﻟﻤﺤﺘﻮ￯ ﻋﻠﻰ ﺃﻧﺸﻄﺔ ﻣﺘﻨﻮﻋﺔ ﺍﻟﻤﺴﺘﻮ￯‪ ،‬ﺗﺘﺴﻢ ﺑﺈﻣﻜﺎﻧﻴﺔ ﺗﻨﻔﻴﺬ ﺍﻟﻄﻠﺒﺔ ﻟﻬﺎ‪ ،‬ﻭﺗﺮﺍﻋﻲ ﻓﻲ ﺍﻟﻮﻗﺖ ﻧﻔﺴﻪ‬ ‫ﹰ‬ ‫ﻭﺃﺷﻜﺎﻻ ﻭﺭﺳﻮ ﹰﻣﺎ ﺗﻮﺿﻴﺤﻴﺔ ﻣﻌﺒﺮﺓ ﺗﻌﻜﺲ‬ ‫ﺻﻮﺭﺍ‬ ‫ﻣﺒﺪﺃ ﺍﻟﻔﺮﻭﻕ ﺍﻟﻔﺮﺩﻳﺔ ﺑﻴﻨﻬﻢ‪ ،‬ﺑﺎﻹﺿﺎﻓﺔ ﺇﻟﻰ ﺗﻀﻤﻴﻨﻪ‬ ‫ﹰ‬ ‫ﻃﺒﻴﻌﺔ ﺍﻟﻔﺼﻞ‪ ،‬ﻣﻊ ﺣﺮﺹ ﺍﻟﻜﺘﺎﺏ ﻋﻠﻰ ﻣﺒﺪﺃ ﺍﻟﺘﻘﻮﻳﻢ ﺍﻟﺘﻜﻮﻳﻨﻲ ﻓﻲ ﻓﺼﻮﻟﻪ ﻭﺩﺭﻭﺳﻪ ﺍﻟﻤﺨﺘﻠﻔﺔ‪.‬‬

‫ﻛﻤﺎ ﺃﻛﺪﺕ ﻓﻠﺴﻔﺔ ﺍﻟﻜﺘﺎﺏ ﺃﻫﻤﻴﺔ ﺇﻛﺴﺎﺏ ﺍﻟﻄﺎﻟﺐ ﺍﻟﻤﻨﻬﺠﻴﺔ ﺍﻟﻌﻠﻤﻴﺔ ﻓﻲ ﺍﻟﺘﻔﻜﻴﺮ ﻭﺍﻟﻌﻤﻞ‪ ،‬ﻭﺗﺰﻭﻳﺪﻩ‬

‫ﺑﻤﻬﺎﺭﺍﺕ ﻋﻘﻠﻴﺔ ﻭﻋﻤﻠﻴﺔ ﺿﺮﻭﺭﻳﺔ‪ ،‬ﻣﺜﻞ‪ :‬ﺍﻷﻧﺸﻄﺔ ﺍﻻﺳﺘﻬﻼﻟﻴﺔ‪ ،‬ﻭﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﻌﻠﻤﻴﺔ ﺍﻷﺧﺮ￯‪ ،‬ﻭﺍﻹﺛﺮﺍﺀ‬ ‫ﺍﻟﻌﻠﻤﻲ‪ ،‬ﺑﺎﻹﺿﺎﻓﺔ ﺇﻟﻰ ﺣﺮﺻﻬﺎ ﻋﻠﻰ ﺭﺑﻂ ﺍﻟﻤﻌﺮﻓﺔ ﺑﻮﺍﻗﻊ ﺣﻴﺎﺓ ﺍﻟﻄﺎﻟﺐ‪ ،‬ﻣﻦ ﺧﻼﻝ ﺭﺑﻄﻬﺎ ﺑﺎﻟﺮﻳﺎﺿﻴﺎﺕ‪،‬‬

‫ﻭﻓﺮﻭﻉ ﺍﻟﻌﻠﻮﻡ ﺍﻷﺧﺮ￯‪ ،‬ﻭﺍﻟﺘﻘﻨﻴﺔ ﻭﺍﻟﻤﺠﺘﻤﻊ‪.‬‬

‫ﻭﺍﻟﻠ ﹶﻪ ﻧﺴﺄﻝ ﺃﻥ ﻳﺤﻘﻖ ﺍﻟﻜﺘﺎﺏ ﺍﻷﻫﺪﺍﻑ ﺍﻟﻤﺮﺟﻮﺓ ﻣﻨﻪ‪ ،‬ﻭﺃﻥ ﻳﻮﻓﻖ ﺍﻟﺠﻤﻴﻊ ﻟﻤﺎ ﻓﻴﻪ ﺧﻴﺮ ﺍﻟﻮﻃﻦ‬

‫ﻭﺗﻘﺪﻣﻪ ﻭﺍﺯﺩﻫﺎﺭﻩ‪.‬‬

‫ﻓﻬﺮس اﻟﻤﺤﺘﻮﻳﺎت‬ ‫اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫‪Chemical Reactions‬‬

‫اﻟﻔﺼﻞ ‪4‬‬ ‫اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪6 ..........................‬‬ ‫‪ 4-1‬ﺍﻟﺘﻔﺎﻋﻼﺕ ﻭﺍﳌﻌﺎﺩﻻﺕ‪8 ...................... .‬‬

‫‪ 4-2‬ﺗﺼﻨﻴﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪15 ...................‬‬ ‫‪ 4-3‬ﺍﻟﺘﻔﺎﻋﻼﺕ ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‪25 .....................‬‬

‫اﻟﻌﺎﻣﺔ‬

‫اﻟﻔﻜﺮة‬

‫ﹸﺤـــﻮﻝ ﺍﻟﺘﻔـﺎﻋــﻼﺕ ﺍﻟﻜﻴﻤﻴـﺎﺋﻴـﺔ‬ ‫ﺗ‬ ‫ﹼ‬ ‫ﻣﻤـﺎ ﻳﺆﺩﻱ‬ ‫ﺍﻟﻤﺘﻔﺎﻋـﻼﺕ ﺇﻟـﻰ ﻧﻮﺍﺗـﺞ‪ ،‬ﹼ‬ ‫ﺇﻟﻰ ﺇﻃﻼﻕ ﻃﺎﻗﺔ ﺃﻭ ﺍﻣﺘﺼﺎﺻﻬﺎ‪.‬‬ ‫‪ 4-1‬اﻟﺘﻔﺎﻋﻼت واﻟﻤﻌﺎدﻻت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺗﹸﻤﺜﱠـﻞ ﺍﻟﺘﻔﺎﻋـﻼﺕ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﺑﻤﻌـﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴـﺔ‬ ‫ﻣﻮﺯﻭﻧﺔ‪.‬‬

‫‪ 4-2‬ﺗﺼﻨﻴﻒ اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻫﻨﺎﻙ ﺃﺭﺑﻌﺔ ﺃﻧﻮﺍﻉ‬ ‫ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻫﻲ‪:‬‬ ‫ﺍﻟﺘﻜﻮﻳﻦ‪ ،‬ﻭﺍﻻﺣﺘﺮﺍﻕ‪ ،‬ﻭﺍﻟﺘﻔﻜﻚ‪،‬‬ ‫ﻭﺍﻹﺣﻼﻝ‪.‬‬

‫‪≥jôëdG πÑb‬‬

‫‪ 4-3‬اﻟﺘﻔﺎﻋﻼت ﻓﻲ اﻟﻤﺤﺎﻟﻴﻞ اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺗﺤﺪﺙ ﺗﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ ﺑﻴﻦ ﺍﻟﻤﻮﺍﺩ ﻓﻲ‬ ‫ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ‪ ،‬ﻣﻨﺘﺠ ﹰﺔ ﺭﻭﺍﺳﺐ‪،‬‬ ‫ﺃﻭ ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯﺍﺕ‪.‬‬

‫‪≥jôëdG ó©H‬‬

‫ﺣﻘﺎﺋﻖ ﻛﻴﻤﻴﺎﺋﻴﺔ‬

‫• ﻟﻜـﻲ ﻳﺸـﺘﻌﻞ ﺍﻟﺨﺸـﺐ ﻳﺠﺐ ﺃﻥ‬ ‫ﻳﺴﺨﻦ ﺇﻟﻰ ‪.260 °C‬‬ ‫• ﻳﻐﻠﻲ ﺍﻟﻤﺎﺀ ﺍﻟﻤﻮﺟﻮﺩ ﻓﻲ ﺍﻟﺨﺸﺐ‬ ‫ﻗﺒﻞ ﺃﻥ ﻳﺤﺘﺮﻕ ﺍﻟﺨﺸﺐ‪.‬‬ ‫• ﻳﺤﺘـﻮﻱ ﺍﻟﺪﺧـﺎﻥ ﺍﻟﻨﺎﺗـﺞ ﻋـﻦ‬ ‫ﺍﺣﺘـﺮﺍﻕ ﺍﻟﺨﺸـﺐ ﻋﻠـﻰ ﺃﻛﺜـﺮ ﻣﻦ‬ ‫ﻣﺎﺩﺓ ﻛﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫ﻛﻴﻒ ﺗﻌﻤﻞ ﺍﻷﺷﻴﺎﺀ ‪ :‬ﻇﺎﻫﺮﺓ ﺍﻟﺘﺄﻟﻖ ﺍﳊﻴﻮﻱ‪34 ........‬‬

‫‪6‬‬

‫اﻟﻤﻮل‬

‫‪The Mole‬‬

‫اﻟﻔﺼﻞ ‪5‬‬

‫اﻟﻌﺎﻣﺔ‬

‫اﻟﻔﻜﺮة‬

‫ﻛﺒﻴﺮﺍ ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ‬ ‫ﻳﻤﺜﻞ ﺍﻟﻤﻮﻝ ﻋﺪﺩﹰﺍ ﹰ‬ ‫ﺍﻟﻤﺘﻨﺎﻫﻴـﺔ ﻓـﻲ ﺍﻟﺼﻐﺮ‪ ،‬ﻭﻳﺴـﺘﻌﻤﻞ ﻓﻲ‬ ‫ﺣﺴﺎﺏ ﻛﻤﻴﺎﺕ ﺍﻟﻤﻮﺍﺩ‪.‬‬ ‫‪ 5-1‬ﻗﻴﺎس اﻟﻤﺎدة‬

‫اﻟﻤﻮل ‪40 ...................................................‬‬ ‫‪ 5-1‬ﻗﻴﺎﺱ ﺍﳌﺎﺩﺓ‪42 .........................................‬‬ ‫‪ 5-2‬ﺍﻟﻜﺘﻠﺔ ﻭﺍﳌﻮﻝ‪47 ................................ .‬‬ ‫‪ 5-3‬ﻣﻮﻻﺕ ﺍﳌﺮﻛﺒﺎﺕ‪55 ................................ .‬‬ ‫‪ 5-4‬ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ‪63 .................‬‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺴﺘﻌﻤﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ‬ ‫ﺍﻟﻤﻮﻝ ﻟﻌﺪ ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻭﺍﻷﻳﻮﻧﺎﺕ‪،‬‬ ‫ﻭﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪ 5-2‬اﻟﻜﺘﻠﺔ واﻟﻤﻮل‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺤﺘﻮﻱ ﺍﻟﻤﻮﻝ ﺩﺍﺋﻤﹰ ﺎ‬ ‫ﻋﻠﻰ ﺍﻟﻌﺪﺩ ﻧﻔﺴﻪ ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ‪،‬‬ ‫ﻏﻴﺮ ﺃﻥ ﻣﻮﻻﺕ ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻟﻬﺎ‬ ‫ﻛﺘﻞ ﻣﺨﺘﻠﻔﺔ‪.‬‬ ‫‪ 5-3‬ﻣﻮﻻت اﻟﻤﺮﻛﺒﺎت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﻤﻜﻦ ﺣﺴﺎﺏ ﺍﻟﻜﺘﻠﺔ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺻﻴﻐﺘﻪ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻛﻤﺎ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻝ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺘﺤﻮﻳﻞ ﺍﻟﻜﺘﻠﺔ ﺇﻟﻰ‬ ‫ﻣﻮﻻﺕ ﺍﻟﻤﺮﻛﺐ‪.‬‬

‫‪ 5-4‬اﻟﺼﻴﻐﺔ ا وﻟﻴﺔ واﻟﺼﻴﻐﺔ‬ ‫اﻟﺠﺰﻳﺌﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻟﺼﻴﻐـﺔ ﺍﻟﺠﺰﻳﺌﻴـﺔ‬ ‫ﻟﻤﺮﻛـﺐ ﻣـﺎ ﻫـﻲ ﺃﻛﺒـﺮ ﻣﻀﺎﻋـﻒ‬ ‫ﻟﺼﻴﻐﺘـﻪ ﺍﻷﻭﻟﻴـﺔ‪ ،‬ﻭﺗﻀـﻢ ﺃﻋـﺪﺍﺩﹰ ﺍ‬ ‫ﺻﺤﻴﺤﺔ ﻓﻘﻂ‪.‬‬ ‫‪ 5-5‬ﺻﻴﻎ ا ﻣﻼح اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﻫﻲ‬ ‫ﻣﺮﻛﺒﺎﺕ ﺃﻳﻮﻧﻴﺔ ﺻﻠﺒﺔ ﻓﻴﻬﺎ ﺟﺰﻳﺌﺎﺕ‬ ‫ﻣﺎﺀ ﻣﺤﺘﺠﺰﺓ‪.‬‬

‫ﺣﻘﺎﺋﻖ ﻛﻴﻤﻴﺎﺋﻴﺔ‬

‫• ﺍﻟﻌﻤﻼﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺴﻌﻮﺩﻳﺔ ﻫﻲ‪:‬‬ ‫‪ 100،50،25،10،5‬ﻫﻠﻼﺕ‪.‬‬ ‫• ﺗﺘﺮﻛﺐ ﺍﻟﻌﻤﻼﺕ ﺍﻟﺴﻌﻮﺩﻳﺔ ﻣﻦ‬ ‫ﻧﺤﺎﺱ ﻭﻧﻴﻜﻞ ﺑﻨﺴﺐ ﻣﺨﺘﻠﻔﺔ‪.‬‬

‫‪40‬‬

‫‪ 5-5‬ﺻﻴﻎ ﺍﻷﻣﻼﺡ ﺍﳌﺎﺋﻴﺔ ‪73 .............................‬‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺀ ﻭﺍﳊﻴﺎﺓ‪ :‬ﺍﻟﺘﺎﺭﻳﺦ ﰲ ﻛﺄﺱ ﻣﺎﺀ‪77 .............‬‬

‫‪5‬‬

‫اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫‪Chemical Reactions‬‬

‫اﻟﻌﺎﻣﺔ‬

‫اﻟﻔﻜﺮة‬

‫ﺤـــﻮﻝ ﺍﻟﺘﻔـﺎﻋــﻼﺕ ﺍﻟﻜﻴﻤﻴـﺎﺋﻴـﺔ‬ ‫ﹸﺗ‬ ‫ﹼ‬ ‫ﻣﻤـﺎ ﻳﺆﺩﻱ‬ ‫ﺍﻟﻤﺘﻔﺎﻋـﻼﺕ ﺇﻟـﻰ ﻧﻮﺍﺗـﺞ‪ ،‬ﹼ‬ ‫ﺇﻟﻰ ﺇﻃﻼﻕ ﻃﺎﻗﺔ ﺃﻭ ﺍﻣﺘﺼﺎﺻﻬﺎ‪.‬‬ ‫‪ 4-1‬اﻟﺘﻔﺎﻋﻼت واﻟﻤﻌﺎدﻻت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﹸﺗﻤﺜﱠـﻞ ﺍﻟﺘﻔﺎﻋـﻼﺕ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﺑﻤﻌـﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴـﺔ‬ ‫ﻣﻮﺯﻭﻧﺔ‪.‬‬

‫‪ 4-2‬ﺗﺼﻨﻴﻒ اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻫﻨﺎﻙ ﺃﺭﺑﻌﺔ ﺃﻧﻮﺍﻉ‬ ‫ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻫﻲ‪:‬‬ ‫ﺍﻟﺘﻜﻮﻳﻦ‪ ،‬ﻭﺍﻻﺣﺘﺮﺍﻕ‪ ،‬ﻭﺍﻟﺘﻔﻜﻚ‪،‬‬ ‫ﻭﺍﻹﺣﻼﻝ‪.‬‬

‫‪≥jôëdG πÑb‬‬

‫‪ 4-3‬اﻟﺘﻔﺎﻋﻼت ﻓﻲ اﻟﻤﺤﺎﻟﻴﻞ اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺗﺤﺪﺙ ﺗﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ ﺑﻴﻦ ﺍﻟﻤﻮﺍﺩ ﻓﻲ‬ ‫ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ‪ ،‬ﻣﻨﺘﺠ ﹰﺔ ﺭﻭﺍﺳﺐ‪،‬‬ ‫ﺃﻭ ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯﺍﺕ‪.‬‬

‫ﺣﻘﺎﺋﻖ ﻛﻴﻤﻴﺎﺋﻴﺔ‬

‫• ﻟﻜـﻲ ﻳﺸـﺘﻌﻞ ﺍﻟﺨﺸـﺐ ﻳﺠﺐ ﺃﻥ‬ ‫ﻳﺴﺨﻦ ﺇﻟﻰ ‪.260 °C‬‬ ‫• ﻳﻐﻠﻲ ﺍﻟﻤﺎﺀ ﺍﻟﻤﻮﺟﻮﺩ ﻓﻲ ﺍﻟﺨﺸﺐ‬ ‫ﻗﺒﻞ ﺃﻥ ﻳﺤﺘﺮﻕ ﺍﻟﺨﺸﺐ‪.‬‬ ‫• ﻳﺤﺘـﻮﻱ ﺍﻟﺪﺧـﺎﻥ ﺍﻟﻨﺎﺗـﺞ ﻋـﻦ‬ ‫ﺍﺣﺘـﺮﺍﻕ ﺍﻟﺨﺸـﺐ ﻋﻠـﻰ ﺃﻛﺜـﺮ ﻣﻦ‬ ‫ﻣﺎﺩﺓ ﻛﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪6‬‬

‫‪≥jôëdG ó©H‬‬

‫‪‹Ó¡à°SG •É°ûf‬‬

‫‪‬‬

‫ﻛﻴﻒ ﻧﺴﺘﺪﻝ ﻋﲆ ﺣﺪﻭﺙ ﺗﻐﲑ ﻛﻴﻤﻴﺎﺋﻲ؟‬

‫ﺍﻟﻜﺎﺷـﻒ )ﺍﻟﺪﻟﻴﻞ( ﻣـﺎﺩﺓ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺗﻀﺎﻑ ﺇﻟـﻰ ﺍﻟﻤﻮﺍﺩ ﻓﻲ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺘﻮﺿﺢ ﻣﺘﻰ ﻳﺤﺪﺙ ﺗﻐﻴﺮ‪.‬‬

‫ﺍﻟﺘﻔﺎﻋـــﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـــﺔ ﺍﻋﻤـــﻞ‬ ‫ﺍﻟﻤﻄﻮﻳـﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻟﺘﺴـﺎﻋﺪﻙ ﻋﻠﻰ ﺗﻨﻈﻴﻢ‬ ‫ﺍﻟﻤﻌﻠﻮﻣـﺎﺕ ﺣـﻮﻝ ﻛﻴﻔﻴـﺔ ﺣـﺪﻭﺙ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺗﺼﻨﻴﻔﻬﺎ‪.‬‬

‫ﻃﻮﻟﻴﺎ‪ ،‬ﻋﻠـﻰ ﺃﻥ ﻳﻈﻞ ﺍﻟﻬﺎﻣﺶ‬ ‫ﺍﻟﺨﻄـﻮﺓ ‪ 1‬ﺍﻃﻮ ﻭﺭﻗـﺔ ﹰﹼ‬ ‫ﻣﺮﺋﻴﺎ‪.‬‬ ‫ﺍﻷﻳﺴﺮ ﹰﹼ‬

‫ﻗﺺ ﺍﻟﺠﺰﺀ ﺍﻟﻌﻠﻮﻱ ﺧﻤﺲ ﻗﻄﻊ‪.‬‬ ‫ﺍﻟﺨﻄﻮﺓ ‪ 2‬ﹼ‬ ‫ﺧطﻮات الﻌمﻞ‪:‬‬ ‫‪O‬‬

‫‪PN‬‬

‫‪IKG JH‬‬ ‫‪L KMI N‬‬ ‫‪LJ M‬‬ ‫‪OK NP L O M‬‬ ‫ﺍﻟﻤﺨﺘﺒﺮ‪.‬‬ ‫ﺍﻟﺴﻼﻣﺔ ﻓﻲ‬ ‫ﻧﻤﻮﺫﺝ‬ ‫‪ .1‬ﺍﻗﺮﺃ‬ ‫‪ .2‬ﻗﺲ ‪ 10.00 ml‬ﻣﻦ ﺍﻟﻤﺎﺀ ﺍﻟﻤﻘﻄﺮ ﻓﻲ ﻣﺨﺒﺎﺭ ﻣﺪﺭﺝ ﺳﻌﺔ‬ ‫‪ ،25.00 ml‬ﻭﺿـﻊ ﺍﻟﻤـﺎﺀ ﻓـﻲ ﻛﺄﺱ ﺳـﻌﺔ ‪،100.0 ml‬‬ ‫ﺍﺳـﺘﻌﻤﻞ ﺍﻟﻤﺎﺻﺔ ﻭﺃﺿـﻒ ﻧﻘﻄﺔ ﻣﻦ ﻣﺤﻠـﻮﻝ ﺍﻷﻣﻮﻧﻴﺎ‬ ‫‪ 0.1 mol/l‬ﺇﻟﻰ ﺍﻟﻤﺎﺀ‪.‬‬ ‫ﺗﺤﺬﻳﺮ‪ :‬ﺑﺨﺎﺭ ﺍﻷﻣﻮﻧﻴﺎ ﻣﻬﻴﺞ ﺟﺪﹰﹼ ﺍ‪.‬‬ ‫‪ .3‬ﺃﺿﻒ ‪ 15‬ﻧﻘﻄﺔ ﻣﻦ ﺍﻟﻜﺎﺷﻒ ﺍﻟﻌﺎﻡ ﺇﻟﻰ ﺍﻟﻤﺤﻠﻮﻝ‪ ،‬ﻭﺣﺮﻛﻪ‪.‬‬ ‫ﻻﺣﻆ ﻟﻮﻧﻪ‪ ،‬ﻭﻗﺲ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺗﻪ ﺑﻤﻘﻴﺎﺱ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬ ‫ﻓﻮﺍﺭﺍ ﻓـﻲ ﺍﻟﻤﺤﻠﻮﻝ‪ ،‬ﻭﻻﺣﻆ ﻣﺎﺫﺍ ﻳﺤﺪﺙ؟‬ ‫‪ .4‬ﺿﻊ ﹰ‬ ‫ﻗﺮﺻﺎ ﹰ‬ ‫ﺃﻱ ﺗﻐﻴﺮ ﻓﻲ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬ ‫ﺳﺠﻞ ﻣﻼﺣﻈﺎﺗﻚ‪ ،‬ﻣﺘﻀﻤﻨ ﹰﺔ ﹼ‬

‫‪F A G B H AC I BD J CEA K DFB L EGC M FHDN G EI O HJFP‬‬

‫ﺃﻱ ﺗﻐﻴﺮﺍﺕ ﻓﻲ ﻟﻮﻥ ﺍﻟﻤﺤﻠﻮﻝ ﺃﻭ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺗﻪ‪.‬‬ ‫‪ .1‬ﺻﻒ ﹼ‬ ‫‪ .2‬ﹼ‬ ‫ﻭﺿﺢ ﻫﻞ ﻧﺘﺞ ﻏﺎﺯ؟ ﻭﺇﺫﺍ ﺣﺪﺙ ﻛﻴﻒ ﺗﻢ ﺍﻻﺳﺘﺪﻻﻝ ﻋﻠﻰ ﺫﻟﻚ؟‬ ‫ﻓﺴﺮ ﺫﻟﻚ‪.‬‬ ‫‪ .3‬ﺣ ﹼﻠﻞ ﻫﻞ ﺣﺪﺙ ﺗﻐﻴﺮ ﻓﻴﺰﻳﺎﺋﻲ ﺃﻡ ﺗﻐﻴﺮ ﻛﻴﻤﻴﺎﺋﻲ؟ ﹼ‬

‫ﺑﻢ ﻳﺨﺒﺮﻙ ﺍﻟﻜﺎﺷﻒ ﺍﻟﻌﺎﻡ ﻋﻦ ﺍﻟﻤﺤﻠﻮﻝ؟‬ ‫اﺳﺘﻘصاﺀ ﹶ‬ ‫ﺻﻤﻢ ﺗﺠﺮﺑﺔ ﻟﺪﻋﻢ ﺗﻮﻗﻌﺎﺗﻚ‪.‬‬ ‫ﹼ‬

‫ﺍﻟﺨﻄﻮﺓ ‪ 3‬ﻋﻨﻮﻥ ﺍﻷﺟﺰﺍﺀ ﺍﻟﺨﻤﺴﺔ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬ ‫ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ‪ ،‬ﺍﻟﺘﻜﻮﻳـﻦ‪ -‬ﺍﻻﺣﺘـﺮﺍﻕ‪-‬‬ ‫ﺍﻟﺘﻔـﻜﻚ‪ -‬ﺍﻹﺣـﻼﻝ ﺍﻟﺒﺴـﻴﻂ ‪ -‬ﺍﻹﺣـﻼﻝ‬ ‫ﺍﻟﻤﺰﺩﻭﺝ‪.‬‬ ‫ﺍﻟﺘﻜﻮﻳﻦ‬ ‫ﺍﻻﺣﺘﺮﺍﻕ‬

‫ﺍﻟﺘﻔﻜﻚ‬

‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫ﺗحلﻴﻞ الﻨﺘاﺋﺞ‬

‫‪E‬‬

‫‪D‬‬

‫‪C‬‬

‫ﺍﻹ‬ ‫ﺣـﻼﻝ ﺍﻟﺒﺴـﻴﻂ‬ ‫ﺍﻹ‬ ‫ﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ‬

‫المطﻮﻳات ﺍﺳـﺘﺨﺪﻡ ﻫـﺬﻩ ﺍﻟﻤﻄﻮﻳـﺔ ﻓﻲ ﺍﻟﺒﻨـﺪ ‪ 4-2‬ﻣﻦ‬ ‫ﻫـﺬﺍ ﺍﻟﻔﺼﻞ ﻓـﻲ ﺃﺛﻨﺎﺀ ﻗﺮﺍﺀﺗـﻚ ﻟﻪ‪ ،‬ﻭﻟﺨـﺺ ﻛﻞ ﻧﻮﻉ ﻣﻦ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺃﻋﻂ ﺃﻣﺜﻠﺔ ﻋﻠﻴﻬﺎ‪.‬‬

‫ﳌﺮﺍﺟﻌﺔ ﳏﺘﻮ￯ ﻫﺬﺍ ﺍﻟﻔﺼﻞ ﻭﺃﻧﺸﻄﺘﻪ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ ﺍﻹﻟﻜﱰﻭﲏ‬

‫‪www.obeikaneducation.com‬‬

‫‪7‬‬

‫‪B‬‬

‫‪A‬‬

‫‪4-1‬‬ ‫ا ﻫﺪاف‬

‫ﺗﺘﻌﺮﻑ ﺃﺩﻟــــﺔ ﺍﻟﺘﻔـــﺎﻋــﻞ‬ ‫ﹼ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﲤ ﹼﺜـﻞ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ‬ ‫ﺑﻤﻌﺎﺩﻻﺕ‪.‬‬ ‫ﺗﺰﻥ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﻟﺘﻐـﲑ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ :‬ﻋﻤﻠﻴﺔ ﺗﺘﻀﻤﻦ‬ ‫ﲢـﻮﻝ ﻣـﺎﺩﺓ ﺃﻭ ﺃﻛﺜـﺮ ﺇﱃ ﻣـﺎﺩﺓ‬ ‫ﺟﺪﻳﺪﺓ‪.‬‬ ‫ﻗﺎﻧـﻮﻥ ﺣﻔـﻆ ﺍﻟﻜﺘﻠـﺔ‪ :‬ﺍﻟﻜﺘﻠـﺔ ﻻ‬ ‫ﺗﻔﻨـﻰ ﻭﻻ ﺗﺴـﺘﺤﺪﺙ ﰲ ﺃﺛﻨـﺎﺀ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫ﺍﳌﺘﻔﺎﻋﻼﺕ‬ ‫ﺍﻟﻨﻮﺍﺗﺞ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ‬ ‫ﺍﳌﻌﺎﻣﻞ‬

‫اﻟﺘﻔﺎﻋﻼت واﻟﻤﻌﺎدﻻت‬

‫‪Reactions and Equations‬‬ ‫اﻟﻔﻜﺮة‬

‫اﻟﺮﺋﻴﺴﺔ‬

‫ﹸﲤ ﱠﺜﻞ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺑﻤﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ‪.‬‬

‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻋﻨﺪﻣﺎ ﺗﺸﱰﻱ ﹰ‬ ‫ﻣﻮﺯﺍ ﺃﺧﴬ ﺍﻟﻠﻮﻥ ﻓﺈﻧﻪ ﻳﺘﺤﻮﻝ ﺧﻼﻝ ﺃﻳﺎﻡ‬ ‫ﻗﻠﻴﻠﺔ ﺇﱃ ﺍﻟﻠﻮﻥ ﺍﻷﺻﻔﺮ‪ ،‬ﻭﻫﺬﺍ ﺍﻟﺘﻐﲑ ﰲ ﺍﻟﻠﻮﻥ ﺩﻟﻴﻞ ﻋﲆ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬

‫الﺘﻔاﻋﻼت الﻜﻴمﻴاﺋﻴﺔ ‪Chemical Reactions‬‬ ‫ﻫﻞ ﺗﻌﻠﻢ ﺃﻥ ﺍﻟﻄﻌﺎﻡ ﺍﻟﺬﻱ ﺗﺄﻛﻠﻪ‪ ،‬ﻭﺍﻷﻟﻴﺎﻑ ﰲ ﻣﻼﺑﺴﻚ‪ ،‬ﻭﺍﻟﺒﻼﺳﺘﻴﻚ ﰲ ﺃﻗﺮﺍﺻﻚ‬ ‫ﺍﳌﺪﳎﺔ ﻟﺪﳞﺎ ﳾﺀ ﻣﺸﱰﻙ؟ ﲨﻴﻊ ﻫﺬﻩ ﺍﳌﻮﺍﺩ ﺗﻨﺘﺞ ﻋﻨﺪﻣﺎ ﹸﻳﻌﺎﺩ ﺗﺮﺗﻴﺐ ﺍﻟﺬﺭﺍﺕ ﻓﻴﻬﺎ‬ ‫ﻟﺘﻜﻮﻳﻦ ﻣﻮﺍﺩ ﺃﺧﺮ￯ ﳐﺘﻠﻔﺔ‪ .‬ﹰ‬ ‫ﻓﻤﺜﻼ ﻳﻌﺎﺩ ﺗﺮﺗﻴﺐ ﺍﻟﺬﺭﺍﺕ ﺧﻼﻝ ﺣﺮﺍﺋﻖ ﺍﻟﻐﺎﺑﺎﺕ‪،‬‬ ‫ﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﻟﺼﻮﺭﺓ ﺍﻟﻮﺍﺭﺩﺓ ﰲ ﺑﺪﺍﻳﺔ ﺍﻟﻔﺼﻞ‪ .‬ﻭﻛﺬﻟﻚ ﺃﻋﻴﺪ ﺗﺮﺗﻴﺐ ﺍﻟﺬﺭﺍﺕ‬ ‫ﻋﻨﺪﻣﺎ ﺃﻟﻘﻲ ﺑﺎﻟﻘﺮﺹ ﺍﻟﻔﻮﺍﺭ ﰲ ﻛﺄﺱ ﺍﳌﺎﺀ ﺧﻼﻝ ﺍﻟﻨﺸﺎﻁ ﺍﻻﺳﺘﻬﻼﱄ‪.‬‬

‫ﺗﺴﻤﻰ ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﺘﻲ ﻳﺘﻢ ﻓﻴﻬﺎ ﺇﻋﺎﺩﺓ ﺗﺮﺗﻴﺐ ﺍﻟﺬﺭﺍﺕ ﰲ ﻣﺎﺩﺓ ﺃﻭ ﺃﻛﺜﺮ ﻟﺘﻜﻮﻳﻦ ﻣﻮﺍﺩ‬ ‫ﳐﺘﻠﻔﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ .‬ﻭﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺗﺆﺛﺮ ﰲ ﲨﻴﻊ ﻧﻮﺍﺣﻲ ﺍﳊﻴﺎﺓ؛ ﻓﻬﻲ‬ ‫ﲢﻠﻞ ﺍﻟﻄﻌﺎﻡ ﺍﻟﺬﻱ ﺗﺄﻛﻠﻪ‪ ،‬ﻣﻨﺘﺠ ﹰﺔ ﺍﻟﻄﺎﻗ ﹶﺔ ﺍﻟﺘﻲ ﲢﺘﺎﺝ ﺇﻟﻴﻬﺎ ﻟﺘﻌﻴﺶ‪ .‬ﻭﺗﻮ ﹼﻓﺮ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﰲ ﳏﺮﻛﺎﺕ ﺍﻟﺴﻴﺎﺭﺍﺕ ﻭﺍﳊﺎﻓﻼﺕ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻼﺯﻣﺔ ﺍﻟﺘﻲ ﲢﺮﻙ ﻫﺬﻩ ﺍﳌﺮﻛﺒﺎﺕ‪ .‬ﻛﲈ‬ ‫ﺃﳖﺎ ﺗﻨﺘﺞ ﺍﻷﻟﻴﺎﻑ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻛﺎﻟﻘﻄﻦ ﰲ ﺍﻟﻨﺒﺎﺗﺎﺕ ﻭﺍﻟﺼﻮﻑ ﰲ ﺍﳊﻴﻮﺍﻧﺎﺕ‪ ،‬ﻭﺍﻷﻟﻴﺎﻑ‬ ‫ﺍﻻﺻﻄﻨﺎﻋﻴﺔ ﻛﺎﻟﻨﺎﻳﻠﻮﻥ ﺍﻟﺬﻱ ﻳﺴﺘﻌﻤﻞ ﰲ ﺍﳌﺼﺎﻧﻊ‪ ،‬ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪.4-1‬‬

‫ﻛﻴﻤﻴﺎﺋﻴـﺎ ﻗﺪ ﺣﺪﺙ؟‬ ‫‪ »FÉ«ª«µdG π``YÉØàdG çhó``M á``dOCG‬ﻛﻴـﻒ ﺗﻌﺮﻑ ﺃﻥ ﺗﻔﺎﻋـﻼ‬ ‫ﹰﹼ‬ ‫ﻛﺜﻴـﺮﺍ ﻣﻨﻬﺎ ﹸﻳﻈﻬﺮ‬ ‫ﺭﻏـﻢ ﺃﻥ ﺑﻌـﺾ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻳﺼﻌﺐ ﺍﻛﺘﺸـﺎﻓﻬﺎ ﺇﻻ ﺃﻥ ﹰ‬ ‫ﺃﺩﻟﺔ ﻓﻴﺰﻳﺎﺋﻴﺔ )ﻣﺤﺴﻮﺳـﺔ( ﻋﻠﻰ ﺣﺪﻭﺛﻬﺎ‪ .‬ﺇﻥ ﺗﻐﻴﺮ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻣﺜ ﹰ‬ ‫ﻼ ﻗﺪ ﻳﺸـﻴﺮ ﺇﻟﻰ‬ ‫ﺣـﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ؛ ﻓﺒﻌﺾ ﺍﻟﺘﻔﺎﻋﻼﺕ ـ ﻛﺘﻠﻚ ﺍﻟﺘﻲ ﺗﺤﺪﺙ ﻓﻲ ﺃﺛﻨﺎﺀ ﺍﺣﺘﺮﺍﻕ‬ ‫ﺍﻟﺨﺸﺐ ـ ﺗﻄﻠﻖ ﻃﺎﻗﺔ ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﺍﺭﺓ ﻭﺿﻮﺀ‪ ،‬ﻭﺑﻌﻀﻬﺎ ﺍﻵﺧﺮ ﻳﻤﺘﺺ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬ ‫ﺍﻟﺸـﻜﻞ ‪ 4-1‬ﻳﻨﺘﺞ ﺍﻟﻨﺎﻳﻠﻮﻥ ﻋﻦ ﺗﻔﺎﻋﻞ‬ ‫ﻛﻴﻤﻴﺎﺋﻲ‪ ،‬ﻭﻳﺴﺘﻌﻤﻞ ﰲ ﻛﺜﲑ‬ ‫ﻣـﻦ ﺍﳌﻨﺘﺠـﺎﺕ‪ ،‬ﻛﺎﳌﻼﺑـﺲ‬ ‫ﻭﺍﻟﺴـﺠـــــﺎﺩ‪ ،‬ﻭﺍﻷﺩﻭﺍﺕ‬ ‫ﺍﻟﺮﻳﺎﺿﻴـﺔ‪ ،‬ﻭﺍﻹﻃـــﺎﺭﺍﺕ‪.‬‬

‫‪8‬‬

‫ﻫﻨﺎﻙ ﺃﻧﻮﺍﻉ ﺃﺧﺮ￯ ﻣﻦ ﺍﻷﺩﻟﺔ ﺍﻟﺘﻲ ﺗﺸﲑ ﺇﱃ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪ ،‬ﺑﺎﻹﺿﺎﻓﺔ‬ ‫ﺇﱃ ﺗﻐﲑ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ‪ ،‬ﻭﻣﻨﻬﺎ ﺗﻐﲑ ﺍﻟﻠﻮﻥ‪ .‬ﺭﺑﲈ ﻻﺣﻈﺖ ﹰ‬ ‫ﻣﺜﻼ ﺃﻥ ﺑﻌﺾ ﺍﳌﺴﺎﻣﲑ‬ ‫ﺍﳌﻠﻘﺎﺓ ﰲ ﺍﻟﻄﺮﻕ ﻳﺘﻐﲑ ﻟﻮﳖﺎ ﻣﻦ ﻓﴤ ﺇﱃ ﺑ ﹼﻨ ﹼﻲ ﰲ ﺯﻣﻦ ﻗﺼﲑ‪ .‬ﺇﻥ ﺗﻐﲑ ﺍﻟﻠﻮﻥ ﻳﺪﻝ‬ ‫ﹰ‬ ‫ﻛﻴﻤﻴﺎﺋﻴﺎ ﻗﺪ ﺣﺪﺙ ﺑﲔ ﺍﳊﺪﻳﺪ ﻭﺍﻷﻛﺴﺠﲔ‪ .‬ﻛﲈ ﺃﻥ ﲢﻮﻝ ﻟﻮﻥ‬ ‫ﺗﻔﺎﻋﻼ‬ ‫ﻋﲆ ﺃﻥ‬ ‫ﹰﹼ‬ ‫ﺍﳌﻮﺯ ﻣﻦ ﺍﻷﺧﴬ ﺇﱃ ﺍﻷﺻﻔﺮ ﻣﺜﺎﻝ ﺁﺧﺮ ﻋﲆ ﺫﻟﻚ‪ .‬ﻭ ﹸﺗﻌﺪ ﺍﻟﺮﺍﺋﺤﺔ‪ ،‬ﻭﺗﺼﺎﻋﺪ‬ ‫ﻭﺗﻜﻮﻥ ﺭﻭﺍﺳﺐ ﺃﺩﻟ ﹰﺔ ﺃﺧﺮ￯ ﻋﲆ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ .‬ﻭﰲ ﻛﻞ ﺻﻮﺭﺓ ﰲ‬ ‫ﺍﻟﻐﺎﺯ‪،‬‬ ‫ﹼ‬ ‫ﺍﻟﺸﻜﻞ ‪ 4-2‬ﺩﻟﻴﻞ ﻋﲆ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-2‬ﻛﻞ ﺻﻮﺭﺓ ﻣﻦ ﻫﺬﻩ ﺍﻟﺼﻮﺭ ﺗﺪﻝ‬ ‫ﻋـﲆ ﺣـﺪﻭﺙ ﺗﻔﺎﻋـﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﺻـﻒ ﻣـﺎ ﺍﻷﺩﻟـﺔ ﻋﻠـﻰ ﺣﺪﻭﺙ‬ ‫ﺗﻔﺎﻋـﻞ ﻛﻴﻤﻴﺎﺋـﻲ ﻓـﻲ ﻛﻞ ﻣـﻦ‬ ‫ﺍﻟﺼﻮﺭ ﺍﻟﻤﺠﺎﻭﺭﺓ؟‬

‫ﺗمﺜﻴﻞ الﺘﻔاﻋﻼت الﻜﻴمﻴاﺋﻴﺔ ‪Representing Chemical Reactions‬‬ ‫ﻳﺴـﺘﺨﺪﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﻣﻌﺎﺩﻻﺕ ﻟﺘﻤﺜﻴﻞ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻭﺗﻮﺿﺢ ﻫﺬﻩ‬ ‫ﹺ‬ ‫ﺍﳌﺘﻔﺎﻋـﻼﺕ ﻭﻫﻲ ﺍﳌﻮﺍﺩ ﺍﻟﺒﺎﺩﺋﺔ ﰲ ﺍﻟﺘﻔﺎﻋﻞ‪ ،‬ﻭﺃﻣﺎ ﺍﻟﻨﻮﺍﺗﺞ ﻓﻬﻲ ﺍﳌﻮﺍﺩ‬ ‫ﺍﳌﻌﺎﺩﻻﺕ‬ ‫ﹸ‬ ‫ﺍﳌﺘﻜﻮﻧـﺔ ﺧـﻼﻝ ﺍﻟﺘﻔﺎﻋـﻞ‪ .‬ﻛﲈ ﻳﺴـﺘﻌﻤﻞ ﻓﻴﻬﺎ ﺳـﻬﻢ ﻟﺘﻮﺿﻴﺢ ﺍﲡـﺎﻩ ﺍﻟﺘﻔﺎﻋﻞ‪،‬‬ ‫ﹼ‬ ‫ﻭﻓﺼـﻞ ﺍﳌﺘﻔﺎﻋـﻼﺕ ﻋـﻦ ﺍﻟﻨﻮﺍﺗـﺞ‪ .‬ﻭﺗﻜﺘـﺐ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻋﻦ ﻳﺴـﺎﺭ ﺍﻟﺴـﻬﻢ‪،‬‬ ‫ﻭﺍﻟﻨﻮﺍﺗـﺞ ﻋﻦ ﻳﻤﻴﻨـﻪ‪ .‬ﻭﻋﻨﺪﻣﺎ ﻳﻜﻮﻥ ﻫﻨﺎﻙ ﺃﻛﺜﺮ ﻣﻦ ﻣﺘﻔﺎﻋﻞ ﺃﻭ ﻧﺎﺗﺞ ﺗﺴـﺘﺨﺪﻡ‬ ‫ﺇﺷـﺎﺭﺓ )‪ (+‬ﻟﻠﻔﺼـﻞ ﺑﲔ ﺍﳌﺘﻔﺎﻋـﻼﺕ ﺃﻭ ﺍﻟﻨﻮﺍﺗﺞ‪ .‬ﻭﻳﺒﲔ ﺍﻟﺘﻌﺒـﲑ ﺍﻟﺘﺎﱄ ﻋﻨﺎﴏ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪:‬‬ ‫ﺍﻟﻨﺎﺗﺞ‪ + 2‬ﺍﻟﻨﺎﺗﺞ‪ → 1‬ﺍﳌﺘﻔﺎﻋﻞ‪ + 2‬ﺍﳌﺘﻔﺎﻋﻞ‪1‬‬

‫ﻭﺗﺴﺘﺨﺪﻡ ﺍﻟﺮﻣﻮﺯ ﰲ ﺍﳌﻌﺎﺩﻻﺕ ﻟﺘﻮﺿﻴﺢ ﺍﳊﺎﻟﺔ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﻟﻠﻤﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ‬ ‫ﻭﺍﻟﻨﺎﲡﺔ ﺍﻟﺘﻲ ﻗﺪ ﺗﻜﻮﻥ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ )‪ (s‬ﺃﻭ ﺍﻟﺴﺎﺋﻠﺔ )‪ (l‬ﺃﻭ ﺍﻟﻐﺎﺯﻳﺔ)‪(g‬‬ ‫ﺃﻭ ﻣﺬﺍﺑﺔ ﰲ ﺍﳌﺎﺀ )‪ (aq‬ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﳉﺪﻭﻝ ‪ .4-1‬ﻭﻣﻦ ﺍﳌﻬﻢ ﺗﻮﺿﻴﺢ ﻫﺬﻩ‬ ‫ﺍﻟﺮﻣﻮﺯ؛ ﻷﳖﺎ ﺗﻌﻄﻲ ﹰ‬ ‫ﺩﻟﻴﻼ ﻋﲆ ﻛﻴﻔﻴﺔ ﺣﺪﻭﺙ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬

‫‪4-1 ∫hó÷G‬‬

‫‪‘ áeóîà°ùŸG RƒeôdG‬‬ ‫‪á«FÉ«ª«µdG ä’OÉ©ŸG‬‬

‫‪õeôdG‬‬

‫‪¢Vô¨dG‬‬

‫‪+‬‬

‫ﻳﻔﺼﻞ ﺑﲔ ﻣﺎﺩﺗﲔ ﺃﻭ ﺃﻛﺜﺮ‬ ‫ﻣﻦ ﺍﳌﺘﻔﺎﻋﻼﺕ ﺃﻭ ﺍﻟﻨﻮﺍﺗﺞ‬

‫→‬

‫ ‬ ‫)‪(s‬‬ ‫)‪(l‬‬ ‫)‪(g‬‬ ‫)‪(aq‬‬

‫ﻳﻔﺼﻞ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻋﻦ‬ ‫ﺍﻟﻨﻮﺍﺗﺞ‬

‫ﻳﻔﺼﻞ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻋﻦ‬ ‫ﺍﻟﻨﻮﺍﺗﺞ‪ ،‬ﻭﻳﺸﲑ ﺇﱃ ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺍﻻﻧﻌﻜﺎﳼ‬ ‫ﻳﺸﲑ ﺇﱃ ﺍﳊﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ‬

‫ﻳﺸﲑ ﺇﱃ ﺍﳊﺎﻟﺔ ﺍﻟﺴﺎﺋﻠﺔ‬ ‫ﻳﺸﲑ ﺇﱃ ﺍﳊﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ‬

‫ﻳﺸﲑ ﺇﱃ ﺍﳌﺤﻠﻮﻝ ﺍﳌﺎﺋﻲ‬

‫‪9‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-3‬ﺍﻟﻌﻠﻢ ﻛﻐﲑﻩ ﻣﻦ ﺍﳌﺠﺎﻻﺕ‪،‬‬ ‫ﻟﻪ ﻟﻐـﺔ ﻣﺘﺨﺼﺼـﺔ ﺗﺴﻤـﺢ‬ ‫ﺑﺘـﺪﺍﻭﻝ ﻣﻌﻠـﻮﻣﺎﺕ ﻣﻌﻴﻨـﺔ‬ ‫ﺑﻄﺮﻳﻘﺔ ﻣﻨﺘﻈﻤﺔ‪ .‬ﻓﺎﻟﺘﻔﺎﻋـﻞ‬ ‫ﺑﲔ ﺍﻷﻟـﻮﻣﻨﻴـﻮﻡ ﻭﺍﻟـﱪﻭﻡ‬ ‫ﻳﻤـﻜـﻦ ﻭﺻـﻔـﻪ ﺑﻤﻌـﺎﺩﻟﺔ‬ ‫ﻟﻔﻈﻴﺔ‪ ،‬ﺃﻭ ﺑﻤﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ‪.‬‬

‫‪ ᫶Ø∏dG ä’OÉ©ªdG‬ﻳﻤﻜﻨﻚ ﺍﺳـﺘﻌﻤﺎﻝ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻠﻔﻈﻴﺔ ﻟﻠﺘﻌﺒﻴﺮ ﻋﻦ ﱟ‬ ‫ﻛﻞ ﻣﻦ‬ ‫ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﺗﺠﺔ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻭﺗﺼﻒ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ‬ ‫ﺃﺩﻧـﺎﻩ ﺍﻟﺘﻔﺎﻋـﻞ ﺑﻴـﻦ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ‪ Al‬ﻭﺍﻟﺒﺮﻭﻡ ﺍﻟﺴـﺎﺋﻞ ‪ Br2‬ﺍﻟﻤﻮﺿﺢ ﻓﻲ ﺍﻟﺸـﻜﻞ‬ ‫‪ .4-3‬ﻓﺎﻟﺴـﺤﺎﺑﺔ ﺍﻟﺤﻤﺮﺍﺀ ﻓﻲ ﺍﻟﺸـﻜﻞ ﻫﻲ ﺑﺮﻭﻡ ﻓﺎﺋﺾ‪ .‬ﺃﻣـﺎ ﻧﺎﺗﺞ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﺬﻱ‬ ‫ﻫﻮ ﺟﺴﻴﻤﺎﺕ ﺻﻠﺒﺔ ﻣﻦ ﺑﺮﻭﻣﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ‪ AlBr3‬ﻓﻴﺴﺘﻘﺮ ﻓﻲ ﻗﻌﺮ ﺍﻟﻜﺄﺱ‪.‬‬ ‫ﺍﻟﻨﺎﺗﺞ )‪ → (1‬ﺍﳌﺘﻔﺎﻋﻞ )‪ + (2‬ﺍﳌﺘﻔﺎﻋﻞ )‪(1‬‬ ‫ﺑﺮﻭﻣﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ → ﺍﻟﱪﻭﻡ ‪ +‬ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‬

‫ﹸﺗﻘﺮﺃ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ ﻋﲆ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﱄ‪" :‬ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﻭﺍﻟﱪﻭﻡ ﻳﺘﻔﺎﻋﻼﻥ‬ ‫ﻹﻧﺘﺎﺝ ﺑﺮﻭﻣﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‪".‬‬ ‫‪‬‬ ‫‪᫪∏Y äGOôØe‬‬

‫ﺍﻟﺼﻴﻐـﺔ‪ :‬ﺗﻌﺒـﲑ ﻳﺴـﺘﺨﺪﻡ ﺍﻟﺮﻣـﻮﺯ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻟﺘﻤﺜﻴﻞ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﲈﺀ ﻫﻲ ‪.H2O‬‬

‫‪ á«FÉ«ª«µdG ä’OÉ``©ªdG‬ﺇﻥ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻠﻔﻈﻴﺔ ﺗﺴـﺎﻋﺪ ﻋﲆ ﻭﺻﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﺇﻻ ﺃﳖـﺎ ﺗﻔﺘﻘـﺮ ﺇﱃ ﻣﻌﻠﻮﻣـﺎﺕ ﻣﻬﻤﺔ‪ .‬ﺃﻣـﺎ ﺍﳌﻌﺎﺩﻟـﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻓﺘﺴـﺘﺨﺪﻡ ﺭﻣﻮﺯ‬ ‫ﺍﻟﻌﻨـﺎﴏ ﻭﺻﻴـﻎ ﺍﳌﺮﻛﺒـﺎﺕ ـ ﺑـﺪ ﹰ‬ ‫ﻻ ﻣـﻦ ﺍﻟﻜﻠـﲈﺕ ـ ﻟﻠﺘﻌﺒـﲑ ﻋـﻦ ﺍﳌﺘﻔﺎﻋـﻼﺕ‬ ‫ﻭﺍﻟﻨﻮﺍﺗـﺞ‪ .‬ﻓﺎﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﲔ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﻭﺍﻟﱪﻭﻡ ﹰ‬ ‫ﻣﺜﻼ ﺗﺴـﺘﺨﺪﻡ‬ ‫ﺭﻣﺰﻱ ﺍﻷﻟﻮﻣﻨﻴـﻮﻡ ﻭﺍﻟﱪﻭﻡ ﻭﺻﻴﻐﺔ ﺑﺮﻭﻣﻴﺪ ﺍﻷﻟﻮﻣﻨﻴـﻮﻡ ﺑﺪ ﹰ‬ ‫ﻻ ﻣﻦ ﺍﻟﻜﻠﲈﺕ‪ ،‬ﻛﲈ‬ ‫ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪.4-4‬‬ ‫ﺗﺸﲑ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺇﱃ ﺃﻥ ﺍﳌﺎﺩﺓ ﹸﲢﻔﻆ ﺧﻼﻝ ﺍﻟﺘﻔﺎﻋﻞ‪ ،‬ﻭﻫﺬﺍ ﻣﺎ ﻳﻨﺺ‬

‫ﻋﻠﻴﻪ ﻗﺎﻧﻮﻥ ﺑﻘﺎﺀ ﺍﻟﻜﺘﻠﺔ‪.‬‬ ‫ﺍﻟﺸـﻜﻞ ‪ 4-4‬ﺍﳌﻌﻠﻮﻣـﺎﺕ ﺍﻟﺘـﻲ ﺗﻨﻘﻠﻬـﺎ ﺍﳌﻌﺎﺩﻟـﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﳏﺪﻭﺩﺓ‪ .‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺍﳌﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﺻﺤﻴﺤـﺔ‪ ،‬ﻭﻟﻜﻨﻬﺎ ﻻ ﺗﻮﺿﺢ‬ ‫ﺍﻟﻌـﺪﺩ ﺍﻟﺼﺤﻴـﺢ ﻟﻠـﺬﺭﺍﺕ ﺍﳌﺘﻔﺎﻋﻠـﺔ‪.‬‬

‫ﺫﺭﺓ ﺍﻟﻮﻣﻨﻴﻮﻡ ﻭﺛﻼﺙ ﺫﺭﺍﺕ ﺑﺮﻭﻡ‬

‫‪10‬‬

‫ﺫﺭﺗﺎ ﺑﺮﻭﻡ‬

‫ﺫﺭﺓ ﺍﻟﻮﻣﻨﻴﻮﻡ‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻟﻠﻤﻌﺎﺩﻻﺕ ﺍﻟﻠﻔﻈﻴﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪.1‬‬

‫ﺑﺮﻭﻣﻴﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ → ﻫﻴﺪﺭﻭﺟﲔ ‪ +‬ﺑﺮﻭﻡ‬

‫‪ .2‬ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ → ﺃﻛﺴﺠﲔ ‪ +‬ﺃﻭﻝ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‬

‫‪ .3‬ﲢﺪﹼ ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ ﻭﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺮﻣﺰﻳﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺍﻵﰐ‪ :‬ﻋﻨﺪ ﺗﺴﺨﲔ ﻛﻠﻮﺭﺍﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪KClO3‬‬

‫ﺍﻟﺼﻠﺒﺔ ﻳﻨﺘﺞ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ﺍﻟﺼﻠﺐ ﻭﻏﺎﺯ ﺍﻷﻛﺴﺠﲔ‪.‬‬

‫ﻟﺘﻤﺜﻴﻞ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺑﻤﻌﺎﺩﻟﺔ ﺑﺸﻜﻞ ﺻﺤﻴﺢ ﳚﺐ ﺃﻥ ﺗﻮﺿﺢ ﺍﳌﻌﺎﺩﻟﺔ ﺃﻥ‬ ‫ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ﺍﳌﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ ﻳﺴﺎﻭﻱ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ﺍﳌﻮﺍﺩ ﺍﻟﻨﺎﲡﺔ‪ .‬ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ‬ ‫ﺗﺴﻤﻰ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ‪ .‬ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ ﺗﻌﺒﲑ ﹶﻳﺴﺘﺨﺪﻡ ﺍﻟﺼﻴﻎﹶ‬ ‫ﺍﳌﺘﻀﻤﻨﺔ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻭﻛﻤﻴﺎﲥﺎ ﺍﻟﻨﺴﺒﻴﺔ‪.‬‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﺘﻮﺿﻴﺢ ﺃﻧﻮﺍﻉ ﺍﳌﻮﺍﺩ‬ ‫ﱠ‬

‫ﻭﺯﻥ المﻌادﻻت الﻜﻴمﻴاﺋﻴﺔ ‪Balancing Chemical Equations‬‬ ‫ﺗﺘﻔﻖ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺍﳌﻮﺯﻭﻧﺔ ﺑﲔ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﻭﺍﻟﱪﻭﻡ ﺍﳌﺒﻴﻨﺔ ﰲ ﺍﻟﺸﻜﻞ ‪،4-5‬‬ ‫ﻣﻊ ﻗﺎﻧﻮﻥ ﺣﻔﻆ ﺍﻟﻜﺘﻠﺔ‪ .‬ﻭﺣﺘﻰ ﺗﺰﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﳚﺐ ﺃﻥ ﲡﺪ ﺍﳌﻌﺎﻣﻼﺕ‬ ‫ﺍﻟﺼﺤﻴﺤﺔ ﻟﻠﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻓﻴﻬﺎ‪ .‬ﺍﳌﻌﺎﻣﻞ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﺬﻱ‬ ‫ﻳﻜﺘﺐ ﻗﺒﻞ ﺍﳌﺘﻔﺎﻋﻞ ﺃﻭ ﺍﻟﻨﺎﺗﺞ‪ .‬ﻭﺗﻜﻮﻥ ﺍﳌﻌﺎﻣﻼﺕ ﻋﺎﺩﺓ ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ‪ ،‬ﻭﻻ‬ ‫ﺗﻜﺘﺐ ﺇﺫﺍ ﻛﺎﻧﺖ ﻗﻴﻤﺘﻬﺎ ﻭﺍﺣﺪﹰ ﺍ‪ .‬ﻭﺗﺼﻒ ﺍﳌﻌﺎﻣﻼﺕ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﳌﻮﺯﻭﻧﺔ ﺃﺑﺴﻂ ﻧﺴﺒﺔ‬ ‫ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻜﻤﻴﺎﺕ ﻛﻞ ﻣﻦ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﻮﺍﺗﺞ‪.‬‬ ‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ‪ ،‬ﺍﺭﺟﻊ ﺇﱃ‬ ‫ﺍﳌﻮﻗﻊ ﺍﻹﻟﻜﺘﺮﻭﻧﻲ‪:‬‬ ‫‪www.obeikaneducation.com‬‬

‫ﺍﻟﺸـﻜﻞ‪ 4-5‬ﻳﺘﺴـﺎﻭ￯ ﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ ﰲ ﻃﺮﰲ‬ ‫ﻛﻞ ﻣـﻦ ﺍﳌﺘﻔﺎﻋـﻼﺕ ﻭﺍﻟﻨﻮﺍﺗـﺞ ﰲ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ‪ .‬ﻭﰲ ﻫﺬﻩ‬

‫ﺍﳊﺎﻟﺔ‪ ،‬ﻳﺘﻄﻠﺐ ﻭﺟﻮﺩ ﺫﺭﰐ ﺃﻟﻮﻣﻨﻴﻮﻡ‬

‫ﻭﺳـﺖ ﺫﺭﺍﺕ ﺑﺮﻭﻡ ﰲ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‬

‫ﺫﺭﺗﺎ ﺃﻟﻮﻣﻨﻴﻮﻡ ﻭﺳﺖ ﺫﺭﺍﺕ ﺑﺮﻭﻡ‬

‫ﺳﺖ ﺫﺭﺍﺕ ﺑﺮﻭﻡ‬

‫ﺫﺭﺗﺎ ﺃﻟﻮﻣﻨﻴﻮﻡ‬

‫‪11‬‬

‫‪ ä’OÉ``©ªdG ¿Rh äGƒ``£N‬ﻳﻤﻜـﻦ ﻭﺯﻥ ﺃﻏﻠـﺐ ﺍﻟﻤﻌـﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺑﺎﺗﺒـﺎﻉ ﺍﻟﺨﻄﻮﺍﺕ‬ ‫ﺍﻟﻤﻮﺿﺤـﺔ ﻓـﻲ ﺍﻟﺠـﺪﻭﻝ ‪ .4-2‬ﻓﻴﻤﻜﻨﻚ ﹰ‬ ‫ﻣﺜﻼ ﺍﺳـﺘﻌﻤﺎﻝ ﻫﺬﻩ ﺍﻟﺨﻄﻮﺍﺕ ﻟﻜﺘﺎﺑـﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﻴﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ ،H2‬ﻭﺍﻟﻜﻠﻮﺭ ‪ Cl2‬ﻹﻧﺘﺎﺝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪.HCl‬‬

‫‪4-2 ∫hó÷G‬‬ ‫‪äGƒ£ÿG‬‬

‫‪1‬‬

‫‪ä’OÉ©ŸG ¿Rh äGƒ£N‬‬

‫ﺇﱃ ﺍﳌﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪á«∏ª©dG‬‬ ‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻏﲑ ﻣﻮﺯﻭﻧﺔ‪.‬‬ ‫ﺗﺄﻛـﺪ ﺃﻥ ﺍﻟﺼﻴـﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻟﻠﻤﺘﻔﺎﻋـﻼﺕ‬ ‫ﻭﺍﻟﻨﻮﺍﺗـﺞ ﺻﺤﻴﺤـﺔ‪ ،‬ﻭﺃﻥ ﺍﻷﺳـﻬﻢ ﺗﻔﺼـﻞ‬ ‫ﺍﳌﺘﻔﺎﻋـﻼﺕ ﻋﻦ ﺍﻟﻨﻮﺍﺗﺞ‪ ،‬ﻭﺇﺷـﺎﺭﺓ )‪ (+‬ﺗﻔﺼﻞ‬ ‫ﺑـﲔ ﻛﻞ ﻣـﻦ ﺍﳌـﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠـﺔ ﻭﺍﳌـﻮﺍﺩ ﺍﻟﻨﺎﲡـﺔ‪،‬‬ ‫ﻭﻭﺟـﻮﺩ ﺍﳊـﺎﻻﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﻟﻠﻤـﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ‬ ‫ﻭﺍﳌﻮﺍﺩ ﺍﻟﻨﺎﲡﺔ‪.‬‬

‫‪2‬‬

‫ﻋﺪﹼ ﺫﺭﺍﺕ ﺍﻟﻌﻨﺎﴏ ﰲ ﺍﳌﺘﻔﺎﻋﻼﺕ‪.‬‬ ‫ﺗﺘﻔﺎﻋﻞ ﺫﺭﺗﺎ ﻫﻴﺪﺭﺟﲔ ﻭﺫﺭﺗﺎ ﻛﻠﻮﺭ‪.‬‬

‫‪3‬‬

‫ﻋﺪﹼ ﺫﺭﺍﺕ ﺍﻟﻌﻨﺎﴏ ﰲ ﺍﻟﻨﻮﺍﺗﺞ ‪.‬‬ ‫ﻳﻨﺘﺞ ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ ﻭﺫﺭﺓ ﻛﻠﻮﺭ‪.‬‬

‫‪4‬‬

‫ﻏﲑ ﺍﳌﻌﺎﻣﻼﺕ ﻟﺘﺠﻌﻞ ﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ‬ ‫ﹼ‬ ‫ﻣﺘﺴﺎﻭ ﹰﻳﺎ ﰲ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪ .‬ﻭﻻ ﺗﻐﲑ ﺃﺑﺪﹰ ﺍ ﺍﻟﺮﻣﺰ‬ ‫ﺍﻟﺴﻔﲇ ﰲ ﺻﻴﻐﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﺘﺰﻥ ﻣﻌﺎﺩﻟﺔ؛ ﻷﻥ‬ ‫ﺫﻟﻚ ﻳﻐﲑ ﻧﻮﻉ ﺍﳌﺎﺩﺓ‪.‬‬

‫‪5‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﻣﻼﺕ ﰲ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﳑﻜﻨﺔ‪ ،‬ﻋﲆ ﺃﻥ‬ ‫ﺗﻜﻮﻥ ﺍﳌﻌﺎﻣﻼﺕ ﺃﺻﻐﺮ ﺃﻋﺪﺍﺩ ﺻﺤﻴﺤﺔ ﳑﻜﻨﺔ‪.‬‬ ‫ﻓﺎﻟﻨﺴﺒﺔ ‪ (2:1:1) 2HCl : 1 Cl2 :1 H2‬ﻫﻲ‬ ‫ﺃﺻﻐﺮ ﻧﺴﺒﺔ ﳑﻜﻨﺔ‪ ،‬ﻷﻧﻪ ﻻ ﻳﻤﻜﻦ ﺍﺧﺘﺼﺎﺭﻫﺎ‬ ‫ﺃﻛﺜﺮ ﻣﻦ ﺫﻟﻚ ﻭﺗﻈﻞ ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ ‪.‬‬

‫‪6‬‬

‫ﺗﺄﻛﺪ ﻣﻦ ﻋﻤﻠﻚ ﺃﻥ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻣﻜﺘﻮﺑﺔ‬ ‫ﺑﺸﻜﻞ ﺻﺤﻴﺢ‪ ،‬ﻭﺃﻥ ﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ‬ ‫ﻣﺘﺴﺎ ﹴﻭ ﰲ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬

‫‪12‬‬

‫ﺟﺪﻭﻝ ﺗﻔﺎﻋﻠﻲ‪ :‬ﳌﻌﺮﻓﺔ ﺍﳌﺰﻳﺪ ﻋﻦ ﻭﺯﻥ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﺭﺟﻊ‬

‫‪∫Éãe‬‬

‫ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬ ‫ﻭﺫﺭﺓ ﻛﻠﻮﺭ‬

‫ﺫﺭﺗﺎ ﻛﻠﻮﺭ‬

‫→‬

‫‪Cl2‬‬

‫‪+‬‬

‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫‪H2‬‬

‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﺟﲔ‬

‫‪HCl‬‬ ‫‪ 1 +‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬

‫‪ 1‬ﺫﺭﺓ ﻛﻠﻮﺭ‬ ‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‪ 2 +‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬ ‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫ﺫﺭﺗﺎ ﻫﻴﺪﺭﻭﺟﲔ‬

‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬

‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬

‫)‪H2(g) + Cl2(g) → 2HCl(g‬‬ ‫‪1:1:2‬‬ ‫‪1 H2 : 1 Cl2 : 2 HCl‬‬

‫)‪2HCl(g‬‬

‫→ )‪Cl2(g‬‬ ‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬

‫‪+‬‬

‫)‪H2(g‬‬ ‫‪2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬

‫‪ 2‬ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﲔ‬ ‫‪ 2‬ﺫﺭﺓ ﻛﻠﻮﺭ‬ ‫ﻳﻮﺟﺪ ﺫﺭﺗﺎ ﻫﻴﺪﺭﻭﺟﲔ ﻭﺫﺭﺗﺎ ﻛﻠﻮﺭ ﰲ ﱟ‬ ‫ﻛﻞ ﻣﻦ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬

‫‪4-1 ∫Éãe‬‬

‫ﻛﺘﺎﺑﺔ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺮﻣﺰﻳﺔ ﺍﳌﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ‬ ‫ﺑﲔ ﳏﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﳏﻠﻮﻝ ﺑﺮﻭﻣﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﻹﻧﺘﺎﺝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺍﻟﺼﻠﺐ ﻭﳏﻠﻮﻝ ﺑﺮﻭﻣﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪.‬‬

‫واﻗﻊ اﻟﻜﻴﻤﻴﺎء ﻓﻲ اﻟﺤﻴﺎة‬ ‫ﻫﻴﺪروﻛﺴﻴﺪ اﻟﻜﺎﻟﺴﻴﻮم‬

‫‪ádCÉ°ùŸG π«∏– 1‬‬

‫ﻟﻘﺪ ﹸﺃﻋﻄﻴﺖ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﻮﺍﺗﺞ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ .‬ﻟﺬﺍ ﺍﺑﺪﺃ ﺑﻤﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻏﲑ ﻣﻮﺯﻭﻧﺔ‪ ،‬ﻣﺴﺘﺨﺪ ﹰﻣﺎ ﺍﳋﻄﻮﺍﺕ ﰲ ﺍﳉﺪﻭﻝ ‪ 4-2‬ﻟﻮﺯﳖﺎ‪.‬‬

‫‪܃∏£ŸG ÜÉ°ùM 2‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻏﲑ ﺍﳌﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ‪ .‬ﺗﺄﻛﺪ ﻣﻦ ﻭﺿﻊ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻋﻦ ﻳﺴﺎﺭ‬ ‫ﺍﻟﺴﻬﻢ‪ ،‬ﻭﺍﻟﻨﻮﺍﺗﺞ ﻋﻦ ﻳﻤﻴﻨﻪ‪ .‬ﻭﺍﻓﺼﻞ ﺍﳌﻮﺍﺩ ﺑﺈﺷﺎﺭﺓ )‪ ،(+‬ﻭﻭﺿﺢ ﺣﺎﻻﲥﺎ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ‪.‬‬ ‫‪NaOH‬‬ ‫‪+ CaBr2‬‬ ‫‪→ Ca(OH)2 + NaBr‬‬ ‫)‪(aq‬‬ ‫)‪(aq‬‬ ‫)‪(s‬‬ ‫)‪(aq‬‬

‫ﹸﻋﺪﹼ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ ﰲ ﺍﳌﺘﻔﺎﻋﻼﺕ‬

‫‪1Na, 1 O, 1 H, 1 Ca, 2 Br‬‬

‫ﻋﺪﹼ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ ﰲ ﺍﻟﻨﻮﺍﺗﺞ‬

‫ﺃﺩﺧﻞ ﺍﳌﻌﺎﻣﻞ ‪ 2‬ﻗﺒﻞ ‪ NaOH‬ﻟﻮﺯﻥ )‪(aq‬‬

‫‪1Ca, 2 O, 2 H, 1 Na, 1 Br‬‬

‫ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﲔ ﻭﺍﳍﻴﺪﺭﻭﺟﲔ‪.‬‬ ‫ﺃﺩﺧﻞ ﺍﳌﻌﺎﻣﻞ ‪ 2‬ﻗﺒﻞ ‪ NaBr‬ﻟﻮﺯﻥ )‪(aq‬‬ ‫ﺫﺭﺍﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﺍﻟﱪﻭﻡ‪.‬‬

‫‪+ NaBr‬‬

‫)‪(aq‬‬

‫‪+ 2NaBr‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﻣﻼﺕ ﰲ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﳑﻜﻨﺔ‪.‬‬

‫ﺗﺄﻛﺪ ﺃﻥ ﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ ﻣﺘﺴﺎ ﹴﻭ‬ ‫ﰲ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬

‫‪áHÉLE’G ˃≤J 3‬‬

‫)‪(s‬‬

‫‪→ Ca(OH)2‬‬

‫)‪(s‬‬

‫‪→ Ca(OH)2‬‬

‫)‪(aq‬‬

‫‪+ CaBr2‬‬

‫)‪(aq‬‬

‫‪2NaOH‬‬

‫‪+ CaBr2‬‬

‫‪2NaOH‬‬

‫)‪(aq‬‬

‫ﻧﺴﺒﺔ ﺍﳌﻌﺎﻣﻼﺕ ‪2 : 1 : 1 : 2‬‬

‫ﺍﻟﻨﻮﺍﺗﺞ‬ ‫ﺍﳌﺘﻔﺎﻋﻼﺕ‬

‫‪2Na, 2O, 2H, 1 Ca, 2 Br‬‬ ‫‪2Na, 2O, 2H, 1 Ca, 2 Br‬‬

‫ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﳉﻤﻴﻊ ﺍﳌﻮﺍﺩ ﻣﻜﺘﻮﺑﺔ ﺑﺸﻜﻞ ﺻﺤﻴﺢ‪ ،‬ﻭﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ ﻣﺘﺴﺎ ﹴﻭ ﰲ‬ ‫ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪ ،‬ﻭﺍﳌﻌﺎﻣﻼﺕ ﻣﻜﺘﻮﺑﺔ ﰲ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﳑﻜﻨﺔ‪ .‬ﻭﺍﳌﻌﺎﺩﻟﺔ ﺍﳌﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻫﻲ‪:‬‬

‫)‪2NaOH(aq) + CaBr2(aq) → Ca(OH)2(s) + 2NaBr(aq‬‬

‫‪á«fÉLôŸG Ö©°û∏d á«FÉŸG ¢VGƒMC’G‬‬ ‫ﻳﺴـﺘﺨﺪﻡ ﳏﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﻜﺎﻟﺴـﻴـﻮﻡ ﺍﳌـﺎﺋـﻲ ﻓــﻲ‬ ‫ﺍﻷﺣـﻮﺍﺽ ﺍﳌﺎﺋﻴـﺔ ﻟﻠﺸـﻌﺐ‬ ‫ﺍﳌﺮﺟﺎﻧﻴـﺔ؛ ﻟﺘﺰﻭﻳـﺪ ﺍﳊﻴﻮﺍﻧﺎﺕ‬ ‫ﻛﺎﳊﻠﺰﻭﻥ ﻭﺍﳌﺮﺟﺎﻥ‪ -‬ﺑﻌﻨﴫ‬‫ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ‪ ،‬ﺣﻴـﺚ ﻳﺘﻔﺎﻋـﻞ‬ ‫ﻫﻴﺪﺭﻭﻛﺴـﻴﺪ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﻣﻊ‬ ‫ﺛـﺎﲏ ﺃﻛﺴـﻴﺪ ﺍﻟﻜﺮﺑـﻮﻥ ﰲ ﺍﳌﺎﺀ‬ ‫ﻹﻧﺘـﺎﺝ ﺃﻳﻮﻧـﺎﺕ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ‬ ‫ﻭﺍﻟﺒﻴﻜﺮﺑﻮﻧﺎﺕ‪.‬‬

‫ﻭﺗﺴﺘﺨﺪﻡ ﺣﻴﻮﺍﻧﺎﺕ ﺍﻟﺸﻌﺐ‬ ‫ﺍﳌﺮﺟﺎﻧﻴـﺔ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﻟﺒﻨـﺎﺀ‬ ‫ﺃﺻﺪﺍﻓﻬـﺎ ﻭﺃﺟﻬﺰﲥﺎ ﺍﳍﻴﻜﻠﻴﺔ‬ ‫ﺑﺼﻮﺭﺓ ﻗﻮﻳﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﻟﻜﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪ .4‬ﻳﺘﻔﺎﻋﻞ ﻛﻠﻮﺭﻳﺪ ﺍﳊﺪﻳﺪ ‪ FeCl3 Ш‬ﻣﻊ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ NaOH‬ﰲ ﺍﳌﺎﺀ ﻹﻧﺘﺎﺝ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﳊﺪﻳﺪ ‪ Fe(OH)3 Ш‬ﺍﻟﺼﻠﺐ ﻭﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪.NaCl‬‬ ‫‪ .5‬ﻳﺘﻔﺎﻋﻞ ﺛﺎﲏ ﻛﱪﻳﺘﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ‪ CS2‬ﺍﻟﺴﺎﺋﻞ ﻣﻊ ﻏﺎﺯ ﺍﻷﻛﺴﺠﲔ ‪ O2‬ﻹﻧﺘﺎﺝ ﻏﺎﺯ ﺛﺎﲏ‬ ‫ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ‪ CO2‬ﻭﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ ‪.SO2‬‬ ‫‪ .6‬ﲢﺪﱟ ‪ :‬ﻳﺘﻔﺎﻋﻞ ﻓﻠﺰ ﺍﳋﺎﺭﺻﲔ ‪ Zn‬ﻣﻊ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ‪ H2SO4‬ﻹﻧﺘﺎﺝ ﻏﺎﺯ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪ H2‬ﻭﳏﻠﻮﻝ‬ ‫ﻛﱪﻳﺘﺎﺕ ﺍﳋﺎﺭﺻﲔ ‪.ZnSO4‬‬

‫‪13‬‬

‫‪á«FÉ«ª«µdG ä’OÉ©ŸG ¿Rh‬‬ ‫ﺍﳌﺘﻔﺎﻋﻼﺕ ﰲ‬ ‫ﺍﳉﺎﻧﺐ ﺍﻷﻳﴪ‬

‫ﺧﻄﻮﺓ ‪1‬‬ ‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ‬ ‫ﻛﻴﻤﻴﺎﺋﻴﺔ ﻏﲑ ﻣﻮﺯﻭﻧﺔ‬

‫ﺍﳌﺘﻔﺎﻋﻼﺕ‬

‫ﺧﻄﻮﺓ ‪4‬‬ ‫ﻋﺪﻝ ﺍﳌﻌﺎﻣﻼﺕ‬

‫ﺍﳋﻄﻮﺗﺎﻥ ‪3،2‬‬ ‫ﻋﺪ ﺍﻟﺬﺭﺍﺕ‬

‫ﺍﻟﻨﻮﺍﺗﺞ ﰲ‬ ‫ﺍﳉﺎﻧﺐ ﺍﻷﻳﻤﻦ‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-6‬ﺇﻥ ﺍﻟﻘـﺪﺭﺓ ﻋـﲆ ﻭﺯﻥ‬ ‫ﺍﳌﻌﺎﺩﻻﺕ ﺃﺳﺎﺳﻴﺔ ﻟﺪﺭﺍﺳﺔ‬ ‫ﺍﻟﻜﻴﻤﻴـﺎﺀ‪ .‬ﺍﺳـﺘﻌﻤﻞ ﻫـﺬﺍ‬ ‫ﺍﳌﺨﻄـﻂ ﳌﺴـﺎﻋﺪﺗﻚ ﻋـﲆ‬ ‫ﺇﺗﻘﺎﻥ ﻫﺬﻩ ﺍﳌﻬﺎﺭﺓ‪ .‬ﻭﻻﺣﻆ‬ ‫ﺃﻥ ﺍﳋﻄﻮﺍﺕ ﺍﳌﺮﻗﻤﺔ ﺗﻘﺎﺑﻞ‬ ‫ﺍﳋﻄﻮﺍﺕ ﰲ ﺍﳉﺪﻭﻝ ‪. 4-2‬‬

‫ﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ‬ ‫ﰲ ﺟﻬﺔ ﺍﻟﻴﺴﺎﺭ‬

‫ﺍﻟﻨﻮﺍﺗﺞ‬

‫ﺧﻄﻮﺓ ‪5‬‬ ‫ﳚﺐ ﺃﻥ ﺍﺧﺘﴫ ﺍﳌﻌﺎﻣﻼﺕ‬ ‫ﻳﺘﺴﺎﻭﻳﺎ‬ ‫ﻷﺑﺴﻂ ﺻﻮﺭﺓ‬

‫ﺧﻄﻮﺓ ‪6‬‬ ‫ﺗﺄﻛﺪ ﻣﻦ ﻋﻤﻠﻚ‬

‫ﻋﺪﺩ ﺫﺭﺍﺕ ﻛﻞ ﻋﻨﴫ‬ ‫ﰲ ﺟﻬﺔ ﺍﻟﻴﻤﲔ‬

‫‪ á∏àµdG ßØM ¿ƒfÉb ≥«≤ëJ‬ﻟﻌﻞ ﻣﻔﻬﻮﻡ ﻗﺎﻧﻮﻥ ﺣﻔﻆ ﺍﻟﻜﺘﻠﺔ ﻣﻦ ﺃﻫﻢ ﺍﻟﻤﻔﺎﻫﻴﻢ ﺍﻷﺳﺎﺳﻴﺔ‬ ‫ﻓـﻲ ﺍﻟﻜﻴﻤﻴـﺎﺀ‪ .‬ﻭﺟﻤﻴـﻊ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺗﺘﺒﻊ ﻫـﺬﺍ ﺍﻟﻘﺎﻧﻮﻥ ﺍﻟـﺬﻱ ﻳﻨﺺ ﻋﻠﻰ ﺃﻥ‬ ‫ﺍﻟﻤﺎﺩﺓ ﻻ ﺗﻔﻨﻰ ﻭﻻ ﺗﺴﺘﺤﺪﺙ‪ .‬ﻭﻟﻬﺬﺍ ﻓﻤﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺃﻥ ﺗﺤﺘﻮﻱ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﻲ ﺗﻤﺜﻞ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻋﻠﻰ ﻣﻌﻠﻮﻣﺎﺕ ﻛﺎﻓﻴﺔ ﺗﻮﺿﺢ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻳﺤﻘﻖ ﻗﺎﻧﻮﻥ ﺣﻔﻆ ﺍﻟﻜﺘﻠﺔ‪.‬‬ ‫ﻟﻘﺪ ﺗﻌﻠﻤﺖ ﻛﻴﻒ ﲢﻘﻖ ﺣﻔﻆ ﺍﻟﻜﺘﻠﺔ ﰲ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ‪ .‬ﻭﺍﳌﺨﻄﻂ ﺍﳌﺒﲔ‬ ‫ﰲ ﺍﻟﺸﻜﻞ ‪ 4-6‬ﻳﻠﺨﺺ ﺧﻄﻮﺍﺕ ﻭﺯﻥ ﺍﳌﻌﺎﺩﻻﺕ‪ .‬ﻭﻟﻌﻠﻚ ﲡﺪ ﺃﻥ ﺑﻌﺾ ﺍﳌﻌﺎﺩﻻﺕ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻳﻤﻜﻦ ﻭﺯﳖﺎ ﺑﺴﻬﻮﻟﺔ‪ ،‬ﰲ ﺣﲔ ﺃﻥ ﻭﺯﻥ ﺑﻌﻀﻬﺎ ﺍﻵﺧﺮ ﻳﻜﻮﻥ ﺃﻛﺜﺮ ﺻﻌﻮﺑﺔ‪.‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪4 -1‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺑﻌﺾ ﺍﻟﺘﻐﻴﺮﺍﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﻗﺪ ﺗﺸﻴﺮ‬ ‫ﺇﻟـﻰ ﺣـﺪﻭﺙ ﺗﻔﺎﻋـﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﺗﻮﻓـﺮ ﺍﻟﻤﻌـﺎﺩﻻﺕ ﺍﻟﻠﻔﻈﻴـﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻭﺍﻟﺮﻣﺰﻳـﺔ ﻣﻌﻠﻮﻣﺎﺕ‬ ‫ﻣﻬﻤﺔ ﻋﻦ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﺗﻮﺿـﺢ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺃﻧﻮﺍﻉ‬ ‫ﺍﻟﻤﺘﻔﺎﻋﻼﺕﻭﺍﻟﻨﻮﺍﺗﺞﻓﻲﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋـﻲ ﻭﻛﻤﻴﺎﺗﻬـﺎ ﺍﻟﻨﺴـﺒﻴﺔ‪.‬‬ ‫ﻳﺘﻀﻤـﻦ ﻭﺯﻥ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺗﻌﺪﻳـﻞ‬ ‫ﺍﻟﻤﻌﺎﻣﻼﺕ ﺣﺘﻰ ﻳﺘﺴـﺎﻭ￯ ﻋﺪﺩ‬ ‫ﺍﻟﺬﺭﺍﺕ ﻓﻲ ﻛﻼ ﻃﺮﻓﻴﻬﺎ‪.‬‬

‫‪14‬‬

‫ﻓﺴﺮ ﻣﺎ ﺃﻫﻤﻴﺔ ﻭﺯﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ؟‬ ‫‪ .7‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﹼ‬ ‫‪ .8‬ﻋﺪﹼ ﺩ ﺛﻼﺛﺔ ﻣﻦ ﺍﻷﺩﻟﺔ ﺍﻟﺘﻲ ﺗﺸﻴﺮ ﺇﻟﻰ ﺣﺪﻭﺙ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫‪ .9‬ﻗﺎﺭﻥ ﺑﻴﻦ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫ﻓﺴﺮ ﻟﻤﺎﺫﺍ ﻳﺠﺐ ﺍﺧﺘﺼﺎﺭ ﺍﻟﻤﻌﺎﻣﻼﺕ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻮﺯﻭﻧﺔ ﺇﻟﻰ ﺃﺑﺴﻂ‬ ‫‪ .10‬ﹼ‬ ‫ﻧﺴﺒﺔ ﻣﻦ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‪.‬‬ ‫‪ .11‬ﺣ ﹼﻠﻞ‪ :‬ﻫﻞ ﻳﻤﻜﻨﻚ ﻋﻨﺪ ﻭﺯﻥ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺗﻌﺪﻳﻞ ﺍﻟﺮﻣﻮﺯ ﺍﻟﺴﻔﻠﻰ ﻓﻲ ﺍﻟﺼﻴﻐﺔ؟‬ ‫ﻗـﻮﻡ‪ :‬ﻫـﻞ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺍﻵﺗﻴـﺔ ﻣﻮﺯﻭﻧـﺔ؟ ﺇﺫﺍ ﻟـﻢ ﺗﻜـﻦ ﻛﺬﻟـﻚ ﻓﺼﺤـﺢ‬ ‫‪ .12‬ﹼ‬ ‫ﺍﻟﻤﻌﺎﻣﻼﺕ ﻟﻮﺯﻧﻬﺎ‪:‬‬

‫‪+ Pb(NO 3 ) 2‬‬ ‫‪→ 2KNO 3‬‬ ‫‪+ PbCrO 4‬‬ ‫)‪(aq‬‬ ‫)‪(aq‬‬ ‫)‪(S‬‬

‫)‪(aq‬‬

‫‪2K 2 Cr 2 O 7‬‬

‫ﻗـﻮﻡ‪ :‬ﻳﺘﻔﺎﻋـﻞ ﻣﺤﻠﻮﻝ ﺣﻤـﺾ ﺍﻟﻔﻮﺳـﻔﻮﺭﻳﻚ ﺍﻟﻤﺎﺋـﻲ ‪ H3PO4‬ﻣﻊ ﻣﺤﻠﻮﻝ‬ ‫‪ .13‬ﹼ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺍﻟﻤﺎﺋﻲ ‪ Ca(OH)2‬ﻹﻧﺘﺎﺝ ﻓﻮﺳﻔﺎﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺍﻟﺼﻠﺒﺔ‬ ‫‪ Ca3(PO4)2‬ﻭﺍﻟﻤﺎﺀ‪ .‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ ﺗﻌﺒﺮ ﻋﻦ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ ﺍﻹﻟﻜﺘﺮﻭﻧﻲ‪:‬‬

‫‪www.obeikaneducation.com‬‬

‫‪4-2‬‬ ‫ا ﻫﺪاف‬

‫ﺗﺼﻨﻴﻒ اﻟﺘﻔﺎﻋﻼت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫‪Classifying Chemical Reactions‬‬

‫ﺗــﺼــﻨــﻒ ﺍﻟــﺘــﻔــﺎﻋــﻼﺕ‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻫﻨﺎﻙ ﺃﺭﺑﻌﺔ ﺃﻧﻮﺍﻉ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻫﻲ‪ :‬ﺍﻟﺘﻜﻮﻳﻦ‪ ،‬ﻭﺍﻻﺣﱰﺍﻕ‪،‬‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫ﲢﺪﺩ ﳑﻴﺰﺍﺕ ﺍﻷﻧﻮﺍﻉ ﺍﳌﺨﺘﻠﻔﺔ ﻭﺍﻟﺘﻔﻜﻚ‪ ،‬ﻭﺍﻹﺣﻼﻝ‪.‬‬ ‫ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻗﺪ ﲢﺘﺎﺝ ﺇﱃ ﻭﻗﺖ ﻃﻮﻳﻞ ﻟﻠﻌﺜﻮﺭ ﻋﲆ ﺭﻭﺍﻳﺔ ﻣﺎ ﰲ ﻣﻜﺘﺒﺔ ﻏﲑ‬ ‫ﻣﻨﻈﻤﺔ‪ .‬ﻟﺬﺍ ﹸﺗﺼﻨﻒ ﺍﻟﻜﺘﺐ ﰲ ﺍﳌﻜﺘﺒﺎﺕ ﰲ ﳎﻤﻮﻋﺎﺕ ﳐﺘﻠﻔﺔ ﻟﺘﺴﻬﻴﻞ ﻋﻤﻠﻴﺔ ﺍﻟﺒﺤﺚ ﻋﻨﻬﺎ‪.‬‬ ‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬ ‫ﺍﻟﻔﻠـﺰ‪ :‬ﻋﻨﴫ ﻳﻜـﻮﻥ ﺻﻠﺒﺎ ﰲ ﻭﻛﺬﻟﻚ ﺗﺼﻨﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺇﱃ ﺃﻧﻮﺍﻉ ﳐﺘﻠﻔﺔ‪.‬‬ ‫ﹰ‬ ‫ﺍﻟﻐﺎﻟـﺐ ﻋﻨـﺪ ﺩﺭﺟـﺔ ﺣـﺮﺍﺭﺓ ﺃﻧﻮاﻉ الﺘﻔاﻋﻼت الﻜﻴمﻴاﺋﻴﺔ ‪Types of Chemical Reactions‬‬ ‫ﹰ‬ ‫ﺍﻟﻐﺮﻓـﺔ‪،‬‬ ‫ﻭﻣﻮﺻـﻼ ﺟﻴـﺪﹰ ﺍ ﻳﺼﻨﻒ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﺘﻨﻈﻴﻢ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻜﺒﲑﺓ ﻣﻦ ﻫﺬﻩ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﻟﻠﺤـﺮﺍﺭﺓ ﻭﺍﻟﻜﻬﺮﺑـﺎﺀ‪،‬‬ ‫ﻳﻮﻣﻴﺎ‪ .‬ﺇﻥ ﻣﻌﺮﻓﺔ ﺃﻧﻮﺍﻉ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻳﻤﻜﻦ ﺃﻥ ﻳﺴﺎﻋﺪﻙ ﻋﲆ‬ ‫ﹰ‬ ‫ﻭﻻﻣﻌﺎ ﺍﻟﺘﻲ ﲢﺪﺙ ﹰﹼ‬ ‫ﺑﺼﻮﺭﺓ ﻋﺎﻣﺔ‪.‬‬ ‫ﺗﻌﺮﻑ ﺃﻧﻮﺍﻋﻬﺎ ﻭﺗﻮﻗﻊ ﻧﻮﺍﺗﺞ ﺍﻟﻜﺜﲑ ﻣﻨﻬﺎ‪ .‬ﻭﻫﻨﺎﻙ‬ ‫ﺗﺬﻛﺮﻫﺎ ﻭﻓﻬﻤﻬﺎ‪ ،‬ﻛﲈ ﺃﻧﻪ ﻳﺴﺎﻋﺪﻙ ﻋﲆ ﹼ‬ ‫ﻋﺪﺓ ﻃﺮﺍﺋﻖ ﻟﺘﺼﻨﻴﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻣﻦ ﺃﺑﺴﻄﻬﺎ ﺗﻠﻚ ﺍﻟﺘﻲ ﺗﺼﻨﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬ ‫ﺇﱃ ﺃﺭﺑﻌﺔ ﺃﻧﻮﺍﻉ‪ ،‬ﻫﻲ‪ :‬ﺍﻟﺘﻜﻮﻳﻦ‪ ،‬ﻭﺍﻻﺣﱰﺍﻕ‪ ،‬ﻭﺍﻟﺘﻔﻜﻚ‪ ،‬ﻭﺍﻹﺣﻼﻝ‪ .‬ﻭﻗﺪ ﺗﻨﺪﺭﺝ ﺑﻌﺾ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻜﻮﻳﻦ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﲢﺖ ﺃﻛﺜﺮ ﻣﻦ ﻧﻮﻉ ﻣﻦ ﻫﺬﻩ ﺍﻷﻧﻮﺍﻉ‪.‬‬ ‫ﺗﻔﺎﻋﻞ ﺍﻻﺣﱰﺍﻕ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻔﻜﻚ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‬ ‫ﺍﻟﺮﺍﺳﺐ‬

‫ﺗﻔاﻋﻼت الﺘﻜﻮﻳﻦ ‪Synthesis Reactions‬‬

‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻜﻮﻳﻦ ﻫﻮ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﺗﺘﺤﺪ ﻓﻴﻪ ﻣﺎﺩﺗﺎﻥ ﺃﻭ ﺃﻛﺜﺮ ﻟﺘﻜﻮﻳﻦ ﻣﺎﺩﺓ ﻭﺍﺣﺪﺓ ﻭﻳﻤﻜﻦ‬ ‫ﲤﺜﻴﻠﻪ ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻟﻌﺎﻣﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪A + B → AB‬‬

‫ﻋﻨﺪﻣﺎ ﻳﺘﻔﺎﻋﻞ ﻋﻨﴫﺍﻥ ﻓﺈﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻳﻜﻮﻥ ﹰ‬ ‫ﺩﺍﺋﲈ ﺗﻔﺎﻋﻞ ﺗﻜﻮﻳﻦ ﻛﲈ ﰲ ﺍﻟﺸﻜﻞ ‪ 4-7‬ﺍﻟﺬﻱ‬ ‫ﻳﻮﺿﺢ ﺗﻔﺎﻋﻞ ﻋﻨﴫ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻣﻊ ﻋﻨﴫ ﺍﻟﻜﻠﻮﺭ ﻟﺘﻜﻮﻳﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪ .‬ﻛﲈ ﻳﻤﻜﻦ ﺃﻥ‬ ‫ﻳﺘﺤﺪ ﻣﺮﻛﺒﺎﻥ ﻟﺘﻜﻮﻳﻦ ﻣﺮﻛﺐ ﻭﺍﺣﺪ‪ .‬ﻓﻤﺜﻼ‪ ،‬ﺍﻟﺘﻔﺎﻋﻞ ﺑﲔ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ CaO‬ﻭﺍﳌﺎﺀ‬ ‫‪ H2O‬ﻟﺘﻜﻮﻳﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca(OH)2‬ﻫﻮ ﺗﻔﺎﻋﻞ ﺗﻜﻮﻳﻦ‪.‬‬ ‫)‪CaO(s) + H2O(l) → Ca(OH)2(s‬‬

‫ﻭﻫﻨﺎﻙ ﻧﻮﻉ ﺁﺧﺮ ﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻜﻮﻳﻦ ﻳﺘﻀﻤﻦ ﺗﻔﺎﻋﻞ ﻣﺮﻛﺐ ﻣﻊ ﻋﻨﴫ‪ ،‬ﻣﺜﻞ ﺗﻔﺎﻋﻞ ﻏﺎﺯ ﺛﺎﲏ‬ ‫ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ ‪ SO2‬ﻣﻊ ﻏﺎﺯ ﺍﻷﻛﺴﺠﲔ ‪ O2‬ﻟﺘﻜﻮﻳﻦ ﻏﺎﺯ ﺛﺎﻟﺚ ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ ‪.SO3‬‬ ‫)‪2SO2(g) + O2(g) →2 SO3(g‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-7‬ﻳﺘﻔﺎﻋـﻞ ﻋﻨـﴫﺍ ﺍﻟﺼﻮﺩﻳـﻮﻡ‬ ‫ﻭﺍﻟﻜﻠـﻮﺭ ﻟﺘﻜﻮﻳـﻦ ﻣﺮﻛﺐ ﻭﺍﺣـﺪ‪ ،‬ﻫﻮ‬ ‫ﻛﻠﻮﺭﻳـﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪) ،‬ﺗﻔﺎﻋـﻞ ﺗﻜﻮﻳﻦ(‬

‫‪15‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-8‬ﺍﻟﻀﻮﺀ ﺍﻟﻨﺎﺗﺞ ﻣﻦ ﻫﺬﻩ ﺍﻟﻠﻌﺒﺔ‬ ‫ﺍﻟﻨﺎﺭﻳﺔ ﻫﻮ ﻧﺘﻴﺠﺔ ﺗﻔﺎﻋﻞ ﺍﺣﱰﺍﻕ‬ ‫ﺑﲔ ﺍﻷﻛﺴﺠﲔ ﻭﻓﻠﺰﺍﺕ ﳐﺘﻠﻔﺔ‪.‬‬

‫‪‬‬

‫ﺗﻔاﻋﻼت اﻻﺣﺘراﻕ ‪Combustion Reactions‬‬

‫‪áª∏µdG π°UCG‬‬

‫ﻳﻤﻜﻦ ﺃﻥ ﻳﺼﻨﻒ ﺗﻔﺎﻋﻞ ﺗﻜﻮﻳﻦ ﺛﺎﻟﺚ ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ ﻋﲆ ﺃﻧﻪ ﺗﻔﺎﻋﻞ ﺍﺣﱰﺍﻕ‬ ‫ﹰ‬ ‫ﺃﻳﻀﺎ‪ .‬ﰲ ﺗﻔﺎﻋﻞ ﺍﻻﺣﱰﺍﻕ ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪ 4-8‬ﻳﺘﺤﺪ ﺍﻷﻛﺴﺠﲔ ﻣﻊ‬ ‫ﻣﺎﺩﺓ ﻛﻴﻤﻴﺎﺋﻴﺔ ﹺ‬ ‫ﻣﻄﻠ ﹰﻘﺎ ﻃﺎﻗﺔ ﻋﲆ ﺷﻜﻞ ﺣﺮﺍﺭﺓ ﻭﺿﻮﺀ‪ .‬ﻭﻳﻤﻜﻦ ﻟﻸﻛﺴﺠﲔ‬ ‫ﺃﻥ ﻳﺘﺤﺪ ﲠﺬﻩ ﺍﻟﻄﺮﻳﻘﺔ ﻣﻊ ﻣﻮﺍﺩ ﻛﺜﲑﺓ ﳐﺘﻠﻔﺔ‪ ،‬ﳑﺎ ﳚﻌﻞ ﺗﻔﺎﻋﻼﺕ ﺍﻻﺣﱰﺍﻕ‬ ‫ﺷﺎﺋﻌﺔ‪ .‬ﻓﻴﺤﺪﺙ ﺗﻔﺎﻋﻞ ﺍﻻﺣﱰﺍﻕ ﹰ‬ ‫ﻣﺜﻼ ﺑﲔ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻭﺍﻷﻛﺴﺠﲔ ﻋﻨﺪﻣﺎ‬ ‫ﱠ‬ ‫ﻳﺴﺨﻦ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪ ،4-9‬ﺣﻴﺚ ﻳﺘﻜﻮﻥ ﺍﳌﺎﺀ ﺧﻼﻝ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‪ ،‬ﻭﺗﻨﻄﻠﻖ ﻛﻤﻴﺔ ﻛﺒﲑﺓ ﻣﻦ ﺍﻟﻄﺎﻗﺔ‪ .‬ﻛﲈ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ﺍﺣﱰﺍﻕ ﻣﻬﻢ ﻋﻨﺪ‬ ‫ﺣﺮﻕ ﺍﻟﻔﺤﻢ ﻟﻠﺤﺼﻮﻝ ﻋﲆ ﻃﺎﻗﺔ‪ ،‬ﺑﺤﺴﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬

‫ﺍﻻﺣﱰﺍﻕ )‪ :(Combustion‬ﺃﺻﻞ ﻫﺬﻩ‬ ‫ﺍﻟﻜﻠﻤﺔ ﻻﺗﻴﻨﻲ‪ ،‬ﻛﻮﻣﺒﻮﺭ‪ ،‬ﻭﺗﻌﻨﻲ ﳛﱰﻕ‪.‬‬

‫) ‪C(s) + O2 (g) → CO2 (g‬‬

‫ ‬

‫‪16‬‬

‫ﰲ ‪ 1775‬ﺃﺛﺒـﺖ ﺃﻧﻄـﻮﲏ‬ ‫ﻻﻓــﻮﺯﻳـﻪ ﺃﻥ ﺗﻔـــﺎﻋـﻼﺕ‬ ‫ﺍﻻﺣـﱰﺍﻕ ﻃـﺎﺭﺩﺓ ﻟﻠﻄﺎﻗـﺔ‪،‬‬ ‫ﻭﺗﺘﻄﻠﺐ ﻭﺟﻮﺩ ﺍﻷﻛﺴﺠﲔ‪.‬‬

‫ ‬

‫ﰲ ﻋـﺎﻡ ‪ 1800‬ﺃﺩﺕ ﺑﻌﺾ ﺃﺑﺤﺎﺙ‬ ‫ﺍﻟﻨﺒـﺎﺕ ﺇﱃ ﺍﻛﺘﺸـﺎﻑ ﻣﻌﺎﺩﻟـﺔ‬ ‫ﻛﻴﻤﻴﺎﺋﻴــﺔ ﻣــﻮﺯﻭﻧﺔ ﻟﻌﻤﻠﻴﺔ ﺍﻟﺒﻨﺎﺀ‬ ‫ﺍﻟﻀﻮﺋﻲ‪.‬‬

‫ﰲ ﻋــﺎﻡ ‪ 1885‬ﺍﺧﱰﻉ‬ ‫ﳏــــﺮﻙ ﺍﻻﺣـــــﱰﺍﻕ‬ ‫ﺍﻟﺪﺍﺧﲇ‪ ،‬ﻭﺍﻟﺬﻱ ﺻﺎﺭ‬ ‫ﻧﻤﻮﺫﺟﺎ‬ ‫ﰲ ﻣــﺎ ﺑﻌﺪ‬ ‫ﹰ‬ ‫ﻟﻠﻤــﺤﺮﻙ ﺍﳊﺪﻳﺚ‪.‬‬

‫‪ 1909-1910‬ﻗـﺎﻡ ﻋﺎﳌـﺎﻥ‬ ‫ﻣﻦ ﺃﳌﺎﻧﻴـﺎ ‪ -‬ﳘﺎ ﻓﺮﺗﺰﻫﺎﺑﺮ‬ ‫ﻭﻛﺎﺭﻝ ﺑـﻮﺵ‪ -‬ﺑﻮﺿـﻊ‬ ‫ﻋﻤﻠﻴﺔ ﻟﺘﺤﻀﲑ ﺍﻷﻣﻮﻧﻴﺎ‪.‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-9‬ﻳﺘﻜﻮﻥ ﺍﳌﺎﺀ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﺍﻻﺣﱰﺍﻕ‬ ‫ﺑﲔ ﻏﺎﺯﻱ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻭﺍﻷﻛﺴﺠﲔ‪.‬‬ ‫ﺣﻠـﻞ ﳌﺎﺫﺍ ﻳﻌﺪ ﻫـﺬﺍ ﺍﻟﺘﻔﺎﻋـﻞ ﺗﻔﺎﻋﻞ‬ ‫ﺍﺣﱰﺍﻕ ﻭﺗﻔﺎﻋﻞ ﺍﲢﺎﺩ ﹰ‬ ‫ﺃﻳﻀﺎ‪.‬‬

‫ﻻﺣﻆ ﺃﻥ ﲨﻴﻊ ﺗﻔﺎﻋﻼﺕ ﺍﻻﺣﱰﺍﻕ ـ ﺍﻟﺘﻲ ﹸﺫﻛﺮﺕ ـ ﻫﻲ ﺗﻔﺎﻋﻼﺕ ﺗﻜﻮﻳﻦ ﹰ‬ ‫ﺃﻳﻀﺎ‪ ،‬ﻟﻜﻦ ﻟﻴﺲ ﻛﻞ ﺗﻔﺎﻋﻼﺕ ﺍﻻﺣﱰﺍﻕ‬ ‫ﺗﻔﺎﻋﻼﺕ ﺗﻜﻮﻳﻦ‪ .‬ﻓﻤﺜ ﹰ‬ ‫ﻼ‪ ،‬ﻳﻨﺘﺞ ﺗﻔﺎﻋﻞ ﺍﺣﱰﺍﻕ ﻏﺎﺯ ﺍﳌﻴﺜﺎﻥ ﺃﻛﺜﺮ ﻣﻦ ﻣﺮﻛﺐ‪ ،‬ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫)‪CH4(g) + 2O2(g) → CO2(g) + 2H2O(g‬‬

‫ﺍﳌﻴﺜﺎﻥ ﻫﻮ ﺍﳌﻜﻮﻥ ﺍﻟﺮﺋﻴﴘ ﻟﻠﻐﺎﺯ ﺍﻟﻄﺒﻴﻌﻲ‪ ،‬ﻭﻳﻨﺘﻤﻲ ﺇﱃ ﳎﻤﻮﻋﺔ ﻣﻦ ﹼ‬ ‫ﺍﳌﺮﻛﺒﺎﺕ ﺗﺴﻤﻰ ﺍﳍﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ‪ ،‬ﻭﻫﻲ ﺍﳌﻜﻮﻥ ﺍﻷﺳﺎﳼ‬ ‫ﻟﻠﻨﻔﻂ‪ .‬ﻭﲢﺘﻮﻱ ﺍﳍﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ ﲨﻴﻌﻬﺎ ﻋﲆ ﻛﺮﺑﻮﻥ ﻭﻫﻴﺪﺭﻭﺟﲔ‪ ،‬ﻭﲢﱰﻕ ﰲ ﺍﻷﻛﺴﺠﲔ ﻹﻧﺘﺎﺝ ﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‬ ‫ﻭﺍﳌﺎﺀ ﻭﻛﻤﻴﺔ ﻛﺒﲑﺓ ﻣﻦ ﺍﻟﻄﺎﻗﺔ‪ ،‬ﻭﻫﺬﺍ ﻣﺎ ﳚﻌﻞ ﻣﻦ ﺍﻟﻨﻔﻂ ﺍﳌﺼﺪﺭ ﺍﻷﺳﺎﳼ ﻟﻠﻄﺎﻗﺔ ﰲ ﺣﻴﺎﺗﻨﺎ ﺍﳌﻌﺎﴏﺓ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﺎﻟﻴﺔ‪ ،‬ﻭﺻﻨﻒ ﻛﻞ ﺗﻔﺎﻋﻞ ﻣﻨﻬﺎ‪:‬‬ ‫‪ .14‬ﺗﻔﺎﻋﻞ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﺍﻟﺼﻠﺐ ‪ Al‬ﻭﺍﻟﻜﱪﻳﺖ ﺍﻟﺼﻠﺐ ‪ S‬ﻹﻧﺘﺎﺝ ﻛﱪﻳﺘﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‬ ‫ﺍﻟﺼﻠﺐ ‪.Al2S3‬‬ ‫‪ .15‬ﺗﻔﺎﻋﻞ ﺍﳌﺎﺀ ﻭﻏﺎﺯ ﺧﺎﻣﺲ ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﱰﻭﺟﲔ ‪ N2O5‬ﻹﻧﺘﺎﺝ ﳏﻠﻮﻝ ﲪﺾ‬ ‫ﺍﻟﻨﻴﱰﻳﻚ ‪.HNO3‬‬ ‫‪ .16‬ﺗﻔﺎﻋﻞ ﻏﺎﺯﻱ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﱰﻭﺟﲔ ‪ NO2‬ﻭﺍﻷﻛﺴﺠﲔ‪ ،‬ﻹﻧﺘﺎﺝ ﻏﺎﺯ ﺧﺎﻣﺲ‬ ‫ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﱰﻭﺟﲔ‪.‬‬ ‫‪ .17‬ﲢﺪﱟ ﺗﻔﺎﻋﻞ ﳏﻠﻮﱄ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ‪ H2SO4‬ﻭﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻹﻧﺘﺎﺝ ﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2SO4‬ﻭﺍﳌﺎﺀ‪.‬‬

‫ﰲ ﻋـﺎﻡ ‪ 1952‬ﻏﻄـﻰ ﺩﺧﺎﻥ‬ ‫ﻛﺜﻴــﻒ ﻣــﻦ ﺛـﺎﲏ ﺃﻛﺴـﻴﺪ‬ ‫ﺍﻟﻜﱪﻳـﺖ ﻭﺑﻌـﺾ ﻧﻮﺍﺗـﺞ‬ ‫ﺍﺣـﱰﺍﻕ ﺍﻟﻔﺤﻢ ﻣﺪﻳﻨﺔ ﻟﻨﺪﻥ‬ ‫ﻣﺪﺓ ﲬﺴـﺔ ﺃﻳﺎﻡ ﻭﺗﺴـﺒﺐ ﰲ‬ ‫‪ 4000‬ﺣﺎﻟﺔ ﻭﻓﺎﺓ‪.‬‬

‫ ‬

‫ ‬

‫‪ 1974-1978‬ﺃﺛﺒـﺖ ﺍﻟﺒﺎﺣﺜـﻮﻥ ﺃﻥ‬ ‫ﺍﻟﻜﻠﻮﺭﻭﻓﻠﻮﺭﻭﻛﺮﺑﻮﻧـﺎﺕ ‪CFCs‬‬ ‫ﺗﺴـﺘﻨﺰﻑ ﻃﺒﻘﺔ ﺍﻷﻭﺯﻭﻥ‪ .‬ﻟﺬﻟﻚ ﺗﻢ‬ ‫ﺣﻈﺮ ﺍﺳﺘﻌﲈﻝ ﻋﻠﺐ ﺍﻟﺮﺵ ﺍﻟﺪﻓﻌﻲ‬ ‫ﺍﻟﺘﻲ ﲢﺘﻮﻱ ﺍﻷﻭﺯﻭﻥ‪.‬‬

‫ﰲ ﻋـــﺎﻡ ‪ 2004‬ﺍﻛﺘﺸﻒ‬ ‫ﺍﻟﻌﻠﲈﺀ ﺃﻥ ﺍﻟﻄﻴﻮﺭ ﺍﳌﻬﺎﺟﺮﺓ‬ ‫ﺗﺴﱰﺷﺪ ﺑﺘﻔﺎﻋﻼﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﲢﺪﺙ ﰲ ﺃﺟﺴﺎﻣﻬﺎ ﻭﺗﺘﺄﺛﺮ‬ ‫ﺑﺎﳌﺠﺎﻝ ﺍﳌﻐﻨﺎﻃﻴﴘ ﻟﻸﺭﺽ‪.‬‬

‫ﰲ ﻋﺎﻡ ‪ 1995‬ﺍﺳـﺘﻌﺎﻥ ﺍﻟﺒﺎﺣﺜﻮﻥ‬ ‫ﺑﺎﻟـﻤﺠﻬـﺮ ﺍﻟـــﺬﺭﻱ ﻹﺣـﺪﺍﺙ‬ ‫ﺗﻔﺎﻋـﻼﺕ ﻛﻴﻤﻴﺎﺋﻴـﺔ‪ ،‬ﻭﻣﻼﺣﻈﺔ‬ ‫ﺁﻟﻴـﺔ ﺣﺪﻭﺛﻬـﺎ ﻋـﲆ ﺍﳌﺴـﺘﻮ￯‬ ‫ﺍﳉﺰﻳﺌﻲ‪ ،‬ﳑﺎ ﻣﻬﺪ ﳍﻨﺪﺳﺔ ﺍﻟﻨﺎﻧﻮ‪.‬‬

‫ﺍﻟﺸﻜﻞ ‪4-10‬‬

‫ﺗﻔﺎﻋﻼﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻦ ﻭﺍﻗﻊ ﺍﻟﺤﻴﺎﺓ‬ ‫ﻋﻤﻞ ﺍﻟﻨﺎﺱ ﻋﲆ ﻣﺮ ﺍﻟﻌﺼﻮﺭ ﻋﲆ ﻓﻬﻢ‬ ‫ﺍﻟﻄﺎﻗﺔ ﺍﻟﻨﺎﲡﺔ ﻋﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻭﺍﻻﺳﺘﻔﺎﺩﺓ ﻣﻨﻬﺎ ﰲ ﺣﻞ ﻣﺸﺎﻛﻠﻬﻢ‪.‬‬

‫ﺗﻔﺎﻋﻞ ﺍﻟﺨﻂ ﺍﻟﺰﻣﻨﻲ‪ :‬ﳌﻌﺮﻓﺔ ﺍﳌﺰﻳﺪ ﻣﻦ ﺍﳌﻌﻠﻮﻣﺎﺕ ﻋﻦ ﻫﺬﻩ‬ ‫ﺍﻻﻛﺘﺸﺎﻓﺎﺕ ﻭﻏﲑﻫﺎ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ‪:‬‬ ‫‪www.obeikaneducation.com‬‬

‫‪17‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-11‬ﺗﻔﻜﻚ ﺃﺯﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫‪ ،NaN3‬ﺍﻟــﺬﻱ ﻳﻨﺘﺞ‬ ‫ﻏــــﺎﺯ ﺍﻟــﻨــﻴــﱰﻭﺟــﲔ‪،‬‬ ‫ﻭﻫــﻮ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟــﺬﻱ‬ ‫ﻳﺴﺘﻌﻤﻞ ﰲ ﻧﻔﺦ ﺃﻛﻴﺎﺱ‬ ‫ﺍﳍـــﻮﺍﺀ ﰲ ﺍﻟﺴﻴﺎﺭﺍﺕ‪.‬‬

‫المطﻮﻳات‬

‫ﺃﺩﺧـﻞ ﻣﻌﻠﻮﻣﺎﺕ ﻣﻦ ﻫﺬﺍ ﺍﻟﺠﺰﺀ ﻓﻲ‬ ‫ﻣﻄﻮﻳﺘﻚ‪.‬‬

‫ﺗﻔاﻋﻼت الﺘﻔﻜﻚ ‪Decomposition Reactions‬‬

‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻔﻜﻚ ﺗﻔﺎﻋﻞ ﻳﺘﻔﻜﻚ ﻓﻴﻪ ﻣﺮﻛﺐ ﻭﺍﺣﺪ ﻹﻧﺘﺎﺝ ﻋﻨﴫﻳﻦ ﺃﻭ ﺃﻛﺜﺮ ﺃﻭ ﻣﺮﻛﺒﺎﺕ‬ ‫ﺟﺪﻳﺪﺓ‪ .‬ﻭﳍﺬﺍ ﻓﺈﻥ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻔﻜﻚ ﻫﻲ ﻋﻜﺲ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻜﻮﻳﻦ‪ .‬ﻭﻳﻤﻜﻦ ﲤﺜﻴﻠﻬﺎ‬ ‫ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻟﻌﺎﻣﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪AB → A + B‬‬

‫ﻭﻏﺎﻟ ﹰﺒﺎ ﻣﺎ ﲢﺘﺎﺝ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻔﻜﻚ ﻟﻜﻲ ﲢﺪﺙ ﺇﱃ ﻣﺼﺪﺭ ﻟﻠﻄﺎﻗﺔ‪ ،‬ﻛﺎﳊﺮﺍﺭﺓ‪ ،‬ﺃﻭ‬ ‫ﺍﻟﻀﻮﺀ‪ ،‬ﺃﻭ ﺍﻟﻜﻬﺮﺑﺎﺀ‪ .‬ﻓﺘﺘﻔﻜﻚ ﻧﱰﺍﺕ ﺍﻷﻣﻮﻧﻴﻮﻡ ﹰ‬ ‫ﻣﺜﻼ ﺇﱃ ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﱰﻭﺟﲔ ﺍﻷﺣﺎﺩﻱ‬ ‫ﻭﻣﺎﺀ‪ ،‬ﻋﻨﺪﻣﺎ ﺗﺴﺨﻦ ﺇﱃ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻋﺎﻟﻴﺔ‪:‬‬ ‫‪‬‬ ‫→ )‪NH4NO3(S‬‬ ‫)‪N2O(g) + 2H2O(g‬‬

‫ﻻﺣﻆ ﺃﻥ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﻳﺘﻀﻤﻦ ﺗﻔﻜﻚ ﻣﺎﺩﺓ ﻣﺘﻔﺎﻋﻠﺔ ﻭﺍﺣﺪﺓ ﺇﱃ ﺃﻛﺜﺮ ﻣﻦ ﻧﺎﺗﺞ‪ .‬ﻭﻣﻦ‬ ‫ﺍﻷﻣﺜﻠﺔ ﺍﳌﺸﻬﻮﺭﺓ ﻋﲆ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻔﻜﻚ ﺗﻔﻜﻚ ﺃﺯﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﻓﻖ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫)‪2NaN3(s) → 2Na(s) + 3N2(g‬‬

‫ﻭﻳﺴﺘﻌﻤﻞ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﰲ ﻧﻔﺦ ﺃﻛﻴﺎﺱ ﺍﳍﻮﺍﺀ )ﺃﻛﻴﺎﺱ ﺍﻟﺴﻼﻣﺔ( ﰲ ﺍﻟﺴﻴﺎﺭﺍﺕ‪ ،‬ﻛﲈ ﻫﻮ‬ ‫ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪ ،4-11‬ﺣﻴﺚ ﻳﻮﺿﻊ ﰲ ﺍﻟﻜﻴﺲ ﻣﻊ ﺍﻷﺯﻳﺪ ﺟﻬﺎﺯ ﻳﻮﻓﺮ ﺇﺷﺎﺭﺓ ﻛﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﻣﻨﺘﺠﺎ‬ ‫ﻟﺒﺪﺀ ﺍﻟﺘﻔﺎﻋﻞ‪ .‬ﻭﻋﻨﺪﻣﺎ ﻳﻨﺸﻂ ﺍﳉﻬﺎﺯ ﻧﺘﻴﺠﺔ ﺍﻻﺻﻄﺪﺍﻡ ﻳﺘﺤﻠﻞ ﺃﺯﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﹰ‬ ‫ﻏﺎﺯ ﺍﻟﻨﻴﱰﻭﺟﲔ ﺍﻟﺬﻱ ﻳﻨﻔﺦ ﺍﻟﻜﻴﺲ ﺑﴪﻋﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪18‬‬

‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﺤﻠﻞ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪ .18‬ﻳﺘﻔﻜﻚ ﺃﻛﺴﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﺍﻟﺼﻠﺐ ‪ Al2O3‬ﻋﻨﺪﻣﺎ ﺗﴪﻱ ﻓﻴﻪ ﺍﻟﻜﻬﺮﺑﺎﺀ ﺇﱃ‬ ‫ﺍﻟﻮﻣﻨﻴﻮﻡ ﺻﻠﺐ ﻭﻏﺎﺯ ﺍﻷﻛﺴﺠﲔ ‪.‬‬ ‫‪ .19‬ﻳﺘﻔﻜﻚ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﻴﻜﻞ ‪ II‬ﺍﻟﺼﻠﺐ ‪ Ni(OH)2‬ﻹﻧﺘﺎﺝ ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﻜﻞ ‪ II‬ﺍﻟﺼﻠﺐ ‪ NiO‬ﻭﺍﳌﺎﺀ‪.‬‬ ‫‪ .20‬ﲢﺪﱟ ‪ :‬ﻳﻨﺘﺞ ﻋﻦ ﺗﺴﺨﲔ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﳍﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﺍﻟﺼﻠﺒﺔ ‪ NaHCO3‬ﻛﺮﺑﻮﻧﺎﺕ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﻟﺼﻠﺒﺔ ‪ Na2CO3‬ﻭﺑﺨﺎﺭ ﺍﳌﺎﺀ ‪ H2O‬ﻭﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ‪.CO2‬‬

‫‪a‬‬

‫‪b‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-12‬ﰲ ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‬ ‫ﲢﻞ ﺫﺭﺍﺕ ﻋﻨﴫ ﳏﻞ ﺫﺭﺍﺕ‬ ‫ﻋــﻨــﴫ ﺁﺧـــﺮ ﰲ ﻣــﺮﻛــﺐ‪.‬‬

‫ﻟﻴﺜﻴﻮﻡ ‪ +‬ﻣﺎﺀ‬

‫ﻧﺤﺎﺱ ‪ +‬ﻧﱰﺍﺕ ﺍﻟﻔﻀﺔ‬

‫ﺗﻔاﻋﻼت اﻹﺣﻼﻝ ‪Replacement Reactions‬‬

‫ﻫﻨﺎﻙ ﺍﻟﻜﺜﲑ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﺘﻀﻤﻦ ﺇﺣﻼﻝ ﻋﻨﴫ ﳏﻞ ﻋﻨﴫ ﺁﺧﺮ ﰲ ﻣﺮﻛﺐ‪،‬‬ ‫ﻭﺗﺴﻤﻰ ﻫﺬﻩ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ‪ .‬ﻭﻫﻨﺎﻙ ﻧﻮﻋﺎﻥ ﻣﻨﻬﺎ‪ ،‬ﳘﺎ‪ :‬ﺍﻹﺣﻼﻝ‬ ‫ﺍﻟﺒﺴﻴﻂ‪ ،‬ﻭﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‪.‬‬

‫‪ §«°ùÑdG∫ÓME’GäÓYÉØJ‬ﻳﺒﻴﻦﺍﻟﺸﻜﻞ )‪ 4-12(a‬ﺍﻟﺘﻔﺎﻋﻞﺑﻴﻦﺍﻟﻠﻴﺜﻴﻮﻡﻭﺍﻟﻤﺎﺀ‪،‬ﺣﻴﺚﺗﺤﻞﻓﻴﻪ‬ ‫ﺫﺭﺓ ﻟﻴﺜﻴﻮﻡ ﻣﺤﻞ ﺫﺭﺓ ﻭﺍﺣﺪﺓ ﻣﻦ ﺫﺭﺗﻲ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﻤﺎﺀ‪ ،‬ﻛﻤﺎ ﺗﻮﺿﺤﻪ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫)‪2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g‬‬

‫ﻭﻳﺴﻤﻰ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﺬﻱ ﲢﻞ ﻓﻴﻪ ﺫﺭﺍﺕ ﻋﻨﴫ ﳏﻞ ﺫﺭﺍﺕ ﻋﻨﴫ ﺁﺧﺮ ﰲ ﻣﺮﻛﺐ ﺗﻔﺎﻋﻞ‬ ‫ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‪ ،‬ﻭﻳﻤﻜﻦ ﲤﺜﻴﻠﻪ ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻟﻌﺎﻣﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪A + BX → AX + B‬‬

‫ﺍﻟﻔﻠـﺰﺍﺕ ﺗﺤﻞ ﻣﺤـﻞ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺃﻭ ﻓﻠﺰ ﺁﺧـﺮ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ ﺍﻟﻠﻴﺜﻴـﻮﻡ ﻭﺍﻟﻤﺎﺀ ﺃﺣﺪ ﺍﻷﻣﺜﻠﺔ‬ ‫ﻋﻠﻰ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴـﻴﻂ‪ ،‬ﺣﻴﺚ ﺗﺤﻞ ﻓﻴﻪ ﺫﺭﺓ ﻓﻠﺰ ﻣﺤﻞ ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺟﺰﻱﺀ‬ ‫ﺍﻟﻤـﺎﺀ‪ .‬ﻭﻳﺤـﺪﺙ ﻧﻮﻉ ﺁﺧﺮ ﻣﻦ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴـﻴﻂ ﻋﻨﺪﻣﺎ ﻳﺤﻞ ﻓﻠﺰ ﻣﺤﻞ ﻓﻠـﺰ ﺁﺧﺮ ﻓﻲ ﻣﺮﻛﺐ‬ ‫ﻣﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺎﺀ‪ .‬ﻭﻳﺒﻴﻦ ﺍﻟﺸﻜﻞ )‪ 4-12 (b‬ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﺇﺣﻼﻝ ﺑﺴﻴﻂ ﻋﻨﺪ ﻭﺿﻊ ﺻﻔﻴﺤﺔ‬ ‫ﻣﻦ ﺍﻟﻨﺤﺎﺱ ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻟﻨﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ‪ .‬ﻓﺎﻟﺒﻠﻮﺭﺍﺕ ﺍﻟﻤﺘﺮﺍﻛﻤﺔ ﻋﻠﻰ ﻗﻄﻌﺔ ﺍﻟﻨﺤﺎﺱ ﻫﻲ‬ ‫ﺫﺭﺍﺕ ﺍﻟﻔﻀﺔ ﺍﻟﺘﻲ ﺣﻠﺖ ﻣﺤﻠﻬﺎ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ‪.‬‬ ‫)‪Cu(s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s‬‬

‫ﻻ ﳛﻞ ﺍﻟﻔﻠﺰ ﹰ‬ ‫ﺩﺍﺋﲈ ﳏﻞ ﻓﻠﺰ ﺁﺧﺮ ﰲ ﻣﺮﻛﺐ ﻣﺬﺍﺏ ﰲ ﺍﳌﺎﺀ؛ ﻭﺫﻟﻚ ﻷﻥ ﺍﻟﻔﻠﺰﺍﺕ ﲣﺘﻠﻒ ﰲ‬ ‫ﻧﺸﺎﻃﻬﺎ‪ ،‬ﺃﻭ ﻗﺪﺭﲥﺎ ﻋﲆ ﺍﻟﺘﻔﺎﻋﻞ ﻣﻊ ﻣﺎﺩﺓ ﺃﺧﺮ￯‪ .‬ﻭﻳﺒﲔ ﺍﻟﺸﻜﻞ ‪ 4-13‬ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺒﻌﺾ ﺍﻟﻔﻠﺰﺍﺕ‪ .‬ﻭﺗﺴﺘﺨﺪﻡ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﰲ ﲢﺪﻳﺪ ﻣﻮﻗﻊ ﺍﻟﻔﻠﺰﺍﺕ ﰲ‬ ‫ﺍﻟﺴﻠﺴﻠﺔ‪ ،‬ﺣﻴﺚ ﻳﻮﺟﺪ ﺃﻧﺸﻂ ﺍﻟﻔﻠﺰﺍﺕ ﰲ ﺃﻋﲆ ﺍﻟﺴﻠﺴﻠﺔ‪ ،‬ﺑﻴﻨﲈ ﻳﻮﺟﺪ ﺃﻗﻠﻬﺎ ﹰ‬ ‫ﻧﺸﺎﻃﺎ ﰲ ﺃﺳﻔﻠﻬﺎ‪.‬‬ ‫ﻭﻗﺪ ﺭﺗﺒﺖ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﰲ ﺳﻠﺴﻠﺔ ﻧﺸﺎﻁ ﺑﻄﺮﻳﻘﺔ ﻣﺸﺎﲠﺔ‪ ،‬ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪. 4-13‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-13‬ﺳﻼﺳـﻞ ﺍﻟﻨﺸـﺎﻁ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋـﻲ ﻛﺎﳌﺒﻴﻨـﺔ ﻫﻨـﺎ‬ ‫ﻟﻠﻔﻠﺰﺍﺕ ﻭﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ‬ ‫ﻫـﻲ ﺃﺩﻭﺍﺕ ﻣﻔﻴـﺪﺓ ﰲ‬ ‫ﲢﺪﻳﺪ ﺇﻣﻜﺎﻧﻴـﺔ ﺣﺪﻭﺙ‬ ‫ﺗﻔﺎﻋــــﻞ ﻛـﻴﻤﻴـﺎﺋــﻲ‪،‬‬ ‫ﻭﲢﺪﻳﺪ ﻧﻮﺍﺗﺞ ﺗﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣــﻼﻝ ﺍﻟﺒﺴـﻴـــﻂ‪.‬‬ ‫ﺍﻟﻔﻠﺰﺍﺕ‬

‫ﻟﻴﺜﻴﻮﻡ‬ ‫ﺭﻭﺑﻴﺪﻳﻮﻡ‬ ‫ﺑﻮﺗﺎﺳﻴﻮﻡ‬ ‫ﻛﺎﻟﺴﻴﻮﻡ‬ ‫ﺻﻮﺩﻳﻮﻡ‬ ‫ﻣﺎﻏﻨﻴﺴﻴﻮﻡ‬ ‫ﺃﻟﻮﻣﻨﻴﻮﻡ‬ ‫ﻣﻨﺠﻨﻴﺰ‬ ‫ﺧﺎﺭﺻﲔ‬ ‫ﺣﺪﻳﺪ‬ ‫ﻧﻴﻜﻞ‬ ‫ﻗﺼﺪﻳﺮ‬ ‫ﺭﺻﺎﺹ‬ ‫ﻧﺤﺎﺱ‬ ‫ﻓﻀﺔ‬ ‫ﺑﻼﺗﲔ‬ ‫ﺫﻫﺐ‬

‫ﺍﻷﻛﺜﺮ ﹰ‬ ‫ﻧﺸﺎﻃﺎ‬

‫ﺍﻷﻗﻞ ﹰ‬ ‫ﻧﺸﺎﻃﺎ‬

‫ﺍﻷﻛﺜﺮ ﹰ‬ ‫ﻧﺸﺎﻃﺎ‬ ‫ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ‬ ‫ﻓﻠﻮﺭ‬ ‫ﻛﻠﻮﺭ‬ ‫ﺑﺮﻭﻡ‬ ‫ﻳﻮﺩ‬ ‫ﺍﻷﻗﻞ ﹰ‬ ‫ﻧﺸﺎﻃﺎ‬

‫‪19‬‬

‫ﻳﻤﻜﻨﻚ ﺍﺳﺘﻌﲈﻝ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺘﺘﻮﻗﻊ ﺇﺫﺍ ﻛﺎﻥ ﺳﻴﺤﺪﺙ ﺗﻔﺎﻋﻞ ﺃﻡ ﻻ‪ .‬ﺇﻥ ﺃﻱ‬ ‫ﻓﻠﺰ ﻳﻤﻜﻨﻪ ﺃﻥ ﳛﻞ ﳏﻞ ﺃﻱ ﻓﻠﺰ ﻳﻘﻊ ﺑﻌﺪﻩ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ ،‬ﻭﻟﻜﻦ ﻻ ﻳﻤﻜﻨﻪ ﺃﻥ‬ ‫ﳛﻞ ﳏﻞ ﺃﻱ ﻓﻠﺰ ﻳﻘﻊ ﻗﺒﻠﻪ‪ .‬ﻓﻤﺜﻼ‪ ،‬ﲢﻞ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ ﳏﻞ ﺫﺭﺍﺕ ﺍﻟﻔﻀﺔ ﰲ ﳏﻠﻮﻝ ﻧﱰﺍﺕ‬ ‫ﺍﻟﻔﻀﺔ‪ ،‬ﻭﻟﻜﻦ ﻟﻮ ﻭﺿﻌﺖ ﹰ‬ ‫ﺳﻠﻜﺎ ﻣﻦ ﺍﻟﻔﻀﺔ ﰲ ﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻟﻨﺤﺎﺱ ‪ II‬ﻓﺈﻥ ﺫﺭﺍﺕ ﺍﻟﻔﻀﺔ ﻻ‬ ‫ﲢﻞ ﳏﻞ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ؛ ﻷﻥ ﺍﻟﻔﻀﺔ ﺗﻘﻊ ﺑﻌﺪ ﺍﻟﻨﺤﺎﺱ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ .‬ﻭﳍﺬﺍ‬ ‫ﻻ ﳛﺪﺙ ﺗﻔﺎﻋﻞ‪ .‬ﻭﻳﺴﺘﺨﺪﻡ ﺍﻟﺮﻣﺰ )‪ (NR‬ﻋﺎﺩﺓ ﻟﻠﺪﻻﻟﺔ ﻋﲆ ﻋﺪﻡ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﻻ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ‪Ag (s) + Cu(NO3)2 (aq) → NR‬‬

‫ﺍﻟﻼﻓﻠﺰﻳﺤﻞﻣﺤﻞﺍﻟﻼﻓﻠﺰﻫﻨﺎﻙﻧﻮﻉﺛﺎﻟﺚﻣﻦﺗﻔﺎﻋﻼﺕﺍﻹﺣﻼﻝﺍﻟﺒﺴﻴﻂﻳﺘﻢﻓﻴﻪﺍﺳﺘﺒﺪﺍﻝﻻﻓﻠﺰﻓﻲ‬ ‫ﻣﺮﻛﺐ ﺑﻼﻓﻠﺰ ﺁﺧﺮ‪ ،‬ﻛﻤﺎ ﻫﻮ ﺷﺎﺋﻊ ﻓﻲ ﺑﻌﺾ ﺗﻔﺎﻋﻼﺕ ﺍﻟﻬﺎﻟﻮﺟﻴﻨﺎﺕ‪ .‬ﻓﺎﻟﻬﺎﻟﻮﺟﻴﻨﺎﺕ ﻛﺎﻟﻔﻠﺰﺍﺕ‪،‬‬ ‫ﻓﻬﻲ ﺗﻈﻬﺮ ﻣﺴﺘﻮﻳﺎﺕ ﻣﺨﺘﻠﻔﺔ ﻣﻦ ﺍﻟﻨﺸﺎﻁ ﻓﻲ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ‪ .‬ﻭﻳﻮﺿﺢ ﺍﻟﺸﻜﻞ‪ 4-13‬ﺳﻠﺴﻠﺔ‬ ‫ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﻠﻬﺎﻟﻮﺟﻴﻨﺎﺕ‪ ،‬ﺍﻟﺘﻲ ﺗﺒﻴﻦ ﺃﻥ ﺍﻟﻔﻠﻮﺭ ﺃﻧﺸﻂ ﺍﻟﻬﺎﻟﻮﺟﻴﻨﺎﺕ‪ ،‬ﻭﺍﻟﻴﻮﺩ ﺃﻗﻠﻬﺎ ﹰ‬ ‫ﻧﺸﺎﻃﺎ‪.‬‬ ‫ﻓﺎﻟﻬﺎﻟﻮﺟﻴﻦ ﺍﻷﻧﺸﻂ ﻳﺤﻞ ﻣﺤﻞ ﺍﻟﻬﺎﻟﻮﺟﻴﻦ ﺍﻷﻗﻞ ﹰ‬ ‫ﻧﺸﺎﻃﺎ ﻓﻲ ﻣﺤﺎﻟﻴﻞ ﻣﺮﻛﺒﺎﺗﻪ‪ .‬ﻓﺎﻟﻔﻠﻮﺭ ﹰ‬ ‫ﻣﺜﻼ ﻳﺤﻞ‬ ‫ﻣﺤﻞ ﺍﻟﺒﺮﻭﻡ ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻰ ﻟﺒﺮﻭﻣﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪.‬‬ ‫)‪F2(g) + 2NaBr(aq) → 2NaF(aq) + Br2(l‬‬ ‫ﻻ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ‪→ NR‬‬

‫)‪2NaF(aq‬‬

‫‪Br2 (l) +‬‬

‫‪ ?äCGôb GPÉe‬ﻭﺿﺢ ﻛﻴﻒ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ؟‬

‫ﻣﺨﺘﺒﺮ ﺣﻞ اﻟﻤﺸﻜﻼت‬ ‫ﺗحلﻴﻞ الﺘﺪﺭﺝ ﻓﻲ الﺨﻮاﺹ‬

‫ﻛﻴﻒ ﹸﺗﻔﴪ ﻧﺸﺎﻁ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ؟ ﺗﻘﻊ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﰲ ﺍﳌﺠﻤﻮﻋﺔ‬ ‫ﺭﻗﻢ ‪ 17‬ﻣﻦ ﺍﳉﺪﻭﻝ ﺍﻟﺪﻭﺭﻱ‪ :‬ﻭﳜﱪﻧﺎ ﻫﺬﺍ ﺑﺄﻥ ﻟﻠﻬﺎﻟﻮﺟﻴﻨﺎﺕ‬ ‫ﺑﻌﺾ ﺍﳋﻮﺍﺹ ﺍﻟﻌﺎﻣﺔ‪ .‬ﻓﺠﻤﻴﻊ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﻻ ﻓﻠﺰﺍﺕ‪،‬‬ ‫ﻭﻳﻮﺟﺪ ﰲ ﺃﻏﻠﻔﺘﻬﺎ ﺍﳋﺎﺭﺟﻴﺔ ﺳﺒﻌﺔ ﺇﻟﻜﱰﻭﻧﺎﺕ‪ .‬ﻭﻣﻊ ﺫﻟﻚ‪،‬‬ ‫ﻓﻠﻜﻞ ﻫﺎﻟﻮﺟﲔ ﻣﺎ ﻳﻤﻴﺰﻩ ﻣﻦ ﺍﳋﻮﺍﺹ‪ ،‬ﻭﻣﻦ ﺫﻟﻚ ﻣﺪ￯ ﻗﺎﺑﻠﻴﺔ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ ﻣﻊ ﻣﺎﺩﺓ ﺃﺧﺮ￯‪.‬‬ ‫اﻟﺘﺤﻠﻴﻞ‬

‫ﺧﻮﺍﺹ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ‬

‫ﻧﺼﻒ ﺍﻟﻘﻄﺮ ﺍﻟﺬﺭﻱ ﻃﺎﻗﺔ ﺍﻟﺘﺄﻳﻦ‬ ‫ﺍﳍﺎﻟﻮﺟﲔ‬ ‫ﺍﻟﻜﻬﺮﻭﺳﺎﻟﺒﻴﺔ‬ ‫)‪(ppm‬‬ ‫)‪(KJ/mol‬‬ ‫ﺍﻟﻔﻠﻮﺭ‬ ‫‪3.98‬‬ ‫‪1681‬‬ ‫‪72‬‬ ‫ﺍﻟﻜﻠﻮﺭ‬ ‫‪3.16‬‬ ‫‪1251‬‬ ‫‪100‬‬ ‫ﺍﻟﱪﻭﻡ‬ ‫‪2.96‬‬ ‫‪1140‬‬ ‫‪114‬‬ ‫ﺍﻟﻴﻮﺩ‬ ‫‪2.66‬‬ ‫‪1008‬‬ ‫‪133‬‬ ‫ﺍﻹﺳﺘﺎﺗﲔ‬ ‫‪2.2‬‬ ‫‪920‬‬ ‫‪140‬‬

‫‪ .2‬ﺻﻒ ﻛﻴﻒ ﺗﺘﻐﲑ ﺧﻮﺍﺹ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﺍﻟﺜﻼﺙ ﻭﻓﻖ‬ ‫ﺗﻔﺤﺺ ﺟﺪﻭﻝ ﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﳌﺒﲔ‪ .‬ﻭﺍﻟﺬﻱ ﻳﺸﺘﻤﻞ ﻋﲆ ﻧﺼﻒ ﺍﻟﻘﻄﺮ‬ ‫ﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﳌﻮﺿﺤﺔ ﰲ ﺍﳉﺪﻭﻝ‪.‬‬ ‫ﺍﻟﺬﺭﻱ‪ ،‬ﻭﻃﺎﻗﺔ ﺍﻟﺘﺄﻳﻦ‪ ،‬ﻭﺍﻟﻜﻬﺮﻭﺳﺎﻟﺒﻴﺔ ﻟﺒﻌﺾ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ‪.‬‬ ‫‪ .3‬ﻣﺎ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﺗﻐﲑ ﺧﻮﺍﺹ ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﺍﳌﻮﺿﺤﺔ ﰲ‬ ‫اﻟﺘﻔﻜﻴﺮ اﻟﻨﺎﻗﺪ‬ ‫ﺍﳉﺪﻭﻝ‪ ،‬ﻭﺑﲔ ﻣﻮﻗﻌﻬﺎ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ؟ ﺍﻟﺸﻜﻞ )‪(4-13‬‬ ‫‪ .1‬ﺍﺭﺳﻢ ﹰ‬ ‫ﺧﻄﺎ ﺑﻴﺎﻧ ﹰﻴﺎ ﻳﻤﺜﻞ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﺍﻟﻌﺪﺩ ﺍﻟﺬﺭﻱ ﻟﻠﻌﻨﺎﴏ ‪ .4‬ﺗﻮﻗﻊ ﻣﻮﻗﻊ ﻋﻨﴫ ﺍﻹﺳﺘﺎﺗﲔ ‪ As‬ﰲ ﺳﻠﺴﻠﺔ ﻧﺸﺎﻁ‬ ‫ﻭﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﳌﻌﻄﺎﺀ ﰲ ﺍﳉﺪﻭﻝ ﺍﳌﺮﻓﻖ )ﻛﻞﹴ ﻋﲆ ﺣﺪﺓ(‬ ‫ﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ‪ .‬ﻓﴪ ﺇﺟﺎﺑﺘﻚ‪.‬‬

‫‪20‬‬

‫‪4-2 ∫Éãe‬‬

‫ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ ﺗﻮﻗﻊ ﻧﻮﺍﺗﺞ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ‪ ،‬ﻭﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﲤﺜﻞ ﹰﹼ‬ ‫ﻛﻼ ﻣﻨﻬﺎ‪:‬‬ ‫‪.a‬‬ ‫‪.b‬‬ ‫‪.c‬‬

‫→ )‪Fe(s) + CuSO4(aq‬‬

‫→ )‪Br2(l) + MgCl2(aq‬‬ ‫→ )‪Mg(s) + AlCl3(aq‬‬

‫‪ádCÉ°ùŸG π«∏– 1‬‬

‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺸﻜﻞ ‪ 4-13‬ﻟﺘﺤﺪﺩ ﻣﺎ ﺇﺫﺍ ﻛﺎﻥ ﻛﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺳﻴﺤﺪﺙ ﺃﻡ ﻻ‪ ،‬ﻭﺣﺪﱢ ﺩ ﻧﻮﺍﺗﺞ ﻛﻞ‬ ‫ﺗﻔﺎﻋﻞ ﻣﺘﻮﻗﻊ‪ ،‬ﻭﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﲤﺜﻞ ﺍﻟﺘﻔﺎﻋﻞ ﻭﺯﳖﺎ‪.‬‬

‫‪܃∏£ŸG ÜÉ°ùM 2‬‬

‫‪ .a‬ﻳﻘﻊ ﺍﳊﺪﻳﺪ ﻗﺒﻞ ﺍﻟﻨﺤﺎﺱ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ .‬ﻭﳍﺬﺍ ﻓﺈﻥ ﺍﻟﺘﻔﺎﻋﻞ ﺳﻴﺤﺪﺙ ﻷﻥ ﺍﳊﺪﻳﺪ ﺃﻧﺸﻂ ﻣﻦ‬ ‫ﺍﻟﻨﺤﺎﺱ‪ .‬ﻭﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺳﻴﺤﻞ ﺍﳊﺪﻳﺪ ﳏﻞ ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﺗﻜﻮﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻋﲆ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﱄ‪:‬‬ ‫)‪Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s‬‬

‫ﻭﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﻣﻮﺯﻭﻧﺔ‪.‬‬ ‫ﹰ‬ ‫ﻧﺸﺎﻃﺎ ﻣﻦ ﺍﻟﻜﻠﻮﺭ؛ ﻷﻧﻪ ﻳﻘﻊ ﺑﻌﺪ ﺍﻟﻜﻠﻮﺭ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ ،‬ﻭﳍﺬﺍ ﻻ ﳛﺪﺙ ﺗﻔﺎﻋﻞ‪.‬‬ ‫‪ .b‬ﺍﻟﱪﻭﻡ ﺃﻗﻞ‬ ‫ﻭﻳﻤﻜﻦ ﲤﺜﻴﻞ ﺫﻟﻚ ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬

‫ﻻ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ‪Br2(l) + MgCl2(aq) → NR‬‬

‫ﻭﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻻ ﺗﺘﻄﻠﺐ ﺍﳌﻌﺎﺩﻟﺔ ﻭﺯ ﹰﻧﺎ‪.‬‬ ‫‪ .c‬ﻳﻘﻊ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ﻗﺒﻞ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﰲ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ ،‬ﻭﳍﺬﺍ ﻓﺈﻥ ﺍﻟﺘﻔﺎﻋﻞ ﺳﻴﺤﺪﺙ ﻷﻥ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ﺃﻧﺸﻂ ﻣﻦ‬ ‫ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‪ .‬ﻭﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺳﻴﺤﻞ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ﳏﻞ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‪ ،‬ﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻏﲑ ﺍﳌﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ‪:‬‬ ‫ﻭﺍﳌﻌﺎﺩﻟﺔ ﺍﳌﻮﺯﻭﻧﺔ ﻫﻲ‪:‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫)‪Mg(s) + AlCl3(aq) → MgCl2(aq) + Al(s‬‬

‫)‪3Mg(s) + 2AlCl3(aq) → 3MgCl2(aq) + 2Al(s‬‬

‫ﺗﺪﻋﻢ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﳌﻮﺿﺤﺔ ﰲ ﺍﻟﺸﻜﻞ ‪ 4-13‬ﺍﻟﺘﻮﻗﻌﺎﺕ‪ .‬ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ؛ ﻷﻥ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ‬ ‫ﻣﺘﺴﺎ ﹴﻭ ﰲ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﺗﻮﻗﻊ ﻣﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ ﺍﻟﺘﺎﻟﻴﺔ ﺳﺘﺤﺪﺙ ﺃﻡ ﻻ‪ ،‬ﻭﺃﻛﻤﻞ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺮﻣﺰﻳﺔ ﺍﳌﻮﺯﻭﻧﺔ ﻟﻜﻞ ﺗﻔﺎﻋﻞ ﻳﺘﻮﻗﻊ ﺣﺪﻭﺛﻪ‪:‬‬ ‫‪.21‬‬

‫→ )‪Zn (s) + NiCl2 (aq‬‬ ‫→ )‪Cl2 (g) + HF (aq‬‬

‫‪.23‬‬

‫→ )‪Fe(s) + Na3PO4(aq‬‬

‫‪.22‬‬

‫‪ .24‬ﲢﺪﹼ‬

‫→ )‪Al(s) + Pb(NO3)2 (aq‬‬

‫‪21‬‬

‫‪ êhOõ``ªdG ∫Ó``ME’G äÓ``YÉØJ‬ﺗﺘﻀﻤـﻦ ﺗﻔﺎﻋـﻼﺕ ﺍﻹﺣـﻼﻝ ﺍﻟﻤـﺰﺩﻭﺝ ﺗﺒﺎﺩﻝ‬ ‫ﺍﻷﻳﻮﻧﺎﺕ ﺑﻴﻦ ﻣﺮﻛﺒﻴﻦ‪ ،‬ﻛﻤﺎ ﻫﻮ ﻣﺒﻴﻦ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻌﺎﻣﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪AX + BY → AY + BX‬‬

‫ﻳﻤﺜﹼﻞ ﺍﻟﺮﻣﺰﺍﻥ ‪ A‬ﻭ ‪ B‬ﰲ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﺃﻳﻮﻧﲔ ﻣﻮﺟﺒﲔ‪ ،‬ﺑﻴﻨﲈ ﻳﻤﺜﻞ ﺍﻟﺮﻣﺰﺍﻥ ‪ X‬ﻭ ‪Y‬‬

‫ﺍﻟﺸﻜﻞ ‪ 4-14‬ﻋﻨﺪﻣـﺎ ﻳﻀﺎﻑ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﺼﻮﺩﻳـﻮﻡ ﺇﱃ ﳏﻠـﻮﻝ‬ ‫ﺍﻟﻨﺤــــــﺎﺱ‬ ‫ﻛﻠﻮﺭﻳـﺪ‬ ‫‪ ،CuCl2 II‬ﺗﺘﺒـﺎﺩﻝ ﺃﻳﻮﻧـﺎﺕ‬ ‫ﻣﻮﺍﻗﻌﻬـــﺎ‪،‬‬ ‫ﻭﻳﻨﺘـﺞ ﻋـﻦ ﺍﻟﺘﻔﺎﻋـﻞ ﻛﻠﻮﺭﻳﺪ‬ ‫ﺍﻟﺼﻮﺩﻳـﻮﻡ ﺍﻟـﺬﻱ ﻳﺒﻘـﻰ ﰲ‬ ‫ﺍﻟـﻤﺤﻠـﻮﻝ‪ ،‬ﻭﻫﻴﺪﺭﻭﻛﺴـﻴﺪ‬ ‫ﺍﻟﻨﺤـــﺎﺱ ‪Cu(OH)2 II‬‬ ‫ﺍﻟـﺬﻱ ﻳﱰﺳـﺐ ﻋـﲆ ﺷـﻜﻞ‬ ‫ﻣـﺎﺩﺓ ﺻﻠﺒـﺔ ﺯﺭﻗـﺎﺀ ﺍﻟﻠـﻮﻥ‪.‬‬

‫ﻭﺻﺎﺭﺍ ﻣﺮﺗﺒﻄﲔ‬ ‫ﺃﻳﻮﻧﲔ ﺳﺎﻟﺒﲔ‪ .‬ﻻﺣﻆ ﺃﻥ ﺍﻷﻳﻮﻧﲔ ﺍﻟﺴﺎﻟﺒﲔ ﻗﺪ ﺗﺒﺎﺩﻻ ﻣﻮﻗﻌﻴﻬﲈ‪،‬‬ ‫ﹶ‬ ‫ﺑﺄﻳﻮﻧﲔ ﻣﻮﺟﺒﲔ ﳐﺘﻠﻔﲔ‪ .‬ﻭﺑﻤﻌﻨﻰ ﺁﺧﺮ‪ ،‬ﺣﻞ ‪ X‬ﳏﻞ ‪ ،Y‬ﻭﺣﻞ ‪ Y‬ﳏﻞ ‪ .X‬ﻭﳍﺬﺍ‬ ‫ﹸ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‪ .‬ﻓﺘﻔﺎﻋﻞ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ‬ ‫ﺍﻟﺴﺒﺐ ﻳﺴﻤﻰ‬ ‫ﹰ‬ ‫ﻣﺜﻼ ﻭﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﺍﳌﻮﺿﺢ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻫﻮ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪.‬‬ ‫‪ .‬ﻻﺣﻆ ﺃﻥ ﺍﻷﻳﻮﻧﲔ ﺍﻟﺴﺎﻟﺒﲔ‬ ‫ﺍﻷﻳﻮﻧﺎﺕ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﻫﻲ‪:‬‬ ‫‪ ،‬ﻋﲆ ﺍﻟﱰﺗﻴﺐ‪.‬‬ ‫ﻏﲑﺍ ﻣﻮﻗﻌﻴﻬﲈ‪ ،‬ﻭﺍﺭﺗﺒﻄﺎ ﺑﺎﻷﻳﻮﻧﲔ ﺍﳌﻮﺟﺒﲔ‬ ‫ﻗﺪ ﹼ‬ ‫ﻛﲈ ﺃﻥ ﺗﻔﺎﻋﻞ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻣﻊ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﺤﺎﺱ ‪ II‬ﻫﻮ ﹰ‬ ‫ﺃﻳﻀﺎ ﺗﻔﺎﻋﻞ‬ ‫ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪.‬‬ ‫ﻏـﲑﺍ ﻣﻮﻗﻌﻴﻬﲈ ﻭﺍﺭﺗﺒﻄـﺎ ﺑﺄﻳﻮﻧﲔ‬ ‫ﻻﺣـﻆ ﺃﻥ ﺍﻷﻳﻮﻧـﲔ ﺍﻟﺴـﺎﻟﺒﲔ‬ ‫ﻗـﺪ ﹼ‬ ‫‪ .‬ﻭﻳﺒـﲔ ﺍﻟﺸـﻜﻞ ‪ 4-14‬ﺃﻥ ﻧﺎﺗﺞ ﻫـﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﻣﺎﺩﺓ‬ ‫ﻣﻮﺟﺒـﲔ ﺁﺧﺮﻳﻦ‬ ‫ﺻﻠﺒﺔ ﻻ ﺗﺬﻭﺏ ﰲ ﺍﳌﺎﺀ ﻫﻲ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ ‪ .Cu(OH)2 II‬ﻭﺗﺴﻤﻰ ﺍﳌﺎﺩﺓ‬ ‫ﹴ‬ ‫ﳏﻠﻮﻝ ﻣﺎ ﺭﺍﺳ ﹰﺒﺎ‪.‬‬ ‫ﺍﻟﺼﻠﺒﺔ ﺍﻟﺘﻲ ﺗﻨﺘﺞ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﰲ‬

‫‪ êhOõ``ªdG ∫Ó``ME’G äÓ``YÉØJ è``JGƒf‬ﺇﺣـﺪ￯ ﺍﻟﻤﻤﻴـﺰﺍﺕ ﺍﻷﺳﺎﺳـﻴﺔ ﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣـﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ ﻫﻲ ﻧـﻮﻉ ﺍﻟﻨﺎﺗﺞ ﺍﻟﻤﺘﻜـﻮﻥ ﻋﻨﺪﻣﺎ ﻳﺤﺪﺙ ﺍﻟﺘﻔﺎﻋـﻞ‪ .‬ﻓﺠﻤﻴﻊ ﻫﺬﻩ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺗﻨﺘﺞ ﻣﺎ ﹰﺀ‪ ،‬ﺃﻭ ﺭﺍﺳ ﹰﺒﺎ‪ ،‬ﺃﻭ ﹰ‬ ‫ﻏﺎﺯﺍ‪.‬‬

‫ﹶ‬ ‫ﺗﻔﺎﻋﲇ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ ﺍﻟﻠﺬﻳﻦ ﻧﻮﻗﺸﺎ ﺳﺎﺑ ﹰﻘﺎ‪ .‬ﺣﻴﺚ ﻳﻨﺘﺞ ﻣﻦ ﺗﻔﺎﻋﻞ‬ ‫ﺍﺭﺟﻊ ﺇﱃ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﻣﻊ ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﻣﺎﺀ‪ ،‬ﻭﻳﻨﺘﺞ ﻣﻦ ﺗﻔﺎﻋﻞ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ ﻣﻊ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﺤﺎﺱ‪ II‬ﺭﺍﺳﺐ‪ .‬ﻭﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ ﺍﻟﺘﻲ ﺗﻨﺘﺞ‬ ‫ﹰ‬ ‫ﻏﺎﺯﺍ ﺗﻔﺎﻋﻞ ﺳﻴﺎﻧﻴﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ KCN‬ﻭﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ‪.HCl‬‬ ‫)‪KCN(aq) + HCl(aq) → KCl(aq) + HCN(g‬‬

‫ﻭﻳﺒﲔ ﺍﳉﺪﻭﻝ ‪ 4-3‬ﺍﳋﻄﻮﺍﺕ ﺍﻷﺳﺎﺳﻴﺔ ﻟﻜﺘﺎﺑﺔ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‪.‬‬

‫‪22‬‬

‫‪êhOõŸG ∫ÓME’G äÓYÉØàd áfhRƒŸG á«FÉ«ª«µdG ä’OÉ©ŸG áHÉàµd á«°SÉ°SC’G äGƒ£ÿG 4-3 ∫hó÷G‬‬ ‫‪äGƒ£ÿG‬‬

‫‪∫É```````ãe‬‬

‫‪ .1‬ﺍﻛﺘﺐ ﺍﻟﺼﻴـﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﻤﺘﻔﺎﻋﻼﺕ‬ ‫ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪Al(NO 3) 3 + H 2SO 4‬‬ ‫‪ Al(NO 3) 3‬ﻓﻴﻪ ‪ NO 3 -‬ﻭ ‪Al 3+‬‬

‫ﻋﲔ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﻮﺟﺒﺔ ﻭﺍﻟﺴﺎﻟﺒﺔ ﰲ ﻛﻞ‬ ‫‪ .2‬ﹼ‬ ‫ﻣﺮﻛﺐ‪.‬‬ ‫‪.3‬‬

‫‪.4‬‬

‫‪.5‬‬

‫‪ H 2SO 4‬ﻓﻴﻪ ‪ SO 4 2-‬ﻭ ‪H +‬‬

‫ﺯﺍﻭﺝ ﺑﲔ ﻛﻞ ﺃﻳﻮﻥ ﻣﻮﺟﺐ ﻭﺍﻷﻳﻮﻥ‬ ‫ﺍﻟﺴﺎﻟﺐ ﰲ ﺍﳌﺮﻛﺐ ﺍﻵﺧﺮ‪.‬‬

‫‪ SO 4 2-‬ﻳﺘﺰﺍﻭﺝ ﻣﻊ ‪Al 3+‬‬

‫ﺍﻛﺘﺐ ﺍﻟﺼﻴـﻎ ﺍﻟﻜﻴﻤﻴـﺎﺋﻴﺔ ﻟﻠﻨﻮﺍﺗﺞ‬ ‫ﻣﺴﺘﺨﺪ ﹰﻣﺎ ﺍﻷﺯﻭﺍﺝ ﰲ ﺍﳋﻄﻮﺓ ‪.3‬‬

‫‪Al 2(SO 4) 3‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ‬ ‫ﻟﺘﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‪.‬‬

‫‪ .6‬ﺯﻥ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬

‫‪ NO 3 -‬ﻳﺘﺰﺍﻭﺝ ﻣﻊ‬

‫‪H+‬‬

‫‪HNO 3‬‬

‫‪+ HNO 3‬‬

‫)‪(aq‬‬

‫)‪(s‬‬

‫‪→ Al 2(SO 4) 3‬‬

‫)‪(aq‬‬

‫‪+ H 2SO 4‬‬

‫)‪(aq‬‬

‫‪Al(NO 3) 3‬‬

‫‪2Al(NO 3) 3‬‬ ‫‪+ 3H 2SO 4‬‬ ‫‪→ Al 2(SO 4) 3 + 6HNO 3‬‬ ‫)‪(aq‬‬ ‫)‪(aq‬‬ ‫)‪(s‬‬ ‫)‪(aq‬‬

‫‪ ?äCGôb GPÉe‬ﺻﻒ ﻣﺎ ﳛﺪﺙ ﻟﻸﻳﻮﻧﺎﺕ ﺍﻟﺴﺎﻟﺒﺔ ﰲ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ ﺍﻵﺗﻴﺔ‪.‬‬

‫ﻣﻌﺎ ﻹﻧﺘﺎﺝ ﻳﻮﺩﻳﺪ ﺍﻟﻔﻀﺔ ‪ AgI‬ﺍﻟﺼﻠﺐ‬ ‫‪ .25‬ﺗﺘﻔﺎﻋﻞ ﺍﳌﺎﺩﺗﺎﻥ ﺍﻟﻠﺘﺎﻥ ﻋﲆ ﺍﻟﻴﺴﺎﺭ ﹰ‬ ‫ﻭﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻟﻠﻴﺜﻴﻮﻡ ‪.LiNO3‬‬

‫‪ .26‬ﻳﺘﻔﺎﻋﻞ ﳏﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺒﺎﺭﻳﻮﻡ ‪ BaCl2‬ﻣﻊ ﳏﻠﻮﻝ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‬ ‫‪ K2CO3‬ﻹﻧﺘﺎﺝ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺒﺎﺭﻳﻮﻡ ﺍﻟﺼﻠﺒﺔ ﻭﳏﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‪.‬‬

‫‪ .27‬ﻳﺘﻔﺎﻋﻞ ﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2SO4‬ﻣﻊ ﳏﻠﻮﻝ ﻧﱰﺍﺕ‬ ‫ﺍﻟﺮﺻﺎﺹ ‪ Pb(NO3)2 II‬ﻹﻧﺘﺎﺝ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺮﺻﺎﺹ ‪ PbSO4 II‬ﺍﻟﺼﻠﺒﺔ‬ ‫ﻭﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪.NaNO3‬‬ ‫‪ .28‬ﲢﺪﱟ ‪ :‬ﻳﺘﻔﺎﻋﻞ ﲪﺾ ﺍﻹﻳﺜﺎﻧﻮﻳﻚ )ﲪﺾ ﺍﳋﻞ( ‪ CH3COOH‬ﻣﻊ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ KOH‬ﻹﻧﺘﺎﺝ ﺇﻳﺜﺎﻧﻮﺍﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ)ﺧﻼﺕ‬ ‫ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ( ‪ CH3COOK‬ﻭﺍﳌﺎﺀ‪.‬‬

‫)‪AgNO3(aq‬‬

‫)‪LiI(aq‬‬

‫‪23‬‬

‫‪4-4 ∫hó÷G‬‬ ‫‪πYÉØàdG ´ƒf‬‬

‫ﺍﻟﺘﻜﻮﻳﻦ‬ ‫ﺍﻻﺣﺘﺮﺍﻕ‬ ‫ﺍﻟﺘﻔﻜﻚ‬

‫‪á«FÉ«ª«µdG äÓYÉØàdG ¢†©Ñd á©bƒàŸG œGƒædG‬‬ ‫‪á∏YÉØàŸG OGƒŸG‬‬

‫• ﻣﺎﺩﺗﺎﻥ ﺃﻭ ﺃﻛﺜﺮ‬

‫‪áeÉ©dG ádOÉ©ŸG‬‬

‫‪á©bƒàŸG œGƒædG‬‬

‫• ﻣﺮﻛﺐ ﻭﺍﺣﺪ‬

‫‪A + B → AB‬‬

‫• ﻓﻠﺰ ﻭ ﺃﻛﺴﺠﻴﻦ • ﺃﻛﺴﻴﺪ ﺍﻟﻔﻠﺰ‬ ‫• ﻻﻓﻠﺰ ﻭﺃﻛﺴﺠﻴﻦ • ﺃﻛﺴﻴﺪ ﺍﻟﻼﻓﻠﺰ‬ ‫• ﻣﺮﻛﺐ ﻭﺃﻛﺴﺠﻴﻦ • ﺃﻛﺴﻴﺪﺍﻥ ﺃﻭ ﺃﻛﺜﺮ‬ ‫ﻣﺮﻛﺐ ﻭﺍﺣﺪ‬

‫ﻓﻠﺰ ﻭﻣﺮﻛﺐ‬ ‫ﺍﻻﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‬ ‫ﻻﻓﻠﺰ ﻭﻣﺮﻛﺐ‬ ‫ﺍﻻﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ ﻣﺮﻛﺒﺎﻥ‬

‫‪A + O2 → AO‬‬

‫ﻋﻨﺼﺮﺍﻥ ﺃﻭ ﺃﻛﺜﺮ ﻭ‪/‬ﺃﻭ ﻣﺮﻛﺒﺎﺕ ﺃﺧﺮ￯‬

‫‪AB → A + B‬‬

‫ﻣﺮﻛﺐ ﺟﺪﻳﺪ ﻭﺍﻟﻔﻠﺰ ﺍﻟﻤﺴﺘﻌﺎﺽ ﻋﻨﻪ‬ ‫ﻣﺮﻛﺐ ﺟﺪﻳﺪ ﻭﺍﻟﻼﻓﻠﺰ ﺍﻟﻤﺴﺘﻌﺎﺽ ﻋﻨﻪ‬

‫ﻣﺮﻛﺒﺎﻥ ﻣﺨﺘﻠﻔﺎﻥ‪ ،‬ﺃﺣﺪﻫﺎ ﺻﻠﺐ‪ ،‬ﺃﻭ ﻣﺎﺀ‪،‬‬ ‫ﺃﻭ ﻏﺎﺯ‪.‬‬

‫‪A + BX → AX + B‬‬ ‫‪AX + BY → AY + BX‬‬

‫ﻳﻠﺨـﺺ ﺍﻟﺠﺪﻭﻝ ‪ 4-4‬ﺃﻧﻮﺍﻉ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻳﻤﻜﻨﻚ ﺍﻻﺳـﺘﻌﺎﻧﺔ ﺑﺎﻟﺠﺪﻭﻝ ﻓﻲ‬ ‫ﻣﻌﺮﻓـﺔ ﺃﻧﻮﺍﻉ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻭﺗﻮﻗﻊ ﻧﻮﺍﺗﺠﻬﺎ‪.‬ﻋﻠﻰ ﺳـﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‪ ،‬ﻛﻴﻒ ﺗﺤﺪﺩ‬ ‫ﻧﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ ﺃﻛﺴـﻴﺪ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﺍﻟﺼﻠﺐ ﻭ ﻏﺎﺯ ﺛﺎﻧﻲ ﺃﻛﺴـﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ﻓﻲ ﺇﻧﺘﺎﺝ‬ ‫ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺍﻟﺼﻠﺒﺔ؟ ﺃﻭ ﹰ‬ ‫ﻻ‪ ،‬ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫ﺛﺎﻧ ﹰﻴﺎ‪ ،‬ﺣﺪﺩ ﻣﺎ ﺍﻟﺬﻱ ﻳﺤﺪﺙ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ‪.‬ﻓﻲ ﻫﺬﻩ ﺍﻟﺤﺎﻟﺔ‪ ،‬ﺗﺘﻔﺎﻋﻞ ﻣﺎﺩﺗﺎﻥ ﻭﻳﺘﻨﺘﺞ ﻋﻨﻬﻤﺎ‬ ‫ﻣﺮﻛﺐ ﻭﺍﺣﺪ‪.‬ﺛﺎﻟ ﹰﺜﺎ‪ ،‬ﺍﺳﺘﻌﻦ ﺑﺎﻟﺠﺪﻭﻝ ﻟﺘﺤﺪﻳﺪ ﻧﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ‪.‬ﺍﻟﺘﻔﺎﻋﻞ ﻫﻮ ﺗﻔﺎﻋﻞ ﺗﻜﻮﻳﻦ‪.‬‬ ‫ﺭﺍﺑﻌﺎ‪،‬ﺍﻓﺤﺺ ﺇﺟﺎﺑﺘﻚ ﺑﻤﻘﺎﺭﻧﺔ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻌﺎﻣﺔ ﻟﻨﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫ﹰ‬ ‫‪→ AB‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪4 -2‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺗﺼﻨﻴـﻒ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ‬ ‫ﻭﺗﻌﺮﻓﻬﺎ‪.‬‬ ‫ﹼ‬ ‫ﻳﺴﻬﻞ ﻓﻬﻤﻬﺎ ﻭﺗﺬﻛﺮﻫﺎ ﹼ‬ ‫ﻳﻤﻜـﻦ ﺍﺳـﺘﺨﺪﺍﻡ ﺳﻠﺴـﻠﺔ‬ ‫ﺍﻟﻨﺸـﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋـﻲ ﻟﻠﻔﻠـﺰﺍﺕ‬ ‫ﻭﺍﻟﻬﺎﻟﻮﺟﻴﻨـﺎﺕ ﻟﺘﻮﻗﻊ ﺣﺪﻭﺙ‬ ‫ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‪.‬‬

‫‪24‬‬

‫‪A + B‬‬

‫‪ .29‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺻﻒ ﺍﻷﻧﻮﺍﻉ ﺍﻷﺭﺑﻌﺔ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺧﻮﺍﺻﻬﺎ‪.‬‬ ‫‪ .30‬ﻭﺿﺢ ﻛﻴﻒ ﺗﻨﻈﻢ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﻠﻔﻠﺰﺍﺕ؟‬ ‫‪ .31‬ﻗﺎﺭﻥ ﺑﻴﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ ﻭﺍﻹﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ‪.‬‬ ‫‪ .32‬ﺻﻒ ﻧﺘﻴﺠﺔ ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ‪.‬‬ ‫‪ .33‬ﺻ ﹼﻨـﻒ‪ :‬ﻣﺎ ﻧـﻮﻉ ﺍﻟﺘﻔﺎﻋـﻞ ﺍﻟﻤﺮﺟﺢ ﺣﺪﻭﺛـﻪ ﻋﻨﺪﻣﺎ ﻳﺘﻔﺎﻋـﻞ ﺍﻟﺒﺎﺭﻳﻮﻡ ﻣﻊ‬ ‫ﺍﻟﻔﻠﻮﺭ؟ ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ‪.‬‬ ‫ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫‪ .34‬ﹼ‬ ‫ﻓﺴﺮ ﺍﻟﺒﻴﺎﻧﺎﺕ‪ :‬ﻫﻞ ﻳﻤﻜﻦ ﻟﻠﺘﻔﺎﻋﻞ ﺍﻵﺗﻲ ﺃﻥ ﻳﺤﺪﺙ؟ ﹼ‬ ‫)‪Ni(s) + 2AgNO3(aq) → Ni(NO3)2 (aq) + 2Ag (s‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪4-3‬‬ ‫ا ﻫﺪاف‬

‫ﺗﺼﻒ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‪.‬‬ ‫ﺗﻜﺘﺐ ﻣﻌﺎﺩﻻﺕ ﺃﻳﻮﻧﻴﺔ ﻛﺎﻣﻠﺔ‬ ‫ﻭﻣﻌﺎﺩﻻﺕ ﺃﻳﻮﻧﻴـﺔ ﺻـﺎﻓـﻴﺔ‬ ‫ﻟﻠﺘﻔﺎﻋـﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋـــﻴﺔ ﻓـﻲ‬ ‫ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‪.‬‬ ‫ﺗﺘﻮﻗﻊ ﻣﺎ ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ ﺳﺘﺆﺩﻱ ﺇﱃ‬ ‫ﺇﻧﺘﺎﺝ ﺭﺍﺳﺐ‪ ،‬ﺃﻭ ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯ‪.‬‬

‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﳌﺤﻠﻮﻝ‪ :‬ﳐﻠﻮﻁ ﻣﺘﺠﺎﻧﺲ ﻗﺪ ﳛﻮﻱ‬ ‫ﻣﻮﺍﺩ ﺻﻠﺒﺔ‪ ،‬ﺃﻭ ﺳﺎﺋﻠﺔ‪ ،‬ﺃﻭ ﻏﺎﺯﻳﺔ‪.‬‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﳌﺤﻠﻮﻝ ﺍﳌﺎﺋﻲ‬ ‫ﺍﳌﺬﺍﺏ‬ ‫ﺍﳌﺬﻳﺐ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‬ ‫ﺍﻷﻳﻮﻥ ﺍﳌﺘﻔﺮﺝ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‬

‫اﻟﺘﻔﺎﻋﻼت ﻓﻲ اﻟﻤﺤﺎﻟﻴﻞ اﻟﻤﺎﺋﻴﺔ‬

‫‪Reactions in Aqueous Solutions‬‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﲢﺪﺙ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ ﺑﲔ ﺍﳌﻮﺍﺩ ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‪،‬‬ ‫ﻭﺗﺆﺩﻱ ﺇﱃ ﺇﻧﺘﺎﺝ ﺭﻭﺍﺳﺐ‪ ،‬ﺃﻭ ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯﺍﺕ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻳﺴﺘﻌﻤﻞ ﻣﺴﺤﻮﻕ ﻧﻜﻬﺔ ﺍﻟﻠﻴﻤﻮﻥ ﰲ ﲢﻀﲑ ﴍﺍﺏ‬ ‫ﻣﻜﻮﻧﺔ‬ ‫ﺍﻟﻠﻴﻤﻮﻥ‪ .‬ﻓﻌﻨﺪﻣﺎ ﻳﻀﺎﻑ ﺍﳌﺴﺤﻮﻕ ﺇﱃ ﺍﳌﺎﺀ ﻓﺈﻥ ﺑﻠﻮﺭﺍﺗﻪ ﺗﺬﻭﺏ ﻓﻴﻪ ﹼ‬ ‫ﹰ‬ ‫ﳏﻠﻮﻻ ﻟﻪ ﻧﻜﻬﺔ ﺍﻟﻠﻴﻤﻮﻥ‪.‬‬

‫المحالﻴﻞ الماﺋﻴﺔ ‪Aqueous Solutions‬‬ ‫ﻋﺮﻓﺖ ﺃﻥ ﺍﳌﺤﻠﻮﻝ ﳐﻠﻮﻁ ﻣﺘﺠﺎﻧﺲ‪ .‬ﱠ‬ ‫ﻭﺃﻥ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺗﺘﻀﻤﻦ ﻣﻮﺍ ﹼﺩ‬ ‫ﹶ‬ ‫ﺃﻱ ﺗﻜﻮﻥ ﻋﲆ ﺷﻜﻞ ﳏﺎﻟﻴﻞ ﻣﺎﺋﻴﺔ‪ .‬ﻭﺍﳌﺤﻠﻮﻝ ﺍﳌﺎﺋﻲ ﳛﺘﻮﻱ ﻋﲆ‬ ‫ﻣﺬﺍﺑﺔ ﰲ ﺍﳌﺎﺀ‪ ،‬ﹾ‬ ‫ﻣﻜﻮﻧﺎﺕ ﺍﳌﺤﻠﻮﻝ –‬ ‫ﻣﺎﺩﺓ ﺃﻭ ﺃﻛﺜﺮ ﻣﺬﺍﺑﺔ ﰲ ﺍﳌﺎﺀ ﺗﺴﻤﻰ ﺍﳌﺬﺍﺏ‪ .‬ﺃﻣﺎ ﺍﳌﺎﺀ – ﺃﻛﱪ ﹼ‬ ‫ﻓﻴﺴﻤﻰ ﺍﳌﺬﻳﺐ‪.‬‬

‫‪ ∫ƒ∏ëŸG ‘ á«Äjõ÷G äÉÑcôŸG‬ﺭﻏﻢ ﺃﻥ ﺍﳌﺎﺀ ﻫﻮ ﺍﳌﺬﻳﺐ ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‬ ‫ﹰ‬ ‫ﲑﺍ ﻣﻦ ﺍﳌﻮﺍﺩ ﺍﳌﺬﺍﺑﺔ ﻛﺎﻟﺴﻜﺮﻭﺯ )ﺳﻜﺮ ﺍﳌﺎﺋﺪﺓ(‪ ،‬ﻭﺍﻹﻳﺜﺎﻧﻮﻝ‬ ‫ﺩﺍﺋﲈ‪ ،‬ﺇﻻ ﺃﻥ ﻫﻨﺎﻙ ﻛﺜ ﹰ‬ ‫)ﺍﻟﻜﺤﻮﻝ(‪ ،‬ﻫﻲ ﻣﺮﻛﺒﺎﺕ ﺗﻮﺟﺪ ﰲ ﺍﳌﺤﻠﻮﻝ ﻋﲆ ﺷﻜﻞ ﺟﺰﻳﺌﺎﺕ‪ ،‬ﻭﻫﻨﺎﻙ ﻣﻮﺍﺩ‬ ‫ﺗﻜﻮﻥ ﺃﻳﻮﻧﺎﺕ ﻋﻨﺪﻣﺎ ﺗﺬﻭﺏ ﰲ ﺍﳌﺎﺀ‪ .‬ﻓﺎﳌﺮﻛﺐ ﺍﳉﺰﻳﺌﻲ ﻛﻠﻮﺭﻳﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ‬ ‫ﺟﺰﻳﺌﻴﺔ ﹼ‬ ‫ﹰ‬ ‫ﻳﻜﻮﻥ ﺃﻳﻮﻧﺎﺕ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻭﺃﻳﻮﻧﺎﺕ ﺍﻟﻜﻠﻮﺭﻳﺪ ﻋﻨﺪﻣﺎ ﻳﺬﻭﺏ ﰲ ﺍﳌﺎﺀ‪ ،‬ﻛﲈ‬ ‫ﻣﺜﻼ ﹼ‬ ‫ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ ‪ .4-15‬ﻭﻳﻤﻜﻦ ﲤﺜﻴﻞ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺄﻳﻦ ﻫﺬﻩ ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫)‪HCl(aq)→ H+(aq) + Cl-(aq‬‬

‫ﺗﺴﻤﻰ ﺍﳌﺮﻛﺒﺎﺕ ﺍﻟﺘﻲ ﺗﻨﺘﺞ ﺃﻳﻮﻧﺎﺕ ﺍﳍﻴﺪﺭﻭﺟﲔ ـ ﻭﻣﻨﻬﺎ ﻛﻠﻮﺭﻳﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ـ‬ ‫ﹰ‬ ‫ﺃﲪﺎﺿﺎ‪ ،‬ﻭﳍﺬﺍ ﻓﺈﻥ ﳏﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ﺍﳌﺎﺋﻲ ﹸﻳﺴﻤﻰ ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ‪.‬‬ ‫ﺍﻟﺸـﻜﻞ ‪ 4-15‬ﻳﺘﻔﻜــﻚ ﲪــﺾ‬ ‫ﺍﳍﻴـﺪﺭﻭﻛﻠـﻮﺭﻳـﻚ ‪HCl‬‬ ‫ﰲ ﺍﻟــﲈﺀ ﺇﻟــﻰ ﺃﻳﻮﻧــﺎﺕ‬ ‫ﻫــﻴﺪﺭﻭﺟــــــﲔ ‪،H+‬‬ ‫ﻭﺃﻳﻮﻧـﺎﺕ ﻛﻠﻮﺭﻳـﺪ ‪.Cl-‬‬

‫‪25‬‬

‫‪ ∫ƒ``∏ëªdG »``a á``«fƒjC’G äÉ``ÑcôªdG‬ﺗﺘﻜـﻮﻥ ﺍﻟﻤﺮﻛﺒـﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ ﻣﻦ ﺃﻳﻮﻧـﺎﺕ ﻣﻮﺟﺒﺔ‬ ‫ﻣﻌﺎ ﺑﺮﻭﺍﺑﻂ ﺃﻳﻮﻧﻴﺔ‪ .‬ﻭﻋﻨﺪﻣﺎ ﺗﺬﻭﺏ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ ﻓﻲ ﺍﻟﻤﺎﺀ‬ ‫ﻭﺃﻳﻮﻧﺎﺕ ﺳﺎﻟﺒﺔ ﻣﺮﺗﺒﻄﺔ ﹰ‬ ‫ﻓـﺈﻥ ﺃﻳﻮﻧﺎﺗﻬـﺎ ﻳﻤﻜﻦ ﺃﻥ ﺗﻨﻔﺼـﻞ ﺑﻌﻀﻬﺎ ﻋﻦ ﺑﻌﺾ‪ .‬ﻭﺗﺴـﻤﻰ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴـﺔ ﺑﺎﻟﺘﻔﻜﻚ‪.‬‬ ‫ـ‬ ‫ﻓﺎﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ ﻟﻜﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﹰ‬ ‫ﻣﺜﻼ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺃﻳﻮﻧﺎﺕ ‪ Na+‬ﻭ ‪.Cl‬‬

‫ﺃﻧﻮاﻉ الﺘﻔاﻋﻼت ﻓﻲ المحالﻴﻞ الماﺋﻴﺔ‬ ‫‪Types of Reactions in Aqueous Solutions‬‬ ‫ﻋﻨﺪ ﻣﺰﺝ ﳏﻠﻮﻟﲔ ﻣﺎﺋﻴﲔ ﳛﻮﻳﺎﻥ ﺃﻳﻮﻧﺎﺕ ﺫﺍﺋﺒﺔ ﻓﺈﻥ ﺍﻷﻳﻮﻧﺎﺕ ﻗﺪ ﻳﺘﻔﺎﻋﻞ ﺑﻌﻀﻬﺎ ﻣﻊ‬ ‫ﺑﻌﺾ‪ .‬ﻭﻛﺜﲑ ﻣﻦ ﻫﺬﻩ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺗﻔﺎﻋﻼﺕ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪ ،‬ﻭﻳﻤﻜﻦ ﺃﻥ ﺗﺆﺩﻱ ﺇﱃ‬ ‫ﺛﻼﺛﺔ ﺃﻧﻮﺍﻉ ﻣﻦ ﺍﻟﻨﻮﺍﺗﺞ ﻫﻲ‪ :‬ﺭﺍﺳﺐ‪ ،‬ﺃﻭ ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯ‪ .‬ﻭﺃﻣﺎ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺬﻳﺐ ـ ﻭﻫﻲ ﰲ‬ ‫ﺍﻟﻐﺎﻟﺐ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ـ ﻓﻼ ﺗﺘﻔﺎﻋﻞ ﻋﺎﺩ ﹰﺓ‪.‬‬ ‫‪ Ö°SGhQ ¿ƒq µJ »àdG äÓ``YÉØàdG‬ﺑﻌﺾ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﺤﺪﺙ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ ﺗﻨﺘﺞ‬ ‫ﺭﻭﺍﺳﺐ‪ .‬ﻓﻌﻨﺪ ﺧﻠﻂ ﻣﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﻣﺤﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﺤﺎﺱ ‪ ،II‬ﻳﺤﺪﺙ‬ ‫ﺗﻔﺎﻋﻞ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ ﻳﺆﺩﻱ ﺇﻟﻰ ﺗﻜﻮﻳﻦ ﺭﺍﺳﺐ ﻣﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ ‪.II‬‬ ‫)‪2NaOH(aq) + CuCl2(aq)→ 2NaCl(aq)+ Cu(OH)2(s‬‬

‫ﻻﺣﻆ ﺃﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻻ ﺗﻮﺿﺢ ﺑﻌﺾ ﺗﻔﺎﺻﻴﻞ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ؛ ﻓﻬﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﺤﺎﺱ ‪ II‬ﻣﺮﻛﺒﺎﺕ ﺃﻳﻮﻧﻴﺔ‪ ،‬ﻭﳍﺬﺍ ﻓﻬﲈ ﻳﻮﺟﺪﺍﻥ ﰲ ﳏﻠﻮﻟﻴﻬﲈ ﻋﲆ‬ ‫ﻛﲈ ﻫﻮ ﻣﺒﲔ ﰲ ﺍﻟﺸﻜﻞ‪ .4-16‬ﻭﻋﻨﺪ ﻣﺰﺝ‬ ‫ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ‬ ‫ﺍﳌﺤﻠﻮﻟﲔ ﺗﺘﺤﺪ ﺃﻳﻮﻧﺎﺕ ‪ Cu2+‬ﻣﻊ ﺃﻳﻮﻧﺎﺕ ‪ OH-‬ﻟﺘﻜﻮﻳﻦ ﺭﺍﺳﺐ ﻣﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫ﻓﺘﺒﻘﻰ ﺫﺍﺋﺒﺔ ﰲ ﺍﳌﺤﻠﻮﻝ‪.‬‬ ‫ﺍﻟﻨﺤﺎﺱ ‪ .Cu(OH)2‬ﺃﻣﺎ ﺃﻳﻮﻧﺎﺕ‬ ‫‪H 2O‬‬

‫‪H2O‬‬ ‫ﺃﻳﻮﲏ‪Na+‬‬ ‫ﺍﻟﺸـﻜﻞ ‪ 4-16‬ﻳﺘﻔـﻜﻚ ‪ NaOH‬ﰲ ﺍﳌـﺎﺀ ﺇﱃ‬ ‫ﻭ‪ ،OH-‬ﻛــﲈ ﻳﺘﻔـﻜــﻚ ‪ CuCl2‬ﺇﱃ ﺃﻳـﻮﲏ‬ ‫‪ Cu2+‬ﻭ ‪.Cl-‬‬

‫‪H2O‬‬

‫‪Cl-‬‬

‫‪-‬‬

‫‪Cl‬‬

‫‪Cu2+‬‬

‫)‪CuCl2(aq‬‬

‫‪Cl-‬‬

‫)‪(aq‬‬ ‫‪CuCl 2‬‬

‫‪Cu2+‬‬

‫‪2+‬‬

‫‪Cu‬‬

‫)‪CuCl2(aq‬‬

‫‪äÓYÉØàe‬‬ ‫‪OH-‬‬

‫‪H2O‬‬

‫‪Na+‬‬

‫)‪NaCl (aq‬‬

‫‪) +‬‬ ‫)‪Cu(OH 2 (s‬‬

‫‪+‬‬

‫‪NaOH(aq) Na‬‬

‫‪H 2O‬‬

‫‪Cu(OH)2 + NaCl‬‬ ‫)‪(s‬‬ ‫)‪(aq‬‬

‫‪Cu(OH)2 + NaCl‬‬ ‫)‪(s‬‬ ‫)‪(aq‬‬

‫‪-‬‬

‫‪OH‬‬

‫)‪(aq‬‬

‫‪NaOH‬‬

‫‪œGƒf‬‬

‫‪26‬‬ ‫‪8-08‬‬ ‫‪2837‬‬ ‫‪1C-8‬‬ ‫‪1‬‬ ‫‬‫‪C10-11C-828378-08‬‬ ‫‪0‬‬ ‫‪C1‬‬

‫‪C10-11C-828378-08‬‬

‫‪Na+‬‬

‫‪ á«fƒjC’G ä’OÉ``©ªdG‬ﻟﺘﻮﺿﻴﺢ ﺗﻔﺎﺻﻴﻞ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﺘﻀﻤﻦ ﺃﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ‬ ‫ﺍﻟﻤﺎﺋﻴـﺔ‪ ،‬ﻳﺴـﺘﺨﺪﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﻮﻥ ﺍﻟﻤﻌـﺎﺩﻻﺕ ﺍﻷﻳﻮﻧﻴﺔ‪ .‬ﻭﻫـﻲ ﺗﺨﺘﻠﻒ ﻋـﻦ ﺍﻟﻤﻌﺎﺩﻻﺕ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻓﻲ ﺃﻥ ﺍﻟﻤﻮﺍﺩ ﺍﻟﺘﻲ ﺗﻜﻮﻥ ﻋﻠﻰ ﺷـﻜﻞ ﺃﻳﻮﻧﺎﺕ ﻓـﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺗﻜﺘﺐ ﻛﺄﻳﻮﻧﺎﺕ‬ ‫ﻓﻠﻜﻲ ﺗﻜﺘﺐ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺍﻷﻳﻮﻧﻴﺔ ﻟﺘﻔﺎﻋﻞ ﻣﺤﻠﻮﻟـﻲ ‪ NaOH‬ﻭ ‪ CuCl2‬ﹰ‬ ‫ﻣﺜﻼ‬ ‫ﻓـﻲ ﺍﻟﻤﻌﺎﺩﻟـﺔ‪ .‬ﹾ‬ ‫ﻳﺠﺐ ﺃﻥ ﺗﻜﺘﺐ ﺍﻟﻤﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﺎﺗﺞ ‪ NaCl‬ﻋﻠﻰ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ‪.‬‬ ‫ﻭ ﹸﺗﺴﻤﻰ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﻲ ﺗﺒﲔ ﺍﳉﺴﻴﲈﺕ ﰲ ﺍﳌﺤﻠﻮﻝ ﺑﺎﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪ .‬ﻻﺣﻆ ﺃﻥ‬ ‫ﺃﻱ ﺃﳖﺎ ﱂ ﺗﺸﺎﺭﻙ ﰲ‬ ‫ﺃﻳﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﺍﻟﻜﻠﻮﺭ ﻣﻮﺍﺩ ﻣﺘﻔﺎﻋﻠﺔ ﻭﻧﺎﲡﺔ ﰲ ﺍﻟﻮﻗﺖ ﻧﻔﺴﻪ‪ ،‬ﹾ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‪ ،‬ﻭﳍﺬﺍ ﺗﺴﻤﻰ ﺃﻳﻮﻧﺎﺕ ﻣﺘﻔﺮﺟﺔ‪ .‬ﻭﻫﻲ ﻋﺎﺩﺓ ﻻ ﺗﻈﻬﺮ ﰲ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻷﻳﻮﻧﻴﺔ‪ .‬ﻭﻋﻨﺪ‬ ‫ﺷﻄﺐ ﻫﺬﻩ ﺍﻷﻳﻮﻧﺎﺕ ﻣﻦ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﲢﺼﻞ ﻋﲆ ﻣﺎ ﻳﺴﻤﻰ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ‬ ‫ﺍﻟﻨﻬﺎﺋﻴﺔ‪ ،‬ﻭﻫﻲ ﺗﺸﺘﻤﻞ ﻋﲆ ﺍﳉﺴﻴﲈﺕ ﺍﳌﺸﺎﺭﻛﺔ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﻓﻘﻂ‪.‬‬

‫ﻻﺣﻆ ﺃﻧﻪ ﱂ ﻳﺘﺒﻖ ﺳﻮ￯ ﺃﻳﻮﻧﺎﺕ ﺍﳍﻴﺪﺭﻭﻛﺴﻴﺪ ﻭﺍﻟﻨﺤﺎﺱ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﻠﻴﺔ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‪ ،‬ﺍﳌﻮﺿﺤﺔ ﺃﺩﻧﺎﻩ‪:‬‬ ‫‪ ?äCGôb GPÉe‬ﻗﺎﺭﻥ ﻓﻴﻢ ﲣﺘﻠﻒ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻷﻳﻮﻧﻴﺔ ﻋﻦ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ؟‬

‫ﺗـﺠــﺮﺑــﺔ‬ ‫ﻳﻜﻮن راﺳ ًﺒﺎ‬ ‫ﻻﺣﻆ ﺗﻔﺎﻋﻼً ّ‬

‫‪?áÑ∏°U IOÉe ¿’ƒ∏ ¿ƒq µj ∞«c‬‬ ‫اﻟﺨﻄﻮات‬

‫‪OK P L‬‬

‫ﺍﻟﻤﺨﺘﺒﺮ‪AK FHBL .‬‬ ‫ﻧﻤﻮﺫﺝ ‪F‬‬ ‫‪G N‬‬ ‫ﺍﻟﺴﻼﻣﺔ‪O‬‬ ‫‪EG‬‬ ‫‪K‬ﻓﻲ ‪GCIM HJDN‬‬ ‫‪IE‬‬ ‫‪JLP‬‬ ‫‪K‬‬ ‫‪M‬‬ ‫‪LH‬‬ ‫‪OI.1 N‬‬ ‫‪PJ‬‬ ‫‪ M‬ﺍﻗﺮﺃ‬

‫‪.2‬‬ ‫‪.3‬‬ ‫‪.4‬‬

‫‪.5‬‬

‫‪DFJ‬‬

‫‪ ،NaOH‬ﻭﺳﺠﻞ ﻣﻼﺣﻈﺎﺗﻚ‪.‬‬ ‫‪ .6‬ﺣﺮﻙ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻨﺎﺗﺞ‪ ،‬ﻭﺳﺠﻞ ﻣﻼﺣﻈﺎﺗﻚ‪.‬‬ ‫‪ .7‬ﺍﺗﺮﻙ ﺍﻟﺮﺍﺳﺐ ﺣﺘﻰ ﻳﺴﺘﻘﺮ‪ ،‬ﺛﻢ ﺍﻓﺼﻞ ﺍﻟﺴﺎﺋﻞ ﻋﻨﻪ ﻓﻲ‬ ‫ﻣﺨﺒﺎﺭ ﻣﺪﺭﺝ ﺳﻌﺔ ‪.100 ml‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪D AE BF ACG BDH CE I‬‬ ‫‪ .8‬ﺗﺨﻠﺺ ﻣﻦ ﺍﻟﺮﺍﺳﺐ ﻛﻤﺎ ﻳﺮﺷﺪﻙ ﻣﻌﻠﻤﻚ‪.‬‬

‫ﻣﻘﻄﺮﺍ ﻓﻲ ﻛﺄﺱ ﺳﻌﺔ ‪.150 ml‬‬ ‫ﻣﺎﺀﺍ‬ ‫ﺿﻊ ‪ 50 ml‬ﹰ‬ ‫ﹰ‬ ‫ﺯﻥ ‪ 4g‬ﻣـﻦ ‪ ،NaOH‬ﺛـﻢ ﺃﺿﻔﻬـﺎ ﺑﺎﻟﺘﺪﺭﻳـﺞ ﺇﻟـﻰ اﻟﺘﺤﻠﻴﻞ‬ ‫‪ .1‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻣﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﻴﻦ ‪NaOH‬‬ ‫ﺍﻟﻜﺄﺱ‪ ،‬ﻭﺣﺮﻛﻬﺎ ﺑﺎﺳﺘﻤﺮﺍﺭ ﺑﺴﺎﻕ ﺍﻟﺘﺤﺮﻳﻚ‪.‬‬ ‫ﻭ‪ .MgSO4‬ﻭﻻﺣﻆ ﺃﻥ ﺃﻏﻠﺐ ﻣﺮﻛﺒﺎﺕ ﺍﻟﻜﺒﺮﻳﺘﺎﺕ ﺗﻮﺟﺪ‬ ‫ﺯﻥ ‪ 6g‬ﻣـﻦ ﻣﻠـﺢ ﺃﺑﺴـﻮﻡ )ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﻤﺎﻏﻨﺴـﻴﻮﻡ‬ ‫ﻋﻠﻰ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ‪.‬‬ ‫‪ ،(MgSO4‬ﻭﺿﻌﻬﺎ ﻓﻲ ﻛﺄﺱ ﺃﺧﺮ￯ ﺳـﻌﺔ ‪.150 ml‬‬ ‫ﻣﻘﻄـﺮﺍ ﺇﻟﻰ ﺍﻟﻤﻠـﺢ‪ ،‬ﻭﺣﺮﻛﻪ ‪ .2‬ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻟﻬﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫ﺛـﻢ ﺃﺿـﻒ ‪ 50 ml‬ﻣﺎﺀ‬ ‫ﹰ‬ ‫‪ .3‬ﺣـﺪﺩ ﺃﻱ ﺍﻷﻳﻮﻧـﺎﺕ ﻣﺘﻔﺮﺟـﺔ‪ ،‬ﺛـﻢ ﺍﻛﺘـﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺑﺴﺎﻕ ﺍﻟﺘﺤﺮﻳﻚ ﺣﺘﻰ ﻳﺬﻭﺏ ﺍﻟﻤﻠﺢ ﺗﻤﺎ ﹰﻣﺎ‪.‬‬ ‫ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ‪.‬‬ ‫ﺃﺿـﻒ ﻣﺤﻠـﻮﻝ ﻣﻠـﺢ ﺃﺑﺴـﻮﻡ ﺑﺒـﻂﺀ ﺇﻟـﻰ ﻣﺤﻠـﻮﻝ‬

‫‪27‬‬

‫‪4-3 ∫Éãe‬‬

‫ﺗﻜﻮﻥ ﺭﺍﺳ ﹰﺒﺎ‪ :‬ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﺘﻔﺎﻋﻞ ﳏﻠﻮﱄ ﻧﱰﺍﺕ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﹼ‬ ‫ﻳﻜﻮﻥ ﺭﺍﺳ ﹰﺒﺎ ﻣﻦ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺒﺎﺭﻳﻮﻡ ‪.BaCO3‬‬ ‫ﺍﻟﺒﺎﺭﻳﻮﻡ ‪ Ba(NO3)2‬ﻭﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2CO3‬ﻭﺍﻟﺬﻱ ﹼ‬

‫‪ádCÉ°ùŸG π«∏– 1‬‬

‫ﻟﻘﺪ ﹸﺃﻋﻄﻴﺖ ﺃﺳﲈﺀ ﻭﺭﻣﻮﺯ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﻮﺍﺗﺞ ﻟﻠﺘﻔﺎﻋﻞ‪ .‬ﻟﻜﺘﺎﺑﺔ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ ﻟﻠﺘﻔﺎﻋﻞ ﳚﺐ ﺃﻥ ﲢﺪﺩ‬ ‫ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻨﺴﺒﻴﺔ ﻟﻠﻤﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﲡﺔ‪ .‬ﻭﻟﻜﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﲢﺘﺎﺝ ﺇﱃ ﺗﻮﺿﻴﺢ ﺍﳊﺎﻻﺕ ﺍﻷﻳﻮﻧﻴﺔ‬ ‫ﻟﻠﻤﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﲡﺔ‪ .‬ﻭﺑﺸﻄﺐ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻣﻦ ﻃﺮﰲ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﻳﻤﻜﻨﻚ ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‪.‬‬ ‫ﺍﻟﺘﻲ ﺗﺸﺘﻤﻞ ﻋﲆ ﻣﻮﺍﺩ ﺃﻗﻞ ﻣﻦ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻷﺧﺮ￯‪.‬‬

‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺼﺤﻴﺤﺔ ﻭﺍﳊﺎﻻﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﻟﻜﻞ ﺍﳌﻮﺍﺩ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﻟﻜﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻪ‪:‬‬ ‫ﺯﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫ﻭﺿــﺢ ﺃﻳــﻮﻧــﺎﺕ ﺍﳌــﻮﺍﺩ‬ ‫ﺍﳌﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﲡﺔ‬

‫ﺍﺣﺬﻑ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻣﻦ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‬ ‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‬

‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﺗﺪﻋﻢ ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﳌﻮﺿﺤﺔ ﰲ ﺍﻟﺸﻜﻞ ‪ 4-13‬ﺍﻟﺘﻮﻗﻌﺎﺕ‪ .‬ﻓﺎﳌﻌﺎﺩﻻﺕ ﻣﻮﺯﻭﻧﺔ؛ ﻷﻥ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ‬ ‫ﻣﺘﺴﺎ ﹴﻭ ﰲ ﻃﺮﻓﻴﻬﺎ‪ .‬ﻭﲢﺘﻮﻱ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻋﲆ ﻋﺪﺩ ﺃﻗﻞ ﻣﻦ ﺍﳌﺮﻛﺒﺎﺕ‪ ،‬ﻭﺗﺒﲔ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺎﻋﻠﺔ‬ ‫ﻟﺘﻜﻮﻳﻦ ﺍﻟﺮﺍﺳﺐ )ﺍﳌﺎﺩﺓ ﺍﻟﺼﻠﺒﺔ(‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﺗﻜﻮﻥ ﺭﺍﺳ ﹰﺒﺎ‪ ،‬ﻣﺴﺘﺨﺪ ﹰﻣﺎ‬ ‫ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺃﻳﻮﻧﻴﺔ ﻛﺎﻣﻠﺔ‪ ،‬ﻭﺃﻳﻮﻧﻴﺔ ﳖﺎﺋﻴﺔ ﻟﻜﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ ﺍﻟﺘﻲ ﻗﺪ ﹼ‬ ‫)‪ (NR‬ﻟﺒﻴﺎﻥ ﻋﺪﻡ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ‪.‬‬ ‫ﺗﻜﻮﻥ ﺭﺍﺳﺐ ﻣﻦ ﻳﻮﺩﻳﺪ ﺍﻟﻔﻀﺔ ‪.AgI‬‬ ‫‪ .35‬ﻋﻨﺪ ﺧﻠﻂ ﳏﻠﻮﱄ ﻳﻮﺩﻳﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ KI‬ﻭﻧﱰﺍﺕ ﺍﻟﻔﻀﺔ ‪ AgNO3‬ﹼ‬

‫‪ .36‬ﻋﻨﺪ ﺧﻠﻂ ﳏﻠﻮﱄ ﻓﻮﺳﻔﺎﺕ ﺍﻷﻣﻮﻧﻴﻮﻡ ‪ (NH4)3PO4‬ﻭﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2SO4‬ﱂ ﻳﺘﻜﻮﻥ ﺃﻱ ﺭﺍﺳﺐ‪ ،‬ﻭﱂ‬ ‫ﻳﺘﺼﺎﻋﺪ ﺃﻱ ﻏﺎﺯ‪.‬‬

‫ﺗﻜﻮﻥ ﺭﺍﺳﺐ ﻣﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ‬ ‫‪ .37‬ﻋﻨﺪ ﺧﻠﻂ ﳏﻠﻮﱄ ﻛﻠﻮﺭﻳﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ‪ AlCl3‬ﻭﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ NaOH‬ﹼ‬ ‫ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ‪.Al(OH)3‬‬ ‫ﺗﻜﻮﻥ ﺭﺍﺳﺐ ﻣﻦ ﻛﱪﻳﺘﺎﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪.CaSO4‬‬ ‫‪ .38‬ﻋﻨﺪ ﺧﻠﻂ ﳏﻠﻮﱄ ﻛﱪﻳﺘﺎﺕ ﺍﻟﻠﻴﺜﻴﻮﻡ ‪ Li2SO4‬ﻭﻧﱰﺍﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca(NO3)2‬ﹼ‬ ‫ﺗﻜﻮﻥ ﺭﺍﺳﺐ ﳛﺘﻮﻱ ﻋﲆ ﺍﳌﻨﺠﻨﻴﺰ‪.‬‬ ‫‪ .39‬ﲢﺪﹼ ‪ :‬ﻋﻨﺪ ﺧﻠﻂ ﳏﻠﻮﱄ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2CO3‬ﻭﻛﻠﻮﺭﻳﺪ ﺍﳌﻨﺠﻨﻴﺰ ‪ MgCl2‬ﹼ‬

‫‪28‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-17‬ﻳﺘﺄﻳـﻦ ﺑﺮﻭﻣﻴـﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪ HBr‬ﰲ‬ ‫ﺍﳌـﺎﺀ ﺇﱃ ‪ H+‬ﻭ‪ .Br-‬ﻭﻫﻴﺪﺭﻭﻛﺴـﻴﺪ‬ ‫ﺍﻟﺼﻮﺩﻳـﻮﻡ ﻳﺘﻔـﻜﻚ ﺇﱃ ‪ Na+‬ﻭ ‪.OH-‬‬ ‫ﺗﺘﻔﺎﻋـﻞ ﺃﻳﻮﻧـﺎﺕ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻭﺃﻳﻮﻧﺎﺕ‬ ‫‬‫ﻭﺗﻜﻮﻥ ﺍﳌﺎﺀ ‪. H2O‬‬ ‫ﺍﳍﻴﺪﺭﻭﻛﺴﻴﺪ ‪ OH‬ﹼ‬ ‫ﺣـﺪﺩ ﺍﻷﻳﻮﻧـﺎﺕ ﺍﻟﺴـﺎﻟﺒﺔ ﻭﺍﻷﻳﻮﻧﺎﺕ‬ ‫ﺍﻟﻤﻮﺟﺒﺔ ﻓﻲ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫‪äÓYÉØàŸG‬‬

‫‪-‬‬

‫‪OH‬‬

‫‪Na+‬‬ ‫‪H2O‬‬

‫‪œGƒædG‬‬

‫)‪(aq‬‬

‫‪NaOH‬‬

‫‪ AÉ``e ¿ƒq µJ »àdG äÓ``YÉØàdG‬ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣـﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ ﻳﺆﺩﻱ ﺇﻟﻰ‬ ‫ﺗﻜﻮﻳﻦ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ‪ ،‬ﻓﻴﺰﺩﺍﺩ ﻋﺪﺩ ﺟﺴـﻴﻤﺎﺕ ﺍﻟﻤﺬﻳﺐ )ﺍﻟﻤﺎﺀ(‪ .‬ﻭﺑﺨﻼﻑ ﺍﻟﺘﻔﺎﻋﻼﺕ‪C10-11C-828378-08-A‬‬ ‫ﺍﻟﺘﻲ ﻳﺘﻜﻮﻥ ﻓﻴﻬﺎ ﺭﺍﺳﺐ‪ ،‬ﻻ ﻳﻼﺣﻆ ﻓﻲ ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺩﻟﻴﻞ ﻋﻠﻰ ﺣﺪﻭﺙ‬ ‫ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ؛ ﻷﻥ ﺍﻟﻤﺎﺀ ﻋﺪﻳﻢ ﺍﻟﻠﻮﻥ ﻭﺍﻟﺮﺍﺋﺤﺔ‪ ،‬ﻛﻤﺎ ﺃﻧﻪ ﹼ‬ ‫ﻳﺸـﻜﻞ ﺃﻏﻠﺐ ﺍﻟﻤﺤﻠﻮﻝ‪.‬‬ ‫ﻓﻌﻨﺪﻣﺎ ﺗﺨﻠﻂ ﻣﺤﻠﻮﻝ ﺣﻤﺾ ﺍﻟﻬﻴﺪﺭﻭﺑﺮﻭﻣﻴﻚ ‪ HBr‬ﹰ‬ ‫ﻣﺜﻼ ﻣﻊ ﻣﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴـﻴﺪ‬ ‫ﺍﻟﺼﻮﺩﻳـﻮﻡ ‪ NaOH‬ﻛﻤﺎ ﻫﻮ ﻣﺒﻴﻦ ﻓﻲ ﺍﻟﺸـﻜﻞ ‪ ،4-17‬ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪،‬‬ ‫ﻭﻳﺘﻜﻮﻥ ﻣﺎﺀ‪ ،‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫)‪HBr(aq) + NaOH(aq) → H2O(l) + NaBr(aq‬‬

‫ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﹸ‬ ‫ﺗﻜﻮﻥ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﺎﺗﺞ ﺑﺮﻭﻣﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻋﲆ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ ﰲ‬ ‫ﺍﳌﺤﻠﻮﻝ ﺍﳌﺎﺋﻲ‪ .‬ﻭﺗﻮﺿﺢ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻫﺬﻩ ﺍﻷﻳﻮﻧﺎﺕ‪:‬‬ ‫→ )‪H +(aq) + Br -(aq) + Na +(aq) + OH -(aq‬‬ ‫)‪H 2O(l) + Na +(aq) + Br −(aq‬‬

‫ﻟﻮ ﺩﻗﻘﺖ ﰲ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﻓﺴﺘﻼﺣﻆ ﺃﻥ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺎﻋﻠﺔ ﻫﻲ ﺃﻳﻮﻧﺎﺕ ﺍﳍﻴﺪﺭﻭﺟﲔ‬ ‫ﻭﺃﻳﻮﻧﺎﺕ ﺍﳍﻴﺪﺭﻭﻛﺴﻴﺪ؛ ﻷﻥ ﹰﹼ‬ ‫ﻛﻼ ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﺃﻳﻮﻧﺎﺕ ﺍﻟﱪﻭﻣﻴﺪ ﺃﻳﻮﻧﺎﺕ‬ ‫ﻣﺘﻔﺮﺟﺔ‪ .‬ﻭﺇﺫﺍ ﺣﺬﻓﺖ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻓﺴﺘﺒﻘﻰ ﻟﺪﻳﻚ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﺘﻲ ﺗﺸﺎﺭﻙ ﰲ‬ ‫ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫ﻭﺗﻜﻮﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻛﺎﻵﰐ‪:‬‬

‫)‪H +(aq) + OH -(aq) → H 2O(l‬‬

‫‪ ?äCGôb GPÉe‬ﺣ ﹼﻠﻞ ﳌﺎﺫﺍ ﺗﺴﻤﻰ ﺃﻳﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﺃﻳﻮﻧﺎﺕ ﺍﻟﱪﻭﻣﻴﺪ ﰲ ﺗﻔﺎﻋﻞ ﳏﻠﻮﻝ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﲪﺾ ﺍﳍﻴﺪﺭﻭﺑﺮﻭﻣﻴﻚ ﺃﻳﻮﻧﺎﺕ ﻣﺘﻔﺮﺟﺔ؟‬

‫‪29‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﻨﺘﺞ ﻣﺎﺀ‪:‬‬ ‫‪ .40‬ﻋﻨﺪ ﺧﻠﻂ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ‪ H2SO4‬ﺑﻤﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ KOH‬ﻳﻨﺘﺞ ﻣﺎﺀ ﻭﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪.K2SO4‬‬ ‫‪ .41‬ﻋﻨﺪ ﺧﻠﻂ ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ‪ HCl‬ﺑﻤﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﻳﻨﺘﺞ ﻣﺎﺀ ﻭﳏﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪.CaCl2‬‬ ‫‪ .42‬ﻋﻨﺪ ﺧﻠﻂ ﲪﺾ ﺍﻟﻨﻴﱰﻳﻚ ‪ HNO3‬ﺑﻤﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻷﻣﻮﻧﻴﻮﻡ ‪ NH4OH‬ﻳﻨﺘﺞ ﻣﺎﺀ ﻭﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻷﻣﻮﻧﻴﻮﻡ‬ ‫‪.NH4NO3‬‬ ‫‪ .43‬ﻋﻨﺪ ﺧﻠﻂ ﻛﱪﻳﺘﻴﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪ H2S‬ﺑﻤﺤﻠﻮﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca(OH)2‬ﻳﻨﺘﺞ ﻣﺎﺀ ﻭﳏﻠﻮﻝ ﻛﱪﻳﺘﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪.CaS‬‬ ‫‪ .44‬ﲢﺪﹼ ‪ :‬ﻋﻨﺪ ﺧﻠﻂ ﲪﺾ ﺍﻟﺒﻨﺰﻭﻳﻚ ‪ C6H5COOH‬ﻭﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ‪ Mg(OH)2‬ﻳﺘﻜﻮﻥ ﻣﺎﺀ ﻭﺑﻨﺰﻭﺍﺕ‬ ‫ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ‪.(C6H5COO)2Mg‬‬

‫‪ äGRÉZ ¿ƒq µJ »àdG äÓ``YÉØàdG‬ﻳﻨﺘﺞ ﻋﻦ ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﻤﺰﺩﻭﺝ‬ ‫ﺗﻜﻮﻳﻦ ﻏﺎﺯﺍﺕ‪ ،‬ﻣﺜﻞ ‪ ،CO2‬ﻭ ‪ ،HCN‬ﻭ ‪ .H2S‬ﻓﻌﻨﺪﻣﺎ ﺗﺨﻠﻂ ﺣﻤﺾ ﺍﻟﻬﻴﺪﺭﻭﻳﻮﺩﻳﻚ‬ ‫‪ HI‬ﺑﻤﺤﻠـﻮﻝ ﻛﺒﺮﻳﺘﻴﺪ ﺍﻟﻠﻴﺜﻴـﻮﻡ ‪ ،Li2S‬ﻳﺘﺼﺎﻋﺪ ﻏﺎﺯ ﻛﺒﺮﻳﺘﻴﺪ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ ،H2S‬ﻛﻤﺎ‬ ‫ﻳﻨﺘﺞ ﻳﻮﺩﻳﺪ ﺍﻟﻠﻴﺜﻴﻮﻡ ‪ LiI‬ﺍﻟﺬﻱ ﻳﻈﻞ ﺫﺍﺋ ﹰﺒﺎ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ‪.‬‬ ‫)‪2HI(aq) + Li 2S(aq) → H 2S(g) + 2LiI(aq‬‬

‫ﻭﺑﺎﺳﺘﺜﻨﺎﺀ ‪ ،H2S‬ﻓﺈﻥ ﲨﻴﻊ ﺍﳌﻮﺍﺩ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﺗﻮﺟﺪ ﻋﲆ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ‪ .‬ﻟﺬﺍ ﻳﻤﻜﻨﻚ‬ ‫ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻋﲆ ﺍﻟﻨﺤﻮ ﺍﻵﰐ‪:‬‬ ‫→ )‪2H +(aq) + 2I −(aq) + 2Li +(aq) + S 2−(aq‬‬

‫)‪H 2S(g) + 2Li +(aq) + 2I −(aq‬‬

‫ﻭﺑﺤﺬﻑ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻳﻤﻜﻨﻚ ﺍﳊﺼﻮﻝ ﻋﲆ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ‪ ،‬ﻭﻫﻲ‪:‬‬ ‫)‪2H +(aq) + S 2-(aq) → H 2S(g‬‬

‫ﻭﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﻨﺘﺞ ﹰ‬ ‫ﻏﺎﺯﺍ ﻣﺎ ﳛﺪﺙ ﰲ ﺍﳌﻄﺒﺦ ﻋﻨﺪﻣﺎ ﲣﻠﻂ ﺍﳋﻞ ﺑﺼﻮﺩﺍ ﺍﳋﺒﺰ‪.‬‬ ‫ﻓﺎﳋﻞ ﳏﻠﻮﻝ ﻣﺎﺋﻲ ﳊﻤﺾ ﺍﻹﻳﺜﺎﻧﻮﻳﻚ‪ ،‬ﻭﺻﻮﺩﺍ ﺍﳋﺒﺰ ﻋﺒﺎﺭﺓ ﻋﻦ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻣﻌ ﹰﺎ ﻳﺘﻔﺎﻋﻼﻥ ﻭﻳﺘﺼﺎﻋﺪ ﻏﺎﺯ ‪ ،CO2‬ﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ‬ ‫ﺍﳍﻴﺪﺭﻭﺟﻴﻨﻴﺔ‪ .‬ﻭﻋﻨﺪ ﺧﻠﻄﻬﲈ ﹰ‬ ‫ﺍﻟﺸﻜﻞ ‪.4-18‬‬

‫ﻭﻫﻨﺎﻙ ﺗﻔﺎﻋﻞ ﺁﺧﺮ ﻣﺸﺎﺑﻪ ﻟﺘﻔﺎﻋﻞ ﺍﳋﻞ ﻣﻊ ﺻﻮﺩﺍ ﺍﳋﺒﺰ‪ ،‬ﳛﺪﺙ ﻋﻨﺪﻣﺎ ﲣﻠﻂ ﺃﻱ ﳏﻠﻮﻝ‬ ‫ﲪﴤ ﺑﻜﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﳍﻴﺪﺭﻭﺟﻴﻨﻴﺔ‪ .‬ﻭﰲ ﺍﳊﺎﻻﺕ ﲨﻴﻌﻬﺎ ﳚﺐ ﺃﻥ ﳛﺪﺙ ﺗﻔﺎﻋﻼﻥ‬ ‫ﻣﺘﺰﺍﻣﻨﺎﻥ ﰲ ﺍﳌﺤﻠﻮﻝ ﻟﻴﻨﺘﺞ ﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‪ .‬ﻭﺃﺣﺪ ﻫﺬﻳﻦ ﺍﻟﺘﻔﺎﻋﻠﲔ ﺗﻔﺎﻋﻞ ﺇﺣﻼﻝ‬ ‫ﻣﺰﺩﻭﺝ‪ ،‬ﻭﺍﻵﺧﺮ ﺗﻔﺎﻋﻞ ﺗﻔﻜﻚ‪ .‬ﻓﻌﻨﺪﻣﺎ ﺗﺬﻳﺐ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﳍﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﹰ‬ ‫ﻣﺜﻼ ﰲ‬ ‫ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪ ،‬ﻭﻳﻨﺘﺞ ﻏﺎﺯ‪.‬‬

‫‪30‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-18‬ﻋﻨﺪﻣـﺎ ﻳﺘﻔﺎﻋـﻞ ﺍﳋـﻞ ﻣـﻊ‬ ‫ﺻـــﻮﺩﺍ ﺍﳋﺒـــﺰ ‪NaHCO3‬‬ ‫ﳛـﺪﺙ ﺗﺼﺎﻋـﺪ ﴎﻳـﻊ ﻟﻐـﺎﺯ‬ ‫ﺛـﺎﲏ ﺃﻛﺴـﻴﺪ ﺍﻟﻜﺮﺑـﻮﻥ ‪CO2‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻣﺎﺩﺓ ﺃﻳﻮﻧﻴﺔ ﺗﺒﻘﻰ ﰲ ﺍﳌﺎﺀ ﻋﲆ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ ﻣﻨﻔﺼﻠﺔ‪ .‬ﺃﻣﺎ ﲪﺾ‬ ‫ﺍﻟﻜﺮﺑﻮﻧﻴﻚ ‪ H2CO3‬ﻓﻴﺘﻔﻜﻚ ﺑﻤﺠﺮﺩ ﺗﻜﻮﻧﻪ ﺇﱃ ﻣﺎﺀ ﻭﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‪.‬‬ ‫)‪HCl(aq) + NaHCO 3(aq) → H 2CO 3(aq) + NaCl(aq‬‬

‫)‪H2CO3(aq)→H2O(l) + CO2(g‬‬

‫ﻭﻳﻤﻜﻦ ﲨﻌﻬﲈ ﻭﲤﺜﻴﻠﻬﲈ ﺑﻤﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﺣﺪﺓ‪:‬‬

‫)‪HCl(aq) +NaHCO3(aq)+H2CO3(aq)→H2CO3(aq‬‬ ‫)‪+NaCl (aq)+H2O(l)+CO2(g‬‬

‫ﻭﺑﺤﺬﻑ ‪ H2CO3‬ﻣﻦ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ ﲢﺼﻞ ﻋﲆ ﻣﺎ ﻳﺴﻤﻰ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ‪.‬‬ ‫)‪HCl(aq) + NaHCO3(aq)→ H2O(l) + CO2(g)+ NaCl(aq‬‬

‫ﻫﺬﺍ‪ ،‬ﻭﻳﻤﻜﻨﻚ ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻛﺎﻵﰐ‪:‬‬

‫→ )‪H+(aq) + Cl-(aq) + Na+(aq) + HCO3-(aq‬‬

‫)‪H2O(l) + CO2(g) + Na+(aq) + Cl-(aq‬‬

‫ﻭﺗﻼﺣﻆ ﺃﻥ ﺃﻳﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻭﺃﻳﻮﻧﺎﺕ ﺍﻟﻜﻠﻮﺭ ﻫﻲ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ‪ ،‬ﻟﺬﺍ‬ ‫ﻳﻤﻜﻦ ﺣﺬﻓﻬﺎ ﻣﻦ ﻃﺮﰲ ﺍﳌﻌﺎﺩﻟﺔ‪ ،‬ﻭﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻛﺎﻵﰐ‪:‬‬ ‫)‪H+ (aq) + HCO3-(aq) → H2O(l)+ CO2(g‬‬

‫‪ ?äCGôb GPÉe‬ﺻﻒ ﻣﺎ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ؟‬

‫‪31‬‬

‫‪4-4 ∫Éãe‬‬

‫ﺗﻜﻮﻥ ﻏﺎﺯﺍﺕ‪ :‬ﺍﻛﺘﺐ ﹰﹼ‬ ‫ﻛﻼ ﻣﻦ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﹼ‬ ‫ﺑﲔ ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﻭﳏﻠﻮﻝ ﻛﱪﻳﺘﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ ،Na2S‬ﻭﺍﻟﺬﻱ ﻳﻨﺘﺞ ﻏﺎﺯ ﻛﱪﻳﺘﻴﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪. H2S‬‬

‫‪ádCÉ°ùŸG π«∏– 1‬‬

‫ﻟﻘﺪ ﹸﺃﻋﻄﻴﺖ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﲔ ﲪﺾ ﺍﳍﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ‪ HCl‬ﻭﻛﱪﻳﺘﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ .Na2S‬ﳚﺐ ﺃﻥ‬ ‫ﺗﻜﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﻭﺗﺰﳖﺎ‪ .‬ﻭﻟﻜﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﳚﺐ ﺃﻥ ﲢﺪﺩ ﺣﺎﻟﺔ ﺍﳌﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ‬ ‫ﻭﺍﻟﻨﺎﲡﺔ‪ .‬ﻭﺑﺤﺬﻑ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻣﻦ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻳﻤﻜﻨﻚ ﻛﺘﺎﺑﺔ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‪.‬‬

‫‪܃∏£ŸG ÜÉ°ùM 2‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻠﺘﻔﺎﻋﻞ‪.‬‬

‫ﺯﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫)‪HCl(aq) + Na2S(aq) → H2S(g) + NaCl(aq‬‬ ‫)‪2HCl(aq) + Na2S(aq) → H2S(g) + 2NaCl(aq‬‬

‫→ )‪2H+(aq) + 2Cl−(aq) + 2Na+(aq) + S (aq‬‬ ‫ﻭﺿﺢ ﺃﻳﻮﻧﺎﺕ ﺍﳌﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﲡﺔ‪.‬‬ ‫)‪H 2S(g) + 2Na +(aq) + 2Cl −(aq‬‬ ‫‪2-‬‬

‫ﺍﺣﺬﻑ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺘﻔﺮﺟﺔ ﻣﻦ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪.‬‬

‫‪2-‬‬

‫→ )‪2H+(aq) + 2Cl−(aq) + 2Na+(aq) + S (aq‬‬ ‫)‪H 2S(g) + 2Na +(aq) + 2Cl −(aq‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﺑﺄﺻﻐﺮ‬ ‫ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ‪.‬‬

‫‪2-‬‬

‫)‪2H+(aq) + S (aq) → H2S(g‬‬

‫‪áHÉLE’G ˃≤J 3‬‬

‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﻠﻴﺔ ﺗﺒﲔ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﺸﺎﺭﻛﺔ ﰲ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﺍﻛﺘﺐ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‪ ،‬ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪ .45‬ﻳﺘﻔﺎﻋﻞ ﲪﺾ ﺍﻟﺒﲑﻛﻠﻮﺭﻳﻚ ‪ HClO4‬ﻣﻊ ﳏﻠﻮﻝ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ Na2CO3‬ﻟﺘﻜﻮﻳﻦ ﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‬ ‫ﻭﺍﳌﺎﺀ ﻭﳏﻠﻮﻝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪.‬‬ ‫‪ .46‬ﻳﺘﻔﺎﻋﻞ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ‪ H2SO4‬ﻣﻊ ﳏﻠﻮﻝ ﺳﻴﺎﻧﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ NaCN‬ﻟﺘﻜﻮﻳﻦ ﻏﺎﺯ ﺳﻴﺎﻧﻴﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪HCN‬‬ ‫ﻭﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪.Na2SO4‬‬ ‫‪ .47‬ﻳﺘﻔﺎﻋﻞ ﲪﺾ ﺍﳍﻴﺪﺭﻭﺑﺮﻭﻣﻴﻚ ‪ HBr‬ﻣﻊ ﳏﻠﻮﻝ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻷﻣﻮﻧﻴﻮﻡ ‪ (NH4)2CO3‬ﻟﺘﻜﻮﻳﻦ ﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ‬ ‫ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﳌﺎﺀ ﻭﺑﺮﻭﻣﻴﺪ ﺍﻷﻣﻮﻧﻴﻮﻡ‪.‬‬ ‫‪ .48‬ﻳﺘﻔﺎﻋﻞ ﲪﺾ ﺍﻟﻨﻴﱰﻳﻚ ‪ HNO3‬ﻣﻊ ﳏﻠﻮﻝ ﻛﱪﻳﺘﻴﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ K2S‬ﻟﺘﻜﻮﻳﻦ ﻏﺎﺯ ﻛﱪﻳﺘﻴﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ‪.H2S‬‬ ‫‪ .49‬ﲢﺪﹼ ‪ :‬ﻳﺘﻔﺎﻋﻞ ﳏﻠﻮﻝ ﻳﻮﺩﻳﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪ KI‬ﻣﻊ ﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻟﺮﺻﺎﺹ ‪ Pb(NO3)2‬ﻟﺘﻜﻮﻳﻦ ﻳﻮﺩﻳﺪ ﺍﻟﺮﺻﺎﺹ‬ ‫‪ PbI2‬ﺍﻟﺼﻠﺐ‪.‬‬

‫‪32‬‬

‫ﺍﳍﻮﺍﺀ‬ ‫ﰲ‬ ‫ﺍﻟﺮﺋﺔ‬

‫ﺍﻟﺸـﻜﻞ ‪ 4-19‬ﺑﻌـﺪ ﺃﻥ ﻳﺪﺧـﻞ ﺃﻳـﻮﻥ‬ ‫ﺍﻟﺒﻴﻜـﺮﺑﻮﻧـﺎﺕ ‪ HCO-3‬ﺧﻠـﻴـﺔ‬ ‫ﺩﻡ ﲪـﺮﺍﺀ‪ ،‬ﻳﺘﻔﺎﻋـﻞ ﻣـﻊ ﺃﻳـﻮﻥ‬ ‫ﺍﳍﻴﺪﺭﻭﺟﲔ ‪ H+‬ﻟﺘﻜﻮﻳﻦ ﻣﺎﺀ ﻭﺛﺎﲏ‬ ‫ﺃﻛﺴـﻴﺪ ﺍﻟﻜﺮﺑـﻮﻥ ‪ ،CO2‬ﺍﻟـﺬﻱ‬ ‫ﳜـﺮﺝ ﻣﻦ ﺍﻟﺮﺋﺘﲔ ﻣﻊ ﻫﻮﺍﺀ ﺍﻟﺰﻓﲑ‪.‬‬ ‫ﺧﻼﻳﺎ ﺩﻡ ﲪﺮﺍﺀ ﰲ ﻭﻋﺎﺀ ﺩﻣﻮﻱ‬

‫ﻣﻬﻦ ﻓﻲ اﻟﻜﻴﻤﻴﺎء‬

‫‪ AÉ«MC’G º∏Y ™e §HôdG‬ﺇﻥ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ ﻛﻞ ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻭﺃﻳﻮﻧﺎﺕ‬ ‫ﺍﻟﺒﻴﻜﺮﺑﻮﻧﺎﺕ ﻹﻧﺘﺎﺝ ﺍﻟﻤﺎﺀ ﻭﺛﺎﻧﻲ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ﻫﻮ ﺗﻔﺎﻋﻞ ﻣﻬﻢ ﻓﻲ ﺟﺴﻤﻚ‪ .‬ﺇﻧﻪ ﻳﺤﺪﺙ ﻓﻲ‬ ‫ﺍﻷﻭﻋﻴﺔ ﺍﻟﺪﻣﻮﻳﺔ ﻓﻲ ﺭﺋﺘﻴﻚ‪ .‬ﻭﻛﻤﺎ ﻫﻮ ﻣﺒﻴﻦ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ 4-19‬ﻓﺈﻥ ﺛﺎﻧﻲ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ﺍﻟﺬﻱ‬ ‫ﻳﻨﺘﺞ ﻓﻲ ﺧﻼﻳﺎ ﺟﺴﻤﻚ ﻳﻨﺘﻘﻞ ﻓﻲ ﺩﻣﻚ ﻋﻠﻰ ﺷﻜﻞ ﺃﻳﻮﻧﺎﺕ ﺍﻟﺒﻴﻜﺮﺑﻮﻧﺎﺕ ‪ .HCO-3‬ﻭﻋﻨﺪﻣﺎ ﺗﻤﺮ‬ ‫ﻫﺬﻩ ﺍﻷﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻷﻭﻋﻴﺔ ﺍﻟﺪﻣﻮﻳﺔ ﻟﺮﺋﺘﻴﻚ ﺗﺘﺤﺪ ﻣﻊ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ H+‬ﻹﻧﺘﺎﺝ ﻏﺎﺯ ‪CO2‬‬ ‫ﺍﻟﺬﻱ ﻳﺨﺮﺝ ﻣﻊ ﻫﻮﺍﺀ ﺍﻟﺰﻓﻴﺮ‪ .‬ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﻳﺤﺪﺙ ﹰ‬ ‫ﺃﻳﻀﺎ ﻓﻲ ﺍﻟﻤﻨﺘﺠﺎﺕ ﺍﻟﺘﻲ ﻳﺪﺧﻞ ﻓﻲ ﺗﺮﻛﻴﺒﻬﺎ‬ ‫ﺻﻮﺩﺍ ﺍﻟﺨﺒﺰ ﺍﻟﻤﺤﺘﻮﻳﺔ ﻋﻠﻰ ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﺍﻟﺘﻲ ﺗﺠﻌﻞ ﺍﻷﺷﻴﺎﺀ ﺍﻟﻤﺨﺒﻮﺯﺓ‬ ‫ﺗﻨﺘﻔﺦ‪ ،‬ﻭﺗﺴﺘﺨﺪﻡ ﻣﻀﺎ ﹰﹼﺩﺍ ﻟﻠﺤﻤﻮﺿﺔ‪ ،‬ﻭﻓﻲ ﻃﻔﺎﻳﺎﺕ ﺍﻟﺤﺮﻳﻖ‪ ،‬ﻭﺻﻨﺎﻋﺔ ﻛﺜﻴﺮ ﻣﻦ ﺍﻟﻤﻨﺘﺠﺎﺕ‪.‬‬

‫ﺍﻟﻤﺨﺘـﺺ ﻓـﻲ ﺍﻟﻜﻴﻤﻴـﺎﺀ ﺍﻟﺤﻴﻮﻳﺔ‪:‬ﻫـﻮ‬ ‫ﻋﺎﻟـﻢ ﻳـﺪﺭﺱ ﺍﻟﻌﻤﻠﻴـﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻓـﻲ‬ ‫ﺍﻟﻤﺨﻠﻮﻗـﺎﺕ ﺍﻟﺤﻴﺔ‪ .‬ﻭﻗﺪ ﻳـﺪﺭﺱ ﻭﻇﺎﺋﻒ‬ ‫ﺟﺴـﻢ ﺍﻹﻧﺴـﺎﻥ‪ ،‬ﺃﻭ ﻳﺒﺤﺚ ﻛﻴﻒ ﻳﺆﺛﺮ ﻛﻞ‬

‫ﻣـﻦ ﺍﻟﻐـﺬﺍﺀ‪ ،‬ﻭﺍﻷﺩﻭﻳﺔ‪ ،‬ﻭﺍﻟﻤـﻮﺍﺩ ﺍﻷﺧﺮ￯‬ ‫ﻓـﻲ ﺍﻟﻤﺨﻠﻮﻗـﺎﺕ ﺍﻟﺤﻴـﺔ‪ .‬ﻟﻠﻤﺰﻳـﺪ ﻣـﻦ‬ ‫ﺍﻟﻤﻌﻠﻮﻣـﺎﺕ ﻋﻦ ﻣﻬـﻦ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺇﺭﺟﻊ ﺇﻟﻰ‬ ‫ﺍﻟﻤﻮﻗﻊ ﺍﻹﻟﻜﺘﺮﻭﻧﻲ‪.‬‬ ‫‪www.obeikaneducation.com‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪4-3‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺩﺍﺋﻤﺎ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ‬ ‫ﺍﻟﻤـﺎﺀ ﻫﻮ ﺍﻟﻤﺬﻳﺐ ﹰ‬ ‫ﺍﻟﻤﺎﺋﻴﺔ‪ ،‬ﻭﻟﻜﻦ ﻫﻨﺎﻙ ﺍﻟﻜﺜﻴﺮ ﻣﻦ ﺍﻟﻤﻮﺍﺩ‬ ‫ﺍﻟﻤﺬﺍﺑﺔ ﺍﻟﻤﺤﺘﻤﻠﺔ‪.‬‬ ‫ﺗﻜـﻮﻥ ﻛﺜﻴﺮ ﻣـﻦ ﺍﻟﻤﺮﻛﺒـﺎﺕ ﺍﻟﺠﺰﻳﺌﻴﺔ‬ ‫ﹼ‬ ‫ﺃﻳﻮﻧـﺎﺕ ﻋﻨﺪﻣـﺎ ﺗـﺬﻭﺏ ﻓـﻲ ﺍﻟﻤـﺎﺀ‪.‬‬ ‫ﻭﻋﻨﺪﻣـﺎ ﺗـﺬﻭﺏ ﺍﻟﻤـﻮﺍﺩ ﺍﻷﻳﻮﻧﻴﺔ ﻓﻲ‬ ‫ﺍﻟﻤﺎﺀ ﻓﺈﻥ ﺃﻳﻮﻧﺎﺗﻬﺎ ﺗﻨﻔﺼﻞ‪.‬‬ ‫ﻗﺪ ﺗﺘﻔﺎﻋﻞ ﺍﻷﻳﻮﻧﺎﺕ ﺑﻌﻀﻬﺎ ﻣﻊ ﺑﻌﺾ‬ ‫ﻋﻨﺪ ﺧﻠـﻂ ﻣﺤﺎﻟﻴﻞ ﺍﻟﻤـﻮﺍﺩ ﺍﻷﻳﻮﻧﻴﺔ‪.‬‬ ‫ﺃﻣـﺎ ﺟﺰﻳﺌـﺎﺕ ﺍﻟﻤﺬﻳـﺐ ﻓـﻼ ﺗﺘﻔﺎﻋﻞ‬ ‫ﻋﺎﺩ ﹰﺓ‪.‬‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﺗﺤﺪﺙ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ‬ ‫ﺍﻟﻤﺎﺋﻴـﺔ ﻫـﻲ ﺗﻔﺎﻋـﻼﺕ ﺍﻹﺣـﻼﻝ‬ ‫ﺍﻟﻤﺰﺩﻭﺝ‪.‬‬

‫‪ .50‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻋـﺪﺩ ﺛﻼﺛـﺔ ﺃﻧـﻮﺍﻉ ﻣﺄﻟﻮﻓـﺔ ﻣـﻦ ﻧﻮﺍﺗـﺞ ﺍﻟﺘﻔﺎﻋـﻼﺕ ﺍﻟﺘﻲ‬ ‫ﺗﺤﺪﺙ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ‪.‬‬ ‫‪ .51‬ﺻﻒ ﺍﻟﻤﺬﻳﺐ ﻭﺍﻟﻤﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ‪.‬‬ ‫ﻣﻴﺰ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻣﻦ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‪.‬‬ ‫‪ .52‬ﹼ‬ ‫‪ .53‬ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﻴﻦ ﺣﻤﺾ‬ ‫ﺍﻟﻜﺒﺮﻳﺘﻴﻚ ‪ H2SO4‬ﻭﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪.CaCO3‬‬ ‫)‪H2SO4(aq) + CaCO3(s)→H2O(l) + CO2(g) + CaSO4(aq‬‬

‫‪ .54‬ﺣ ﹼﻠﻞ‪ :‬ﺃﻛﻤﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻵﺗﻴﺔ ﻭﺯﻧﻬﺎ‪:‬‬

‫→ )‪HBr(aq)+KCN(aq‬‬

‫ﺍﻟﺘﺎﻟﻲ؟ﻓﺴﺮ ﺫﻟﻚ‪.‬‬ ‫‪ .55‬ﺗﻮ ﹼﻗﻊ ﻣﺎ ﻧﻮﻉ ﺍﻟﻨﺎﺗﺞ ﺍﻟﺬﻱ ﺳﻴﺘﻜﻮﻥ ﻋﻠﻰ ﺍﻷﺭﺟﺢ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﱢ‬ ‫→ )‪Ba(OH)2(aq)+2HCl(aq‬‬

‫‪ .56‬ﺻﻎ ﻣﻌﺎﺩﻻﺕ‪ :‬ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ ﻋﻨﺪﻣﺎ ﻳﺨﻠﻂ ﺣﻤﺾ ﺍﻟﻨﻴﺘﺮﻳﻚ ‪HNO3‬‬ ‫ﺑﻤﺤﻠـﻮﻝ ﻣﺎﺋﻲ ﻣـﻦ ﻛﺮﺑﻮﻧـﺎﺕ ﺍﻟﺒﻮﺗﺎﺳـﻴﻮﻡ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴـﺔ ‪،KHCO3‬‬

‫ﻭﻳﻨﺘﺞ ﻣﺤﻠﻮﻝ ﻧﺘﺮﺍﺕ ﺍﻟﺒﻮﺗﺎﺳـﻴﻮﻡ ‪ .KNO3‬ﺍﻛﺘـﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻭﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ‪.‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪33‬‬

‫ﻛﻴﻒ ﺗﻌﻤﻞ ا ﺷﻴﺎء؟‬

‫‪…ƒ«◊G ≥dCÉàdG‬‬ ‫ﻋﻨﺪﻣﺎ ﻳﺘﺠﻤﻊ ﺍﻟﲑﺍﻉ )ﺧﻨﺎﻓﺲ ﻣﻀﻴﺌﺔ( ﰲ ﺍﻟﻈﻼﻡ‪ ،‬ﻳﻌﻠﻦ ﺃﺣﺪ ﺍﻟﺬﻛﻮﺭ ﻋﻦ ﻭﺟﻮﺩﻩ ﺑﺈﺭﺳﺎﻝ ﺇﺷﺎﺭﺓ ﻣﻦ ﺍﻟﻀﻮﺀ ﺍﻷﺻﻔﺮ‬ ‫ﺍﳌﺨﴬ‪ ،‬ﻓﺘﺠﻴﺐ ﺃﻧﺜﻰ ﻗﺮﻳﺒﺔ ﻣﻦ ﺍﻷﺭﺽ ﻧﺪﺍﺀﻩ‪ ،‬ﻓﻴﻬﺒﻂ ﰲ ﺍﲡﺎﻫﻬﺎ‪ .‬ﻭﻗﺪ ﻳﻨﺘﺞ ﻋﻦ ﺫﻟﻚ ﺗﺰﺍﻭﺝ ﻧﺎﺟﺢ‪ ،‬ﺃﻭ ﻗﺪ ﹸﻳﻠﺘﻬﻢ ﺑﴩﺍﻫﺔ‬ ‫ﺇﺫﺍ ﺧﺪﻋﺘﻪ ﺃﻧﺜﻰ ﻣﻦ ﻧﻮﻉ ﺁﺧﺮ ﻣﻦ ﺍﻟﲑﺍﻉ‪ .‬ﺇﻥ ﺇﻧﺘﺎﺝ ﺍﻟﲑﺍﻋﺔ ﻟﻠﻀﻮﺀ ﻫﻮ ﻧﺘﻴﺠﺔ ﻋﻤﻠﻴﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺗﺴﻤﻰ ﺍﻟﺘﺄﻟﻖ)ﺍﻟﺘﻸﻟﺆ(‬ ‫ﺍﳊﻴﻮﻱ‪ .‬ﻭﻫﻲ ﺍﺳﱰﺍﺗﻴﺠﻴﺔ ﻳﺴﺘﺨﺪﻣﻬﺎ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﳌﺨﻠﻮﻗﺎﺕ ﺍﳊﻴﺔ ﰲ ﺑﻴﺌﺎﺕ ﻛﺜﲑﺓ ﳐﺘﻠﻔﺔ‪ .‬ﻓﻜﻴﻒ ﺗﻌﻤﻞ؟‬ ‫ﺍﳋﻨﺎﻓﺲ ﺍﳌﻀﻴﺌﺔ ﻟﻴﺲ ﺫﺑﺎ ﹰﺑﺎ ﻋﲆ ﺍﻹﻃﻼﻕ‪ ،‬ﻭﻟﻜﻨﻬﺎ ‪1‬‬

‫ﳎﻤﻮﻋﺔ ﻣﻦ ﺍﳋﻨﺎﻓﺲ ﺍﻟﺘﻲ ﺗﺮﺳﻞ ﻭﻣﻀﺎﲥﺎ ﻟﻠﺘﺰﺍﻭﺝ‪،‬‬ ‫ﻛﲈ ﺃﳖﺎ ﺗﺴﺘﺨﺪﻡ ﺿﻮﺀﻫﺎ ﳋﺪﺍﻉ ﻓﺮﻳﺴﺘﻬﺎ‪ .‬ﻭﻳﻨﺒﻌﺚ‬ ‫ﺍﻟﻀﻮﺀ ﺍﻷﺻﻔﺮ ﺍﳌﺨﴬ ﻣﻦ ﺧﻼﻳﺎ ﰲ ﺟﺬﻋﻬﺎ ﺍﻷﺳﻔﻞ‪،‬‬ ‫ﻭﺗﺒﻠﻎ ﻃﻮﻝ ﻣﻮﺟﺘﻪ ﻣﻦ ‪ 510‬ﺇﱃ ‪.670 nm‬‬

‫ﺍﻛﺘﺸﺎﻓﺎﺕ ﻣﻀﻴﺌﺔ ﺃﺩ￯ ﺍﻟﺒﺤﺚ ﰲ ﳎﺎﻝ ﺍﻟﺘﺄﻟﻖ ﺍﳊﻴﻮﻱ ﺇﱃ ﺍﻛﺘﺸﺎﻑ ‪2‬‬

‫ﺍﻟﱪﻭﺗﲔ ﺍﳊﻴﻮﻱ ﺍﻷﺧﴬ ﺍﳌﺸﻊ‪ ،‬ﺍﻟﺬﻱ ﻳﻮﺟﺪ ﰲ ﺑﻌﺾ ﺃﻧﻮﺍﻉ‬ ‫ﻗﻨﺎﺩﻳﻞ ﺍﻟﺒﺤﺮ‪.‬‬ ‫ﻭﻳﺸﻊ ﻫﺬﺍ ﺍﻟﱪﻭﺗﲔ ﺿﻮ ﹰﺀﺍ ﺃﺧﴬ ﻋﻨﺪ ﺗﻌﺮﺿﻪ ﻟﻸﺷﻌﺔ ﻓﻮﻕ‬ ‫ﺍﻟﺒﻨﻔﺴﺠﻴﺔ‪ .‬ﻭﻗﺪ ﻗﺎﻡ ﺍﻟﻌﻠﲈﺀ ﺑﺈﺩﺧﺎﻝ ﺍﻟﱪﻭﺗﲔ ﺍﳌﺸﻊ ﰲ ﳐﻠﻮﻗﺎﺕ‬ ‫ﳐﺘﻠﻔﺔ‪ ،‬ﻛﺎﳉﺮﺫﺍﻥ‪ ،‬ﻷﻏﺮﺍﺽ ﺍﻟﺒﺤﺚ ﺍﻟﻌﻠﻤﻲ ﰲ ﳎﺎﻻﺕ ﺍﻟﴪﻃﺎﻥ‪،‬‬ ‫ﻭﻧﻈﺮﺍ ﺇﱃ ﺃﳘﻴﺔ ﻫﺬﺍ ﺍﻻﻛﺘﺸﺎﻑ ﻓﻘﺪ‬ ‫ﻭﺍﳌﻼﺭﻳﺎ‪ ،‬ﻭﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳋﻠﻮﻳﺔ‪.‬‬ ‫ﹰ‬ ‫ﻣﻨﺢ ﻣﻜﺘﺸﻔﻮ ﺍﻟﱪﻭﺗﲔ ﺍﳌﺸﻊ ﺟﺎﺋﺰﺓ ﻧﻮﺑﻞ ﰲ ﺍﻟﻜﻴﻤﻴﺎﺀ‪.‬‬

‫ﺍﻟﺘﺄﻟﻖ ﺍﳊﻴﻮﻱ ﻳﻨﺘﺞ ﻭﻣﻴﺾ ﺍﻟﲑﺍﻉ ﻋﻦ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪3 .‬‬

‫ﻭﺍﳌﺘﻔﺎﻋﻼﺕ ﻫﻲ ﺍﻷﻛﺴﺠﲔ‪ ،‬ﻭﺍﻟﻠﻮﺳﻔﺮﻳﻦ )ﻣﺎﺩﺓ‬ ‫ﻭﻳﴪﻉ‬ ‫ﻣﺸﻌﺔ ﻟﻠﻀﻮﺀ ﺗﻮﺟﺪ ﰲ ﺑﻌﺾ ﺍﳌﺨﻠﻮﻗﺎﺕ(‪.‬‬ ‫ﱢ‬ ‫ﺇﻧﺰﻳﻢ ﻳﺴﻤﻰ ﺍﻟﻠﻮﺳﻔﺮﻳﺰ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﺬﻱ ﻳﺆﺩﻱ ﺇﱃ ﺇﻧﺘﺎﺝ‬ ‫ﺍﻷﻭﻛﺴﻴﻠﻮﺳﻔﺮﻳﻦ ﻭﻃﺎﻗﺔ ﻋﲆ ﺷﻜﻞ ﺿﻮﺀ‪.‬‬

‫‪‘ áHÉàµdG‬‬

‫اﻟﻜﻴﻤﻴﺎء‬

‫ﺃﻧﻮﺍﻋﺎ ﻣﺨﺘﻠﻔﺔ ﻣﻦ ﺍﻟﻤﺨﻠﻮﻗﺎﺕ ﺍﻟﺤﻴﺔ ﺗﺴﺘﺨﺪﻡ ﺍﻟﺘﺄﻟﻖ ﺍﻟﺤﻴﻮﻱ‪ ،‬ﻭﺍﻋﻤﻞ ﻛﺘﻴ ﹰﺒﺎ‬ ‫ﺍﺑﺤﺚ‪ :‬ﺣﺪﱢ ﺩ ﹰ‬ ‫ﹰ‬ ‫ﻳﻮﺿﺢ ﻟﻤﺎﺫﺍ ﻳﻜﻮﻥ ﺍﻟﺘﺄﻟﻖ ﺍﻟﺤﻴﻮﻱ ﻓﻌﺎﻻ ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺨﻠﻮﻗﺎﺕ؟ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺍﻟﻤﺰﻳﺪ ﻣﻦ‬ ‫ﺍﻟﻤﻌﻠﻮﻣﺎﺕ‪ ،‬ﺇﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ ﺍﻹﻟﻜﺘﺮﻭﻧﻲ ‪www.obeikaneducation.com‬‬

‫‪34‬‬

‫‪π°üØdG‬‬

‫‪π°üØdG á©LGôe π«dO‬‬ ‫الفكرة‬

‫ﲢﻮﻝ ﻣﻼﻳﲔ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻣﻦ ﺣﻮﻟﻚ ﻭﰲ ﺩﺍﺧﻞ ﺟﺴﻤﻚ ﺍﳌﻮﺍﺩ ﺍﳌﺘﻔﺎﻋﻠﺔ ﺇﱃ ﻣﻮﺍﺩ‬ ‫ﻧﺎﲡﺔ‪ ،‬ﳑﺎ ﻳﺆﺩﻱ ﺇﱃ ﺇﻃﻼﻕ ﺍﻟﻄﺎﻗﺔ ﺃﻭ ﺍﻣﺘﺼﺎﺻﻬﺎ‪.‬‬

‫العامة‬

‫‪ 4-1‬اﻟﺘﻔﺎﻋﻼت و اﻟﻤﻌﺎدﻻت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﲤﺜﻞ ﺍﻟﺘﻔﺎﻋﻼﺕ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺑﻤﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﻮﺯﻭﻧﺔ‪.‬‬ ‫• ﺑﻌﺾ ﺍﻟﺘﻐﲑﺍﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﺃﺩﻟﺔ ﺗﺸﲑ ﺇﱃ ﺣﺪﻭﺙ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫• ﺗﻮﻓﺮ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻠﻔﻈﻴﺔ ﻭﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻣﻌﻠﻮﻣﺎﺕ ﻣﻬﻤﺔ ﻋﻦ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫اﻟﻤﻔﺮدات‬ ‫• ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫• ﲢﺪﺩ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺃﻧﻮﺍﻉ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﻮﺍﺗﺞ ﰲ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻭﻛﻤﻴﺎﲥﺎ‬ ‫• ﺍﳌﺘﻔﺎﻋﻼﺕ‬ ‫ﺍﻟﻨﺴﺒﻴﺔ‪.‬‬ ‫• ﺍﻟﻨﻮﺍﺗﺞ‬ ‫• ﻳﺘﻀﻤﻦ ﻭﺯﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺗﻌﺪﻳﻞ ﺍﳌﻌﺎﻣﻼﺕ ﺣﺘﻰ ﻳﺘﺴﺎﻭ￯ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ﻛﻞ ﻣﻦ ﻃﺮﰲ‬ ‫• ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﳌﻮﺯﻭﻧﺔ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬ ‫• ﺍﳌﻌﺎﻣﻞ‬ ‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫‪ 4-2‬ﺗﺼﻨﻴﻒ اﻟﻤﻌﺎدﻻت اﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻫﻨﺎﻙ ﺃﺭﺑﻌﺔ ﺃﻧﻮﺍﻉ‬ ‫ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ :‬ﺍﻟﺘﻜﻮﻳﻦ‪،‬‬ ‫ﻭﺍﻻﺣﱰﺍﻕ‪ ،‬ﻭﺍﻟﺘﻔﻜﻚ‪ ،‬ﻭﺍﻹﺣﻼﻝ‪.‬‬ ‫اﻟﻤﻔﺮدات‬

‫•‬ ‫•‬ ‫•‬ ‫•‬ ‫•‬ ‫•‬

‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻜﻮﻳﻦ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻻﺣﱰﺍﻕ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻔﻜﻚ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ‬ ‫ﺍﻟﺮﺍﺳﺐ‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫ﺴﻬﻞ ﺗﺼﻨﻴﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻓﻬﻤﻬﺎ ﻭﺗﺬﻛﺮﻫﺎ‪ ،‬ﻭﺗﻌﺮﻓﻬﺎ‪.‬‬ ‫• ﹸﻳ ﹼ‬ ‫• ﺗﺴﺘﺨﺪﻡ ﺳﻠﺴﻠﺘﻲ ﺍﻟﻨﺸﺎﻁ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﻠﻔﻠﺰﺍﺕ ﻭﺍﳍﺎﻟﻮﺟﻴﻨﺎﺕ ﰲ ﺗﻮﻗﻊ ﺣﺪﻭﺙ‬ ‫ﺗﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴﻴﻂ‪.‬‬

‫‪ 4-3‬اﻟﺘﻔﺎﻋﻼت ﻓﻲ اﻟﻤﺤﺎﻟﻴﻞ اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﲢﺪﺙ ﺗﻔﺎﻋﻼﺕ‬ ‫ﺍﻹﺣﻼﻝ ﺍﳌﺰﺩﻭﺝ ﺑﲔ ﺍﳌﻮﺍﺩ ﰲ ﺍﳌﺤﺎﻟﻴﻞ‬ ‫ﺍﳌﺎﺋﻴﺔ‪ ،‬ﻭﺗﺆﺩﻱ ﺇﱃ ﺇﻧﺘﺎﺝ ﺭﻭﺍﺳﺐ‪ ،‬ﺃﻭ‬ ‫ﻣﺎﺀ‪ ،‬ﺃﻭ ﻏﺎﺯﺍﺕ‪.‬‬ ‫اﻟﻤﻔﺮدات‬

‫•‬ ‫•‬ ‫•‬ ‫•‬ ‫•‬ ‫•‬

‫ﺍﳌﺤﻠﻮﻝ ﺍﳌﺎﺋﻲ‬ ‫ﺍﳌﺬﺍﺏ‬ ‫ﺍﳌﺬﻳﺐ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‬ ‫ﺍﻷﻳﻮﻥ ﺍﳌﺘﻔﺮﺝ‬ ‫ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫•‬ ‫•‬ ‫•‬ ‫•‬

‫ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ‪ ،‬ﺍﳌﺬﻳﺐ ﹰ‬ ‫ﺩﺍﺋﲈ ﻫﻮ ﺍﳌﺎﺀ‪ ،‬ﻭﻟﻜﻦ ﻫﻨﺎﻙ ﺃﻛﺜﺮ ﻣﻦ ﺍﺣﺘﲈﻝ ﳌﺎﺩﺓ ﻣﺬﺍﺑﺔ‪.‬‬ ‫ﺍﻟﻜﺜﲑ ﻣﻦ ﺍﳌﺮﻛﺒﺎﺕ ﺍﳉﺰﻳﺌﻴﺔ ﺗﻜﻮﻥ ﺃﻳﻮﻧﺎﺕ ﻋﻨﺪﻣﺎ ﺗﺬﻭﺏ ﰲ ﺍﳌﺎﺀ‪ .‬ﻋﻨﺪﻣﺎ ﺗﺬﻭﺏ ﺑﻌﺾ‬ ‫ﺍﳌﺮﻛﺒﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ ﰲ ﺍﳌﺎﺀ ﺗﻨﻔﺼﻞ ﺃﻳﻮﻧﺎﲥﺎ‪.‬‬ ‫ﻋﻨﺪ ﻣﺰﺝ ﳏﻠﻮﻟﲔ ﳛﺘﻮﻳﺎﻥ ﻋﲆ ﺃﻳﻮﻧﺎﺕ ﺫﺍﺋﺒﺔ‪ ،‬ﻗﺪ ﺗﺘﻔﺎﻋﻞ ﺍﻷﻳﻮﻧﺎﺕ ﺑﻌﻀﻬﺎ ﻣﻊ ﺑﻌﺾ‪،‬‬ ‫ﺃﻣﺎ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺬﻳﺐ ﻻ ﺗﺘﻔﺎﻋﻞ ﻋﺎﺩﺓ‪.‬‬ ‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻲ ﲢﺪﺙ ﰲ ﺍﳌﺤﺎﻟﻴﻞ ﺍﳌﺎﺋﻴﺔ ﻫﻲ ﺗﻔﺎﻋﻼﺕ ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‪.‬‬

‫‪35‬‬

‫‪π°üØdG‬‬

‫‪á©LGôe‬‬

‫‪4-1‬‬ ‫‪º«gÉتdG ¿É≤JEG‬‬

‫ﻋﺮﻑ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫‪ .57‬ﹼ‬

‫ﻣﻴﺰ ﺑﲔ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻭﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫‪ .58‬ﹼ‬ ‫‪ .59‬ﻭﺿﺢ ﺍﻟﻔﺮﻕ ﺑﲔ ﺍﳌﺘﻔﺎﻋﻼﺕ ﻭﺍﻟﻨﻮﺍﺗﺞ‪.‬‬

‫‪ .60‬ﻫﻞ ﻳﺸﲑ ﲢﻮﻝ ﻣﺎﺩﺓ ﺇﱃ ﻣﺎﺩﺓ ﺟﺪﻳﺪﺓ ﹰ‬ ‫ﺩﺍﺋﲈ ﺇﱃ ﺣﺪﻭﺙ‬ ‫ﻓﴪ ﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ؟ ﱢ‬

‫‪ .61‬ﺣـﺪﱢ ﺩ ﺍﳌﺘﻔﺎﻋـﻼﺕ ﰲ ﺍﻟﺘﻔﺎﻋـﻞ ﺍﻵﰐ‪ :‬ﻋﻨـﺪ ﺇﺿﺎﻓـﺔ‬ ‫ﺍﻟﺒﻮﺗﺎﺳـﻴﻮﻡ ﺇﱃ ﳏﻠـﻮﻝ ﻧـﱰﺍﺕ ﺍﳋﺎﺭﺻـﲔ‪ ،‬ﻳﺘﻜـﻮﻥ‬ ‫ﺍﳋﺎﺭﺻﲔ ﻭﳏﻠﻮﻝ ﻧﱰﺍﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‪.‬‬ ‫‪ .62‬ﺯﻥ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻵﺗﻴﺔ‪:‬‬

‫)‪H2S(g) + O2(g) → SO2(g)+H2O(g‬‬

‫‪ .63‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻟﻔﻈﻴﺔ ﻟﻠﻤﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪Cu(s) +O2(g)→CuO(s) .a‬‬

‫‪K(s)+ H2 O(l) → KOH(aq) + H2(g) .b‬‬

‫‪CaCl2(aq)+Na2SO4(aq) → CaSO4(s)+NaCl(aq) .c‬‬

‫‪ .64‬ﺯﻥ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺘﲔ ﺍﻵﺗﻴﺘﲔ‪:‬‬

‫‪(NH4)2Cr2O7(s) → Cr2O3(s)+N2(g) +H2(g) .a‬‬

‫‪CO2(g) + H2O(l) → C6H12O6(s)+O2(g) .b‬‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬

‫‪ .d‬ﻧﱰﺍﺕ ﺍﻟﻔﻀﺔ )‪ + (aq‬ﻛﱪﻳﺘﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪(aq‬‬

‫ﻛﱪﻳﺘﻴﺪ ﺍﻟﻔﻀﺔ )‪ + (s‬ﻧﱰﺍﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪(aq‬‬

‫‪ .67‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﲔ ﺍﻟﻠﻴﺜﻴﻮﻡ ﺍﻟﺼﻠﺐ‬ ‫ﻭﻏﺎﺯ ﺍﻟﻜﻠﻮﺭ ﻹﻧﺘﺎﺝ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻠﻴﺜﻴﻮﻡ ﺍﻟﺼﻠﺐ‪.‬‬ ‫‪ .68‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ ﺛﻢ ﺯﳖﺎ‪:‬‬ ‫‪ .a‬ﻣﺎﺀ )‪ + (l‬ﺛﺎﻟﺚ ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ)‪(g‬‬ ‫ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ )‪(aq‬‬

‫‪.b‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﳊﺪﻳﺪ ‪ + (s) III‬ﻣﺎﻏﻨﺴﻴﻮﻡ )‪(s‬‬

‫‪.c‬‬

‫ﺃﻛﺴﺠﲔ)‪ + (g‬ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﻴﻜﻞ ‪(s) II‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ )‪ + (aq‬ﺣﺪﻳﺪ )‪(s‬‬

‫ﺃﻛﺴﻴﺪ ﺍﻟﻨﻴﻜﻞ ‪ + (s) II‬ﲬﺎﳼ ﺃﻛﺴﻴﺪ ﺛﻨﺎﺋﻲ ﺍﻟﻜﻠﻮﺭ‪.‬‬

‫‪ .69‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪ .a‬ﻋﻨﺪ ﺣﺮﻕ ﻏﺎﺯ ﺍﻟﺒﻴﻮﺗﺎﻥ ‪ C4H10‬ﰲ ﺍﳍﻮﺍﺀ ﻳﻨﺘﺞ ﻣﺎﺀ‬ ‫ﻭﻏﺎﺯ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‪.‬‬

‫‪ .b‬ﻳﺘﻔﺎﻋﻞ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ﺍﻟﺼﻠﺐ ﻣﻊ ﻏﺎﺯ ﺍﻟﻨﻴﱰﻭﺟﲔ‬ ‫ﻹﻧﺘﺎﺝ ﻧﻴﱰﻳﺪ ﺍﳌﺎﻏﻨﺴﻴﻮﻡ ﺍﻟﺼﻠﺐ‪.‬‬

‫‪ .c‬ﻋﻨﺪ ﺗﺴﺨﲔ ﻏﺎﺯ ﺛﺎﲏ ﻓﻠﻮﺭﻳﺪ ﺍﻷﻛﺴﺠﲔ ‪ OF2‬ﻳﻨﺘﺞ‬ ‫ﻏﺎﺯ ﺍﻷﻛﺴﺠﲔ ﻭﻏﺎﺯ ﺍﻟﻔﻠﻮﺭ‪.‬‬

‫‪4-2‬‬ ‫‪º«gÉتdG ¿É≤JEG‬‬

‫‪ .65‬ﻳﺘﺤﻠﻞ ﻳﻮﺩﻳﺪ ﺍﳍﻴﺪﺭﻭﺟﲔ ﺇﱃ ﻏﺎﺯ ﺍﳍﻴﺪﺭﻭﺟﲔ ﻭﻏﺎﺯ ﺍﻟﻴﻮﺩ ﰲ ‪ .70‬ﺍﺫﻛﺮ ﺃﻧﻮﺍﻉ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻷﺭﺑﻌﺔ‪ ،‬ﻭﺃﻋﻂ ﻣﺜﺎ ﹰ‬ ‫ﻻ‬ ‫ﺗﻔﺎﻋﻞ ﺍﻟﺘﻔﻜﻚ‪ .‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﺗﺒﲔ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫ﻭﺍﺣﺪﹰ ﺍ ﻋﻠﻰ ﻛﻞ ﻣﻨﻬﺎ‪.‬‬ ‫‪ .66‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪ .71‬ﻣﺎ ﻧﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ ﻣﺎﺩﺗﻴﻦ ﻧﺎﺗﺠﻬﻤﺎ ﻣﺮﻛﺐ ﻭﺍﺣﺪ؟‬

‫‪‬‬ ‫‪.a‬‬ ‫ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪ .72 (s‬ﺃﻱ ﻓﻠـﺰ ﺳـﻴﺤﻞ ﻣﺤـﻞ ﺍﻟﻔﻠـﺰ ﺍﻵﺧـﺮ ﻓـﻲ ﺗﻔﺎﻋـﻼﺕ‬ ‫ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ)‪ + (g‬ﺃﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ)‪(s‬‬ ‫ﺍﻹﺣـﻼﻝ ﻓـﻲ ﻛﻞ ﻣـﻦ ﺍﻷﺯﻭﺍﺝ ﺍﻵﺗﻴـﺔ )ﻣﺴـﺘﺨﺪ ﹰﻣﺎ‬ ‫‪ .b‬ﻳﻮﺩﻳﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ )‪→ (s‬ﺃﻟﻮﻣﻨﻴﻮﻡ)‪ + (s‬ﻳﻮﺩ )‪(s‬‬ ‫ﺳﻠﺴﻠﺔ ﺍﻟﻨﺸﺎﻁ(‪:‬‬ ‫→ ﺃﻛﺴﻴﺪ ﺍﳊﺪﻳﺪ ‪ +(s) II‬ﺃﻛﺴﺠﲔ)‪(g‬‬ ‫‪.c‬‬ ‫‪ .c‬ﺍﻟﺮﺻﺎﺹ ﻭﺍﻟﻔﻀﺔ‬ ‫‪ .a‬ﺍﻟﻘﺼﺪﻳﺮ ﻭﺍﻟﺼﻮﺩﻳﻮﻡ‬

‫‪36‬‬

‫ﺃﻛﺴﻴﺪ ﺍﳊﺪﻳﺪ‪(s) III‬‬

‫‪ .b‬ﺍﻟﻔﻠﻮﺭ ﻭﺍﻟﻴﻮﺩ‬

‫‪ .d‬ﺍﻟﻨﺤـﺎﺱ ﻭﺍﻟﻨﻴﻜﻞ‬

‫‪π°üØdG á©LGôe‬‬ ‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ .73‬ﺻﻨﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻮﺍﺭﺩﺓ ﻓﻲ ﺳﺆﺍﻝ ‪. 68‬‬ ‫‪ .74‬ﺻﻨﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻮﺍﺭﺩﺓ ﻓﻲ ﺳﺆﺍﻝ ‪.70‬‬

‫‪ .75‬ﺍﻛﺘـﺐ ﻣﻌﺎﺩﻟـﺔ ﻛﻴﻤﻴﺎﺋﻴـﺔ ﻣﻮﺯﻭﻧـﺔ ﻟﺘﻔﺎﻋـﻞ ﺍﺣﺘـﺮﺍﻕ‬ ‫ﺍﻟﻤﻴﺜﺎﻧﻮﻝ ﺍﻟﺴﺎﺋﻞ ‪.CH3OH‬‬

‫‪4-3‬‬ ‫‪º«gÉتdG ¿É≤JEG‬‬ ‫‪ .80‬ﺃﻛﻤﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻠﻔﻈﻴﺔ ﺍﻵﺗﻴﺔ‪:‬‬

‫ﻣﺬﺍﺏ ‪ +‬ﻣﺬﻳﺐ‬

‫‪ .76‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﻜﻞ ﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﻜﻮﻳﻦ ﺍﻵﺗﻲ‪ .81 :‬ﻣﺎ ﺃﻧﻮﺍﻉ ﺍﻟﻨﻮﺍﺗﺞ ﺍﻟﺸـﺎﺋﻌﺔ ﻋﻨﺪﻣﺎ ﺗﺤﺪﺙ ﺍﻟﺘﻔﺎﻋﻼﺕ ﻓﻲ‬ ‫ﻣﺤﺎﻟﻴﻞ ﻣﺎﺋﻴﺔ؟‬ ‫ﺑﻮﺭﻭﻥ ‪ +‬ﻓﻠﻮﺭ‬ ‫‪.a‬‬ ‫ﺟﺮﻣﺎﻧﻴﻮﻡ ‪ +‬ﻛﺒﺮﻳﺖ ‪ .82‬ﻗﺎﺭﻥ ﺑﻴﻦ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻷﻳﻮﻧﻴﺔ‪.‬‬ ‫‪.b‬‬ ‫ﻛﺎﻟﺴﻴﻮﻡ ‪ +‬ﻧﻴﺘﺮﻭﺟﻴﻦ ‪ .83‬ﻣـﺎ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺍﻷﻳﻮﻧﻴـﺔ ﺍﻟﻨﻬﺎﺋﻴـﺔ؟ ﻭﻓﻴـﻢ ﺗﺨﺘﻠـﻒ ﻋـﻦ‬ ‫‪.c‬‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ؟‬ ‫‪ .77‬ﺍﻻﺣﺘـﺮﺍﻕ ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟـﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺭﻣﺰﻳﺔ ﻻﺣﺘﺮﺍﻕ ﻛﻞ‬ ‫ﻣﻦ ﺍﻟﻤﻮﺍﺩ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪.84‬ﻋﺮﻑ ﺍﻷﻳﻮﻥ ﺍﻟﻤﺘﻔﺮﺝ‪.‬‬ ‫‪ .a‬ﺍﻟﺒﺎﺭﻳﻮﻡ ﺍﻟﺼﻠﺐ‬ ‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ .b‬ﺍﻟﺒﻮﺭﻭﻥ ﺍﻟﺼﻠﺐ‬

‫‪ .c‬ﺍﻷﺳﻴﺘﻮﻥ ﺍﻟﺴﺎﺋﻞ ‪C3H6O‬‬ ‫‪ .d‬ﺍﻷﻭﻛﺘﺎﻥ ﺍﻟﺴﺎﺋﻞ ‪C8H18‬‬

‫‪ .78‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺘﻔﻜﻚ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪.a‬‬

‫‪.b‬‬

‫‪.c‬‬

‫ﺑﺮﻭﻣﻴﺪ ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ‬

‫ﺃﻛﺴﻴﺪ ﺍﻟﻜﻮﺑﻠﺖ‪II‬‬

‫ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺒﺎﺭﻳﻮﻡ‬

‫‪ .79‬ﺍﻛﺘـﺐ ﻣﻌـﺎﺩﻻﺕ ﻛﻴﻤﻴﺎﺋﻴـﺔ ﻟﺘﻔﺎﻋﻼﺕ ﺍﻹﺣﻼﻝ ﺍﻟﺒﺴـﻴﻂ‬ ‫ﺍﻵﺗﻴﺔ ﺍﻟﺘـﻲ ﺗﺤﺪﺙ ﻓﻲ ﺍﻟﻤـﺎﺀ‪) :‬ﻭﺇﺫﺍ ﻟﻢ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ‬ ‫ﻓﺎﻛﺘﺐ ﻻ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ )‪ (NR‬ﻓﻲ ﻣﻜﺎﻥ ﺍﻟﻨﻮﺍﺗﺞ(‪.‬‬ ‫‪.a‬‬

‫‪.b‬‬ ‫‪.c‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ ‪ +‬ﻧﻴﻜﻞ‬

‫ﺑﺮﻭﻣﻴﺪ ﺍﻟﻨﺤﺎﺱ‪ + II‬ﻛﺎﻟﺴﻴﻮﻡ‬ ‫ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ‪ +‬ﻣﺎﻏﻨﺴﻴﻮﻡ‬

‫‪ .85‬ﺃﻛﻤﻞ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪.a‬‬

‫→ )‪Na (s)+ H2O(l‬‬

‫‪.b‬‬

‫→ )‪K(s) + H2O (l‬‬

‫‪.c‬‬

‫→ )‪CuCl2(s)+ Na2SO4(aq‬‬

‫‪ .86‬ﺍﻛﺘـﺐ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻷﻳﻮﻧﻴـﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻭﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‬ ‫ﻟﻜﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪.a‬‬ ‫‪.b‬‬ ‫‪.c‬‬ ‫‪.d‬‬

‫)‪H3PO4(aq) + 3KOH(aq) → 3H2O(p) + K3PO4(aq‬‬ ‫)‪HCl(aq)+NH4OH(aq) → H2O(l)+NH4Cl(aq‬‬ ‫)‪HCN(g)+KNO3(aq‬‬

‫→ )‪HNO3(aq)+KCN(aq‬‬

‫)‪2HClO(aq)+Ca(OH)2(aq) → 2H2O(l)+Ca(ClO)2(aq‬‬

‫‪37‬‬

‫‪π°üØdG á©LGôe‬‬ ‫‪áeÉY á©LGôe‬‬

‫‪ .87‬ﺗﻮﻗﻊ ﻫﻞ ﻛﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ ﺳﻴﺤﺪﺙ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ‬ ‫ﺍﻟﻤﺎﺋﻴـﺔ‪ .‬ﻭﺇﺫﺍ ﺗﻮﻗﻌـﺖ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻻ ﻳﺤـﺪﺙ ﻓﺎﻛﺘﺐ ﻻ‬ ‫ﻳﺤـﺪﺙ ﺗﻔﺎﻋـﻞ )‪) :(NR‬ﻣﻼﺣﻈـﺔ‪ :‬ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﺒﺎﺭﻳﻮﻡ‬ ‫ﻭﺑﺮﻭﻣﻴﺪ ﺍﻟﻔﻀﺔ ﻳﺘﺮﺳﺒﺎﻥ ﻓﻲ ﺍﻟﻤﺤﺎﻟﻴﻞ ﺍﻟﻤﺎﺋﻴﺔ(‪.‬‬ ‫‪.a‬‬

‫‪.b‬‬ ‫‪.c‬‬

‫→ )‪NaOH +(NH4) SO4(aq‬‬

‫→ )‪Cu SO4 +Ba(NO3)2(aq‬‬ ‫→ )‪Mg Br2 +AgNO3(aq‬‬

‫‪ .88‬ﺗﻜـﻮﻥ ﺭﺍﺳـﺐ‪ :‬ﺇﺿﺎﻓﺔ ﺣﻤـﺾ ﺍﻟﻬﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﺇﻟﻰ‬ ‫ﻛﺄﺳـﻴﻦ‪ ،‬ﺇﺣﺪﺍﻫﻤﺎ ﻓﻴﻬـﺎ ﻣﺤﻠﻮﻝ ﻛﻠﻮﺭﻳـﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪،‬‬ ‫ﻭﻓﻲ ﺍﻷﺧﺮ￯ ﻣﺤﻠﻮﻝ ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ﻳﺆﺩﻱ ﺇﻟﻰ ﺗﺮﺳﺐ‬ ‫ﻣﺎﺩﺓ ﺑﻴﻀﺎﺀ ﻓﻲ ﺇﺣﺪ￯ ﺍﻟﻜﺄﺳﻴﻦ‪.‬‬ ‫‪ .a‬ﺃﻱ ﺍﻟﻜﺄﺳﻴﻦ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺭﺍﺳﺐ؟‬ ‫‪ .b‬ﻣﺎ ﺍﻟﺮﺍﺳﺐ؟‬ ‫‪ .c‬ﺍﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺗﻮﺿﺢ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫‪ .d‬ﺻﻨﻒ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫‪ .89‬ﻣﻴـﺰ‪ :‬ﺑﻴﻦ ﻣﺮﻛـﺐ ﺃﻳﻮﻧﻲ ﻭﻣﺮﻛﺐ ﺟﺰﻳﺌـﻲ ﻣﺬﺍﺑﻴﻦ ﻓﻲ‬ ‫ﺍﻟﻤﺎﺀ‪.‬ﻭﻫـﻞ ﺗﺘﺄﻳﻦ ﺍﻟﻤﻮﺍﺩ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺟﻤﻴﻌﻬﺎ ﻋﻨﺪ ﺇﺫﺍﺑﺘﻬﺎ‬ ‫ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫ﻓﻲ ﺍﻟﻤﺎﺀ؟ ﱢ‬

‫ﺗﺘﺮﺳﺐ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ‪.‬‬

‫→ )‪KNO3(aq) + CsCl(aq‬‬ ‫→)‪Li3PO4 (aq) +MgSO4(aq‬‬ ‫→)‪K2S (aq) + HCl (aq‬‬

‫ﺗﻘﻮﻳﻢ ﺇﺿاﻓﻲ‬ ‫‪‘ áHÉàµdG‬‬

‫‪ .93‬ﻛﻴﻤﻴﺎﺀ ﺍﻟﻤﻄﺒﺦ‪ :‬ﺍﻋﻤﻞ ﻣﻠﺼ ﹰﻘﺎ ﻳﺼﻒ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﺍﻟﺘﻲ ﺗﺤﺪﺙ ﻓﻲ ﺍﻟﻤﻄﺒﺦ‪.‬‬ ‫‪ .94‬ﻭﺯﻥ ﺍﻟﻤﻌـﺎﺩﻻﺕ‪ :‬ﺍﻋﻤـﻞ ﻟﻮﺣﺔ ﺗﺼﻒ ﻓﻴﻬـﺎ ﺧﻄﻮﺍﺕ ﻭﺯﻥ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ؟‬

‫‪äGóæà°ùªdG á∏Ä°SCG‬‬

‫ﺍﻟﺬﻭﺑﺎﻧﻴﺔ‪ .‬ﻳﺴﺘﺨﺪﻡ ﺍﻟﻌﻠﻤﺎﺀ ﺟﺪﻭ ﹰ‬ ‫ﻻ ﻟﻘﻮﺍﻋﺪ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ﻟﺘﺤﺪﻳﺪ ﻣﺎ‬ ‫ﺇﺫﺍ ﻛﺎﻥ ﺳﻴﺘﻜﻮﻥ ﺭﺍﺳﺐ ﻓﻲ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﻳﺒﻴﻦ ﺍﻟﺠﺪﻭﻝ ‪ 4-5‬ﻗﻮﺍﻋﺪ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ ﻓﻲ ﺍﻟﻤﺎﺀ‪.‬‬ ‫ﺍﻟﺠﺪﻭﻝ ‪ 4-5‬ﻗﻮﺍﻋﺪ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ ﻓﻲ ﺍﻟﻤﺎﺀ‬

‫ﺍﻟﻤﺮﻛﺐ ﺍﻷﻳﻮﻧﻲ‬

‫ﺍﻷﻣﻼﺡ ﺍﻟﺬﺍﺋﺒﺔ‬

‫‪óbÉædG ô«µØàdG‬‬

‫ﻃﺒـﻖ‪ :‬ﺻﻒ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻴﻦ ﻣﺤﻠﻮﻟﻲ ﻛﺒﺮﻳﺘﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫‪ .90‬ﹼ‬ ‫ﻭﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﻨﺤـﺎﺱ ‪ II‬ﺍﻟﺬﻱ ﻳﺆﺩﻱ ﺇﻟﻰ ﺇﻧﺘﺎﺝ ﺭﺍﺳـﺐ‬ ‫ﻣﻦ ﻛﺒﺮﻳﺘﻴﺪ ﺍﻟﻨﺤﺎﺱ ‪.II‬‬

‫‪ .91‬ﺗﻮ ﹼﻗـﻊ‪ :‬ﻭﺿﻌﺖ ﻗﻄﻌﺔ ﻣﻦ ﻓﻠـﺰ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﻓﻲ ﻣﺤﻠﻮﻝ‬ ‫‪ KCl‬ﺍﻟﻤﺎﺋـﻲ‪ ،‬ﻭﻭﺿﻌﺖ ﻗﻄﻌﺔ ﺃﺧـﺮ￯ ﻣﻦ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‬ ‫ﻓـﻲ ﻣﺤﻠﻮﻝ ‪ AgNO3‬ﺍﻟﻤﺎﺋﻲ‪ .‬ﻓﻬﻞ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ ﻓﻲ‬ ‫ﻛﻞ ﻣﻦ ﺍﻟﺤﺎﻟﺘﻴﻦ؟ ﻭﻟﻤﺎﺫﺍ؟‬

‫ﻃﺒـﻖ‪ :‬ﺍﻛﺘـﺐ ﺍﻟﻤﻌﺎﺩﻟـﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﻷﻳﻮﻧﻴـﺔ ﺍﻟﻨﻬﺎﺋﻴﺔ‬ ‫‪ .92‬ﹼ‬ ‫ﻟـﻜﻞ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ‪) .‬ﺇﺫﺍ ﻛﺎﻥ ﻻ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ‬ ‫ﻓﺎﻛﺘﺐ ‪ NR‬ﻓﻲ ﻣﻜﺎﻥ ﺍﻟﻨﻮﺍﺗﺞ(‪ .‬ﻓﻮﺳﻔﺎﺕ ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ‬

‫‪38‬‬

‫اﻟﻜﻴﻤﻴﺎء‬

‫ﺍﻷﻣﻼﺡ ﻏﻴﺮ‬ ‫ﺍﻟﺬﺍﺋﺒﺔ‬

‫ﺍﻟﻘﺎﻋﺪﺓ‬

‫ﺃﻳﻮﻧﺎﺕ ﻋﻨﺎﺻﺮ ﺍﻟﻤﺠﻤﻮﻋﺔ ﺍﻷﻭﻟﻰ )ﻣﺜﻞ‪K+‬‬ ‫‪+‬‬ ‫‪+ +‬‬ ‫ﺃﻣﻼﺣﺎ ﺫﺍﺋﺒﺔ‪.‬‬ ‫ﺗﻜﻮﻥ‬ ‫ﹰ‬ ‫‪ ،(Na ، Li‬ﻭ ‪ NH4‬ﹼ‬ ‫ﺟﻤﻴﻊ ﺃﻣﻼﺡ ﺍﻟﻨﺘﺮﺍﺕ ﺫﺍﺋﺒﺔ‪.‬‬ ‫ﻣﻌﻈﻢ ﺍﻟﻬﺎﻟﻴﺪﺍﺕ ﺗﺬﻭﺏ ﻓﻲ ﺍﻟﻤﺎﺀ ﻣﺎ ﻋﺪﺍ ﻫﺎﻟﻴﺪﺍﺕ‬ ‫ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪Hg22+،Ag+،Cu+،Pb2+ :‬‬ ‫ﻣﻌﻈﻢ ﺍﻟﻜﺒﺮﻳﺘﺎﺕ ﺫﺍﺋﺒﺔ ﺑﺎﺳﺘﺜﻨﺎﺀ ﻛﺒﺮﻳﺘﺎﺕ ‪،Ba2+‬‬ ‫ﻭ‪ ،Sr2+‬ﻭ‪ ، pb2+‬ﺃﻣﺎ ﻛﺒﺮﻳﺘﺎﺕ ‪ ،Ag+‬ﻭ‪،Ca2+‬‬ ‫ﻭ‪ Hg22+‬ﻓﻬﻲ ﻗﻠﻴﻠﺔ ﺍﻟﺬﻭﺑﺎﻥ‪.‬‬

‫ﺍﻟﻬﻴﺪﺭﻭﻛﺴـﻴﺪﺍﺕ‪ ،‬ﻭﺍﻟﻜﺒﺮﻳﺘﻴـﺪﺍﺕ‪ ،‬ﻭﺍﻷﻛﺎﺳـﻴﺪ‬ ‫ﻋـﺎﺩﺓ ﻏﻴـﺮ ﺫﺍﺋﺒـﺔ‪ ،‬ﺑﺎﺳـﺘﺜﻨﺎﺀ ﻣﺮﻛﺒﺎﺗﻬـﺎ ﻣـﻊ ﻋﻨﺎﺻﺮ‬ ‫ﺍﻟﻤﺠﻤﻮﻋـﺔ ﺍﻷﻭﻟـﻰ‪ ،‬ﻭﺃﻳﻮﻧـﺎﺕ ‪ ، NH4+‬ﺃﻣـﺎ‬ ‫ﻋﻨﺎﺻـﺮ ﺃﻳﻮﻧـﺎﺕ ﺍﻟﻤﺠﻤﻮﻋـﺔ ﺍﻟﺜﺎﻧﻴـﺔ ﻓﻬـﻲ ﻗﻠﻴﻠـﺔ‬ ‫ﺍﻟﺬﻭﺑﺎﻥ‪.‬‬ ‫ﺍﻟﻜﺮﻭﻣﺎﺕ ﻭﺍﻟﻔﻮﺳﻔﺎﺕ ﻋﺎﺩﺓ ﻏﻴﺮ ﺫﺍﺋﺒﺔ‪ ،‬ﺑﺎﺳﺘﺜﻨﺎﺀ‬ ‫ﻣﺮﻛﺒﺎﺗﻬﺎ ﻣﻊ ﻋﻨﺎﺻﺮ ﺍﻟﻤﺠﻤﻮﻋﺔ ﺍﻷﻭﻟﻰ‪ ،‬ﻭﺃﻳﻮﻧﺎﺕ‬ ‫‪. NH4+‬‬

‫ﺃﻛﻤﻞ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻵﺗﻴﺔ ﺑﺎﺳـﺘﺨﺪﺍﻡ ﻗﻮﺍﻋﺪ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ﺍﻟﻮﺍﺭﺩﺓ ﻓﻲ‬ ‫ﺍﻟﺠـﺪﻭﻝ ﺃﻋﻼﻩ‪ .‬ﻭﺑﻴﻦ ﻫﻞ ﻳﺘﻜﻮﻥ ﺭﺍﺳـﺐ ﺃﻡ ﻻ‪ ،‬ﻭﺣﺪﺩﻩ‪) .‬ﻭﺇﺫﺍ‬ ‫ﻛﺎﻥ ﻻ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ ﻓﺎﻛﺘﺐ ‪:(NR‬‬ ‫→ )‪Ca(NO3)2(aq) + Na2CO3(aq‬‬ ‫‪.95‬‬ ‫→ )‪Mg(s) +Na OH(aq‬‬ ‫‪.96‬‬ ‫→ )‪PbS(s) +LiNO3(aq‬‬ ‫‪.97‬‬

‫‪≤e QÉÑàNG‬‬ ‫‪Oó©àe øe QÉ«àN’G á∏Ä°SCG‬‬

‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺠﺪﻭﻝ ﺃﺩﻧﺎﻩ ﻟﻺﺟﺎﺑﺔ ﻋﻦ ﺍﻷﺳﺌﻠﺔ ﻣﻦ ‪ 1‬ﺇﻟﻰ ‪.3‬‬ ‫ﺍﻟﻤﺮﻛﺐ‬ ‫‪NaClO3‬‬

‫‪Na2SO4‬‬ ‫‪NiCl2‬‬

‫‪Ni(OH)2‬‬ ‫‪AgNO3‬‬

‫ﺍﻟﺨﻮﺍﺹ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ ﻟﺒﻌﺾ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻷﻳﻮﻧﻴﺔ‬

‫ﺍﻻﺳﻢ‬

‫ﻛﻠﻮﺭﺍﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﻴﻜﻞ ‪II‬‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﻴﻜﻞ‪II‬‬ ‫ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ‬

‫ﺍﻟﺤﺎﻟﺔ ﻋﻨﺪ ‪ C‬ﹾ‪25‬‬

‫‪ .1‬ﺇﺫﺍ ﺧﻠﻂ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﻦ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﻴﻜﻞ ‪ II‬ﺑﻤﺤﻠﻮﻝ‬ ‫ﻣﺎﺋﻲ ﻣﻦ ﻫﻴﺪﺭﻭﻛﺴـﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪ ،‬ﻓﻬﻞ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ‬ ‫ﻣﺎﺋﻲ؟‬

‫‪ .a‬ﻻ؛ ﻷﻥ ﻫﻴﺪﺭﻭﻛﺴـﻴﺪ ﺍﻟﻨﻴـﻜﻞ ‪ II‬ﺍﻟﺼﻠـﺐ ﻳـﺬﻭﺏ ﻓـﻲ‬ ‫ﺍﻟﻤﺎﺀ‪.‬‬

‫‪ .b‬ﻻ؛ ﻷﻥ ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﺼﻮﺩﻳـﻮﻡ ﺍﻟﺼﻠﺒـﺔ ﺗﺬﻭﺏ ﻓﻲ‬ ‫ﺍﻟﻤﺎﺀ‪.‬‬

‫‪ .c‬ﻧﻌﻢ؛ ﻷﻥ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪ II‬ﺍﻟﺼﻠﺒﺔ ﺳﺘﺘﺮﺳﺐ‬ ‫ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ‪.‬‬

‫ﻳﺬﻭﺏ ﻓﻲ ﺍﻟﻤﺎﺀ‬ ‫ﻧﻌﻢ‬ ‫ﻧﻌﻢ‬ ‫ﻧﻌﻢ‬ ‫ﻻ‬ ‫ﻧﻌﻢ‬

‫ﺻﻠﺐ‬ ‫ﺻﻠﺐ‬ ‫ﺻﻠﺐ‬ ‫ﺻﻠﺐ‬ ‫ﺻﻠﺐ‬

‫ﺩﺭﺟﺔ ﺍﻹﻧﺼﻬﺎﺭ )‪ C‬ﹾ(‬ ‫‪248‬‬ ‫‪884‬‬ ‫‪1009‬‬ ‫‪230‬‬

‫‪212‬‬

‫‪.a‬‬

‫→ )‪Ni(OH)2(s) + HCl(aq‬‬

‫‪.b‬‬

‫→ )‪Ni(OH)2(s) + 2HCl(aq‬‬

‫‪.c‬‬

‫→ )‪Ni(OH)2(s) +2H2O(l‬‬

‫‪.d‬‬

‫→ )‪Ni(OH)2(s) +2H2O(l‬‬

‫)‪NiO(aq)+ H2(g) + HCl(aq‬‬ ‫)‪NiCl2(aq)+ 2H2O(l‬‬

‫)‪NiCl2 (aq) + 2H2O(l‬‬

‫)‪NiCl2(aq) +3 H2O(l) + O2(g‬‬

‫‪ .d‬ﻧﻌﻢ؛ ﻷﻥ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﻴﻜﻞ ‪ II‬ﺍﻟﺼﻠﺐ ﺳﻴﺘﺮﺳﺐ ‪ .4‬ﻣﺎ ﻧﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻤﻮﺻﻮﻑ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻵﺗﻴﺔ؟‬ ‫ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ‪.‬‬

‫‪ .2‬ﻣﺎﺫﺍ ﻳﺤﺪﺙ ﻋﻨﺪ ﺧﻠﻂ ﻣﺤﻠﻮﻝ )‪AgClO3(aq‬ﺑﻤﺤﻠﻮﻝ‬ ‫‪NaNO3‬؟‬ ‫‪ .a‬ﻻ ﻳﺤﺪﺙ ﺗﻔﺎﻋﻞ ﻣﺮﺋﻲ‪.‬‬

‫‪ .b‬ﺗﺘﺮﺳﺐ ‪ NaClO3‬ﺍﻟﺼﻠﺒﺔ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ‪.‬‬ ‫‪ .c‬ﻳﻨﻄﻠﻖ ﻏﺎﺯ ‪ NO2‬ﺧﻼﻝ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬ ‫‪ .d‬ﻳﻨﺘﺞ ﻓﻠﺰ ‪ Ag‬ﺍﻟﺼﻠﺐ‪.‬‬

‫‪ .3‬ﻋﻨـﺪ ﺇﺿﺎﻓـﺔ ﺣﻤـﺾ ﺍﻟﻬﻴﺪﺭﻭﻛﻠﻮﺭﻳـﻚ ‪ HCl‬ﺇﻟـﻰ‬ ‫ﻫﻴﺪﺭﻭﻛﺴــﻴﺪ ﺍﻟﻨﻴﻜـﻞ ‪ II‬ﺍﻟﺼﻠــﺐ ﻓﺈﻥ ﺍﻟﻬﻴﺪﺭﻭﻛﺴـﻴﺪ‬ ‫ﻳﺨﺘﻔﻲ‪ .‬ﻣﺎ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻲ ﺗﺼﻒ ﻣﺎ ﺣﺪﺙ ﻓﻲ ﺍﻟﻜﺄﺱ؟‬

‫‪ .a‬ﺗﻜﻮﻳﻦ‬ ‫‪ .c‬ﺗﻔﻜﻚ‬

‫‪ .b‬ﺍﺣﺘﺮﺍﻕ‬

‫‪ .d‬ﺇﺣﻼﻝ ﺑﺴﻴﻂ‬

‫‪ .5‬ﺃﻱ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻵﺗﻴﺔ ﺳـﺘﺤﺪﺙ ﺑﻴﻦ ﺍﻟﻬﺎﻟﻮﺟﻴﻨﺎﺕ ﻭﺃﻣﻼﺡ‬ ‫ﺍﻟﻬﺎﻟﻴﺪﺍﺕ ؟‬ ‫‪.a‬‬

‫)‪F2(g) + FeI2(aq) → FeF2(aq) +I2(l‬‬

‫‪.c‬‬

‫)‪Cl2(s) + SrF2(aq) → SrCl2(aq) +F2(g‬‬

‫‪.b‬‬

‫‪.d‬‬

‫)‪I2(s) + MnBr2(aq)→ MnI2(aq) +Br2(g‬‬ ‫)‪Br2(l) + CoCl2(aq) → CoBr2(aq) + Cl2(g‬‬

‫‪39‬‬

‫اﻟﻤﻮل‬

‫‪The Mole‬‬ ‫اﻟﻌﺎﻣﺔ‬

‫اﻟﻔﻜﺮة‬

‫ﻛﺒﻴﺮﺍ ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ‬ ‫ﻳﻤﺜﻞ ﺍﻟﻤﻮﻝ ﻋﺪ ﹰﺩﺍ ﹰ‬ ‫ﺍﻟﻤﺘﻨﺎﻫﻴـﺔ ﻓـﻲ ﺍﻟﺼﻐﺮ‪ ،‬ﻭﻳﺴـﺘﻌﻤﻞ ﻓﻲ‬ ‫ﺣﺴﺎﺏ ﻛﻤﻴﺎﺕ ﺍﻟﻤﻮﺍﺩ‪.‬‬ ‫‪ 5-1‬ﻗﻴﺎس اﻟﻤﺎدة‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺴﺘﻌﻤﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ‬ ‫ﺍﻟﻤﻮﻝ ﻟﻌﺪ ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻭﺍﻷﻳﻮﻧﺎﺕ‪،‬‬ ‫ﻭﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪ 5-2‬اﻟﻜﺘﻠﺔ واﻟﻤﻮل‬

‫ﺩﺍﺋﻤﺎ‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺤﺘﻮﻱ ﺍﻟﻤﻮﻝ ﹰ‬ ‫ﻋﻠﻰ ﺍﻟﻌﺪﺩ ﻧﻔﺴﻪ ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ‪،‬‬ ‫ﻏﻴﺮ ﺃﻥ ﻣﻮﻻﺕ ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻟﻬﺎ‬ ‫ﻛﺘﻞ ﻣﺨﺘﻠﻔﺔ‪.‬‬

‫‪ 5-3‬ﻣﻮﻻت اﻟﻤﺮﻛﺒﺎت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﻤﻜﻦ ﺣﺴﺎﺏ ﺍﻟﻜﺘﻠﺔ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺻﻴﻐﺘﻪ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻛﻤﺎ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻝ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺘﺤﻮﻳﻞ ﺍﻟﻜﺘﻠﺔ ﺇﻟﻰ‬ ‫ﻣﻮﻻﺕ ﺍﻟﻤﺮﻛﺐ‪.‬‬

‫‪ 5-4‬اﻟﺼﻴﻐﺔ ا وﻟﻴﺔ واﻟﺼﻴﻐﺔ‬ ‫اﻟﺠﺰﻳﺌﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻟﺼﻴﻐـﺔ ﺍﻟﺠﺰﻳﺌﻴـﺔ‬ ‫ﻟﻤﺮﻛـﺐ ﻣـﺎ ﻫـﻲ ﺃﻛﺒـﺮ ﻣﻀﺎﻋـﻒ‬ ‫ﻟﺼﻴﻐﺘـﻪ ﺍﻷﻭﻟﻴـﺔ‪ ،‬ﻭﺗﻀـﻢ ﺃﻋـﺪﺍ ﹰﺩﺍ‬ ‫ﺻﺤﻴﺤﺔ ﻓﻘﻂ‪.‬‬ ‫‪ 5-5‬ﺻﻴﻎ ا ﻣﻼح اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﻫﻲ‬ ‫ﻣﺮﻛﺒﺎﺕ ﺃﻳﻮﻧﻴﺔ ﺻﻠﺒﺔ ﻓﻴﻬﺎ ﺟﺰﻳﺌﺎﺕ‬ ‫ﻣﺎﺀ ﻣﺤﺘﺠﺰﺓ‪.‬‬

‫ﺣﻘﺎﺋﻖ ﻛﻴﻤﻴﺎﺋﻴﺔ‬

‫• ﺍﻟﻌﻤﻼﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺴﻌﻮﺩﻳﺔ ﻫﻲ‪:‬‬ ‫‪ 100،50،25،10،5‬ﻫﻠﻼﺕ‪.‬‬ ‫• ﺗﺘﺮﻛﺐ ﺍﻟﻌﻤﻼﺕ ﺍﻟﺴﻌﻮﺩﻳﺔ ﻣﻦ‬ ‫ﻧﺤﺎﺱ ﻭﻧﻴﻜﻞ ﺑﻨﺴﺐ ﻣﺨﺘﻠﻔﺔ‪.‬‬

‫‪40‬‬

‫‪‹Ó¡à°SG •É°ûf‬‬

‫ﻣﺎ ﻣﻘﺪﺍﺭ ﺍﳌﻮﻝ؟‬

‫ﻳﺴﻬﻞ ﻋﺪ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻜﺒﻴﺮﺓ ﺑﺎﺳﺘﻌﻤﺎﻝ ﻭﺣﺪﺍﺕ ﺍﻟﻌﺪ ﻛﺎﻟﻌﻘﺪ‪،‬‬ ‫ﻭﺍﻟﺪﺭﺯﻥ‪.‬ﻭﻳﺴﺘﻌﻤﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﻭﺣﺪﺓ ﻋﺪ ﺗﺪﻋﻰ ﺍﻟﻤﻮﻝ‪.‬‬

‫ﺧطﻮات الﻌمﻞ‪:‬‬ ‫‪P‬‬

‫ﺗحلﻴﻞ الﻨﺘاﺋﺞ‬ ‫‪.1‬‬

‫‪.3‬‬

‫اﺳــﺘﻘصاﺀ ﻗـﺎﺭﻥ ﻧﺘﺎﺋﺠـﻚ ﺑﻨﺘﺎﺋـﺞ ﺃﺣـﺪ ﺯﻣﻼﺋـﻚ ﻓﻲ‬ ‫ﺍﻟﺼﻒ‪.‬ﻫـﻞ ﻛﺘﻠﺔ ﻣﻮﻝ ﻣﻦ ﺍﻟﺠﺴـﻢ ﺍﻟﺬﻱ ﺍﺧﺘﺮﺗﻪ ﺗﺴـﺎﻭﻱ‬ ‫ﻛﺘﻠﺔ ﺍﻟﻤﻮﻝ ﻟﻠﺠﺴﻢ ﺍﻟﺬﻱ ﺍﺧﺘﺎﺭﻩ ﺯﻣﻴﻠﻚ؟ ﺻﻤﻢ ﺍﺳﺘﻘﺼﺎﺀ‬ ‫ﺗﺤﺪﺩ ﻓﻴﻪ ﺇﻥ ﻛﺎﻥ ﻫﻨﺎﻙ ﻋﻼﻗﺔ ﻣﺎ ﺑﻴﻦ ﺍﻟﻤﻮﻝ ﻭﺍﻟﻜﺘﻠﺔ‪.‬‬

‫ﺍﻟﺨﻄـﻮﺓ ‪ 3‬ﺩﺑـﺲ ﺍﻷﻭﺭﺍﻕ‬ ‫ﻣﻌـﺎ ﻣـﻦ ﺍﻟﻤﻨﺘﺼـﻒ‬ ‫ﺍﻟﺜـﻼﺙ ﹰ‬ ‫ﻋﻠـﻰ ﻃـﻮﻝ ﺣﺎﻓﺘﻬـﺎ ﺍﻟﺨﺎﺭﺟﻴـﺔ‬

‫ﺍﻟﻤﻮﻻﺕ‪/‬ﻋﺪﺩ‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ‬

‫ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ‬

‫‪.2‬‬

‫ﺍﺣﺴـﺐ ﻛـﻢ ﻳﻤﺘـﺪ ﻣـﻮﻝ ) ‪ ( 6.02 × 1023 Particles‬ﻣﻦ‬ ‫ﺍﻟﺠﺴـﻢ ﺍﻟـﺬﻱ ﺍﺧﺘﺮﺗـﻪ ﺇﺫﺍ ﺭﺗﺒﺘـﻪ ﺑﺸـﻜﻞ ﻣﺘﺮﺍﺹ؟ﻋﺒﺮﻋـﻦ‬ ‫ﺇﺟﺎﺑﺘﻚ ﺑﻮﺣﺪﺓ ﺍﻟﻤﺘﺮ‪.‬‬ ‫ﺍﺣﺴـﺐ ﺍﻟﻤﺴﺎﻓﺔ ﻓﻲ ﺍﻟﺨﻄﻮﺓ ‪ 1‬ﺑﻮﺣﺪﺓ ﺍﻟﺴﻨﺔ ﺍﻟﻀﻮﺋﻴﺔ‬ ‫ﻋﻠﻤﺎ ﺑﺄﻥ ) ‪.( ly=9.46 × 1015m‬‬ ‫)‪ (ly‬ﹰ‬ ‫ﻗﺎﺭﻥ ﺍﻟﻤﺴـﺎﻓﺔ ﺍﻟﺘﻲ ﺣﺴﺒﺘﻬﺎ ﻓﻲ ﺍﻟﺨﻄﻮﺓ ﺍﻟﺜﺎﻧﻲ ﻣﻊ ﻫﺬﻩ‬ ‫ﺍﻟﻤﺴﺎﻓﺎﺕ ﺍﻟﻬﺎﺋﻠﺔ‪:‬‬ ‫‪ .a‬ﺍﻟﻤﺴـﺎﻓﺔ ﺇﻟﻰ ﺃﻗﺮﺏ ﻧﺠﻢ )ﻏﻴﺮ ﺍﻟﺸـﻤﺲ( = ‪4.3 ly‬‬ ‫‪ .b‬ﺍﻟﻤﺴﺎﻓﺔ ﺇﻟﻰ ﻣﺮﻛﺰﻣﺠﺮﺗﻨﺎ = ‪30.000ly‬‬ ‫‪ .c‬ﺍﻟﻤﺴﺎﻓﺔ ﺇﻟﻰ ﺃﻗﺮﺏ ﻣﺠﺮﺓ = ‪2 × 10 6 ly‬‬

‫ﺍﻟﺨﻄـﻮﺓ ‪ 2‬ﻋﻨـﻮﻥ ﻛﻞ ﻭﺭﻗـﺔ‬ ‫ﺑﻮﺻﻒ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ‪.‬‬

‫ﺍﻟﻤﻮﻻﺕ‪/‬ﻋﺪﺩ‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ‬

‫ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ‬

‫‪IKO J PL KM N‬‬ ‫‪L‬‬ ‫‪O NP O‬‬ ‫‪،‬ﻣﺸﺒﻚ ﺍﻟﻮﺭﻕ‪ ،‬ﺃﻭﻗﻄﻌﺔ‬ ‫ﻟﺘﻘﻴﺲ ﻃﻮﻟﻪ ﻣﺜﻞ‬ ‫ﺟﺴﻤ‪M‬ﺎ‬ ‫‪ .1‬ﺍﺧﺘﺮ‬ ‫ﹰ‬ ‫ﺣﻠـﻮ￯‪ ،‬ﺑﻨﺎ ﹰﺀ ﻋﻠﻰ ﺍﻷﺟﺴـﺎﻡ ﺍﻟﺘﻲ ﻳﺰﻭﺩﻙ ﺑﻬﺎ ﻣﻌﻠﻤﻚ‪.‬‬ ‫‪ .2‬ﺍﺳﺘﻌﻤﻞ ﺍﻟﻤﺴﻄﺮﺓ ﻓﻲ ﻗﻴﺎﺱ ﻃﻮﻝ ﺍﻟﺠﺴﻢ ﻷﻗﺮﺏ ‪0.1 cm‬‬

‫‪EGK FH L G IM HJN‬‬

‫ﻣﻌﺎﻣـﻼﺕ ﺍﻟﺘﺤﻮﻳـﻞ ﻗـﻢ ﺑﻌﻤـﻞ‬ ‫ﺍﻟﻤﻄﻮﻳـﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻟﻤﺴـﺎﻋﺪﺗﻚ ﻓﻲ‬ ‫‪‬‬ ‫ﺗﻨﻈﻴﻢ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﻋﻦ ﻣﻌﺎﻣﻼﺕ‬ ‫ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﺍﻟﺨﻄـﻮﺓ ‪ 1‬ﺍﺣﻀـﺮ ﺛـﻼﺙ ﺃﻭﺭﺍﻕ‪ .‬ﺍﺛـﻦ ﻛﻞ ﻭﺭﻗـﺔ‬ ‫ﻋﺮﺿﻴـﺎ ﻣـﻦ ﺍﻟﻤﻨﺘﺼـﻒ‪.‬‬ ‫ﻗﺲ ﻭﺍﺭﺳـﻢ ﺧﻄﺎ ﻋﻠﻰ ﺑﻌﺪ‬ ‫‪ 3cm‬ﻣﻦ ﺍﻟﻄﺮﻑ ﺍﻷﻳﺴـﺮ‪.‬‬ ‫ﻗـﺺ ﺍﻟﻮﺭﻗـﺔ ﻋﻠـﻰ ﻃـﻮﻝ‬ ‫ﻫـﺬﺍ ﺍﻟﺨﻂ‪ ،‬ﻭﻛـﺮﺭ ﺫﻟﻚ ﻣﻊ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪D AE BF ACG BDH CE I DF J‬‬ ‫ﺍﻟﻮﺭﻗﺘﻴﻦ ﺍﻷﺧﺮﻳﻴﻦ‪.‬‬

‫المطﻮﻳات ﺍﺳـﺘﻌﻤﻞ ﻫـﺬﻩ ﺍﻟﻤﻄﻮﻳـﺔ ﻓـﻲ ﺍﻟﺒﻨـﻮﺩ‬ ‫‪ ،5-2‬ﻣـﻦ ﻫـﺬﺍ ﺍﻟﻔﺼـﻞ‪ .‬ﺩﻭﻥ ﻣﻌﻠﻮﻣﺎﺗـﻚ ﻋـﻦ ﻣﻌﺎﻣـﻼﺕ‬ ‫ﺍﻟﺘﺤﻮﻳﻞ‪،‬ﻭﻟﺨﺺ ﺍﻟﺨﻄﻮﺍﺕ ﺍﻟﺘﻲ ﻳﺘﻀﻤﻨﻬﺎ ﻛﻞ ﺗﺤﻮﻳﻞ‪.‬‬

‫‪5-1‬‬

‫ﳌﺮﺍﺟﻌﺔ ﳏﺘﻮ￯ ﻫﺬﺍ ﺍﻟﻔﺼﻞ ﻭﺃﻧﺸﻄﺘﻪ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ ﺍﻹﻟﻜﱰﻭﲏ‬

‫‪www.obeikaneducation.com‬‬

‫‪41‬‬ ‫‪41‬‬

‫‪5-1‬‬ ‫ا ﻫﺪاف‬

‫ﺗﻔـﴪ ﻛﻴـﻒ ﻳﺴـﺘﺨﺪﻡ ﺍﳌـﻮﻝ‬ ‫ﺑﺸﻜﻞ ﻏﲑ ﻣﺒﺎﴍ ﻟﻌﺪﹼ ﺟﺴﻴﲈﺕ‬ ‫ﺍﳌﺎﺩﺓ‪.‬‬ ‫ﺗﺮﺑﻂ ﺍﳌـﻮﻝ ﺑﻮﺣﺪﺓ ﻋـﺪﹼ ﻳﻮﻣﻴﺔ‬ ‫ﺷﺎﺋﻌﺔ‪.‬‬ ‫ﲢــﻮﻝ ﺑــﲔ ﺍﳌـــﻮﻻﺕ ﻭﻋــﺪﺩ‬ ‫ﺍﳉﺴﻴﲈﺕ‪.‬‬

‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﳉﺰﻱﺀ‪ :‬ﺫﺭﺗﺎﻥ ﺃﻭ ﺃﻛﺜﺮ ﻣﺮﺗﺒﻄﺘﺎﻥ‬ ‫ﻣﻌﺎ ﻟﺘﻜﻮﻳﻦ ﻭﺣﺪﺓ ﻭﺍﺣﺪﺓ‪.‬‬ ‫ﹰ‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﳌﻮﻝ‬ ‫ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‬

‫ﻗﻴﺎس اﻟﻤﺎدة‬ ‫‪Measuring Matter‬‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺴﺘﻌﻤﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﺍﳌﻮﻝ ﻟﻌﺪﹼ ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻭﺍﳉﺰﻳﺌﺎﺕ‪ ،‬ﻭﺍﻷﻳﻮﻧﺎﺕ‪،‬‬ ‫ﻭﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻫﻞ ﺃﻗﺎﻡ ﺻﻔﻚ ﻳﻮ ﹰﻣﺎ ﻣﺴﺎﺑﻘ ﹰﺔ ﳌﻌﺮﻓﺔ ﻋﺪﺩ ﺍﻟﻘﻄﻊ ﺍﻟﻨﻘﺪﻳﺔ‬ ‫ﺃﻭ ﻋﺪﺩ ﻗﻄﻊ ﺍﳊﻠﻮ￯ ﺍﳌﻮﺟﻮﺩﺓ ﰲ ﻋﻠﺒﺔ؟ ﻟﻌﻠﻚ ﻻﺣﻈﺖ ﺃﻧﻪ ﻛﻠﲈ ﺻﻐﺮﺕ ﺍﳌﺎﺩﺓ‬ ‫ﻳﺼﺒﺢ ﺍﻟﻌﺪﹼ ﺃﺻﻌﺐ‪.‬‬

‫ﻋﺪ الﺠسﻴمات ‪Counting Particles‬‬ ‫ﹼ‬

‫ﺇﺫﺍ ﺃﺭﺩﺕ ﺃﻥ ﺗﺸﱰﻱ ﺑﺎﻗﺔ ﺃﺯﻫﺎﺭ ﰲ ﺇﺣﺪ￯ ﺍﳌﻨﺎﺳﺒﺎﺕ ﻓﺮﺑﲈ ﻻ ﺗﻄﻠﺐ ‪ 12‬ﺃﻭ ‪،24‬‬ ‫ﺯﻭﺟﺎ ﻣﻦ ﺍﻟﻘﻔﺎﺯﺍﺕ‪ ،‬ﺃﻭ ﺭﺯﻣﺔ‬ ‫ﺑﻞ ﺳﺘﻄﻠﺐ ﺩﺭﺯ ﹰﻧﺎ ﻭﺍﺣﺪﹰ ﺍ ﺃﻭ ﺩﺭﺯﻧﲔ‪ .‬ﻭﻗﺪ ﺗﺸﱰﻱ ﹰ‬ ‫ﻣﻦ ﻭﺭﻕ ﺍﻟﻄﺒﺎﻋﺔ‪ ،‬ﺃﻭ ﺩﺭﺯ ﹰﻧﺎ ﻣﻦ ﺃﻗﻼﻡ ﺍﻟﺮﺻﺎﺹ‪ .‬ﻛﻞ ﻣﻦ ﻫﺬﻩ ﺍﻟﻮﺣﺪﺍﺕ ﺍﳌﺒﻴﻨﺔ ﰲ‬ ‫ﺍﻟﺸﻜﻞ ‪ 5-1‬ﻭﻫﻲ ﺍﻟﺰﻭﺝ‪ ،‬ﻭﺍﻟﺪﺭﺯﻥ‪ ،‬ﻭﺍﻟﺮﺯﻣﺔ ﲤﺜﹼﻞ ﻋﺪ ﹰﺩﺍ ﳏﺪ ﹰﺩﺍ ﻣﻦ ﺍﻷﺷﻴﺎﺀ‪ .‬ﻭﻛﻠﻬﺎ‬ ‫ﺗﺴﻬﻞ ﻋﻤﻠﻴﺔ ﺍﻟﻌﺪ‪ .‬ﻓﻤﻦ ﺍﻟﺴﻬﻞ ﴍﺍﺀ ﺍﻟﻮﺭﻕ ﻭﺑﻴﻌﻪ ﺑﺎﻟﺮﺯﻣﺔ )‪ 500‬ﻭﺭﻗﺔ( ﺑﺪ ﹰ‬ ‫ﻻ ﻣﻦ‬ ‫ﴍﺍﺋﻪ ﻭﺑﻴﻌﻪ ﺑﺎﻟﻮﺭﻗﺔ‪.‬‬ ‫ﻛﻞ ﻭﺣﺪﺍﺕ ﺍﻟﻌﺪﹼ ﺍﳌﺒﻴﻨﺔ ﰲ ﺍﻟﺸﻜﻞ ‪ 5-1‬ﻣﻨﺎﺳﺒﺔ ﻟﻨﻮﻉ ﻣﻌﲔ ﻣﻦ ﺍﻷﺷﻴﺎﺀ؛ ﺍﻋﺘﲈ ﹰﺩﺍ ﻋﲆ‬ ‫ﺣﺠﻤﻬﺎ ﻭﺍﺳﺘﺨﺪﺍﻣﻬﺎ‪ .‬ﻭﺑﻐﺾ ﺍﻟﻨﻈﺮ ﻋﻦ ﻛﻮﻥ ﺍﻟﴚﺀ ﻗﻔﺎﺯﺍﺕ‪ ،‬ﺃﻭ ﹰ‬ ‫ﺑﻴﻀﺎ‪ ،‬ﺃﻭ ﺃﻗﻼﻡ‬ ‫ﺭﺻﺎﺹ ﺃﻭ ﻭﺭ ﹰﻗﺎ‪ ،‬ﻓﺈﻥ ﺍﻟﻌﺪﺩ ﺍﻟﺬﻱ ﲤﺜﻠﻪ ﺍﻟﻮﺣﺪﺓ ﻳﺒﻘﻰ ﹰ‬ ‫ﺩﺍﺋﲈ ﺛﺎﺑ ﹰﺘﺎ‪ .‬ﳛﺘﺎﺝ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ‬ ‫ﺇﱃ ﻃﺮﻳﻘﺔ ﻣﻼﺋﻤﺔ ﻭﺻﺤﻴﺤﺔ ﻟﻌﺪ ﹼ‬ ‫ﺍﻟﺬﺭﺍﺕ ﻭﺍﳉﺰﻳﺌﺎﺕ ﻭﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﰲ ﻋﻴﻨﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﹴ‬ ‫ﳌﺎﺩﺓ ﻣﺎ‪ .‬ﺇﻻ ﺃﻥ ﺍﻟﺬﺭﺍﺕ ﻣﺘﻨﺎﻫﻴﺔ ﺍﻟﺼﻐﺮ‪ ،‬ﻭﻫﻨﺎﻙ ﺍﻟﻜﺜﲑ ﻣﻨﻬﺎ ﺣﺘﻰ‬ ‫ﺟﺪﺍ‪ ،‬ﳑﹼﺎ ﳚﻌﻞ ﻋﺪﻫﺎ ﺑﺸﻜﻞ ﻣﺒﺎﴍ ﻣﺴﺘﺤﻴ ﹰ‬ ‫ﻼ‪ .‬ﻟﺬﻟﻚ ﻗﺎﻡ‬ ‫ﰲ ﺍﻟﻌﻴﻨﺎﺕ ﺍﻟﺼﻐﲑﺓ ﹰﹼ‬ ‫ﹰ‬ ‫ﺿﺨﲈ ﻣﻦ ﺃﻱ ﺟﺴﻢ‪.‬‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﺑﺈﳚﺎﺩ ﻭﺣﺪﺓ ﻋﺪﱟ ﹸﺗﺴﻤﻰ ﺍﳌﻮﻝ‪ ،‬ﺍﻟﺬﻱ ﻳﻤﺜﻞ ﻋﺪ ﹰﺩﺍ‬ ‫ﺍﻟﺸﻜﻞ ‪ 5-1‬ﻭﺣـــﺪﺍﺕ ﳐﺘﻠﻔﺔ‬ ‫ﺗﺴﺘﺨﺪﻡ ﻟﻌﺪﹼ ﺃﺟﺴﺎﻡ‬ ‫ﳐﺘﻠﻔﺔ‪ .‬ﺍﻟﺰﻭﺝ ﻋﺒﺎﺭﺓ ﻋﻦ‬ ‫ﺟﺴﻤﲔ‪ .‬ﻭﺍﻟﺪﺭﺯﻥ ‪،12‬‬ ‫ﻭﺍﻟﺮﺯﻣﺔ ‪.500‬‬ ‫ﻋﺪﹼ ﺩ ﻭﺣﺪﺍﺕ ﺑﻄﺮﻳﻘﺔ‬ ‫ﺃﺧﺮ￯ ﻣﺄﻟﻮﻓﺔ ﻟﺪﻳﻚ‪.‬‬

‫‪42‬‬

‫‪ ∫ƒ``ªdG‬ﹸﺗﺴـﻤﻰ ﻭﺣﺪﺓ ﺍﻟﻨﻈﺎﻡ ﺍﻟﺪﻭﻟﻲ ﺍﻷﺳﺎﺳـﻴﺔ ﺍﻟﻤﺴـﺘﺨﺪﻣﺔ ﻟﻘﻴـﺎﺱ ﻛﻤﻴﺔ ﺍﻟﻤﺎﺩﺓ‬ ‫ﺍﻟﻤـﻮﻝ‪ .‬ﻭﻳﻌـﺮﻑ ﺍﻟﻤﻮﻝ ﺑﺄﻧﻪ ﻋـﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑـﻮﻥ – ‪ 12‬ﻓﻲ ﻋﻴﻨـﺔ ﻛﺘﻠﺘﻬﺎ ‪.12g‬‬ ‫ﻭﺧﻼﻝ ﺳـﻨﻮﺍﺕ ﻋﺪﻳـﺪﺓ ﻣﻦ ﺍﻟﺘﺠﺎﺭﺏ ﺗﻢ ﺍﻻﺗﻔﺎﻕ ﻋﻠـﻰ ﺃﻥ ﺍﻟﻤﻮﻝ ﺍﻟﻮﺍﺣﺪ ﻣﻦ ﺃﻱ‬ ‫‪23‬‬ ‫ﺷـﻲﺀ ﻳﺤﺘﻮﻱ ﻋﻠـﻰ ‪ 6.02 × 10‬ﻣﻦ ﺍﻟﺠﺴـﻴﻤﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻟﻬﺬﺍ ﺍﻟﺸـﻲﺀ‪ ،‬ﻣﺜﻞ‬ ‫ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻭﺍﻟﺠﺰﻳﺌﺎﺕ‪ ،‬ﻭﺍﻷﻳﻮﻧﺎﺕ‪ ،‬ﻭﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻓﺈﺫﺍ ﻛﺘﺒﺖ ﺍﻟﻌﺪﺩ‬ ‫ﻓﺴﻮﻑ ﻳﺒﺪﻭ ﻛﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪602‚213‚670‚000‚000‚000‚000‚000‬‬ ‫‪23‬‬ ‫ﹰ‬ ‫ﺗﻜﺮﻳﲈ ﻟﻠﻔﻴﺰﻳﺎﺋﻲ ﺍﻹﻳﻄﺎﱄ‬ ‫ﻭ ﹸﻳﺴﻤﻰ ﺍﻟﻌﺪﺩ ‪ 6.0221367 × 10‬ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‪،‬‬ ‫ﻭﺍﳌﺤﺎﻣﻲ ﺃﻣﻴﺪﻭ ﺃﻓﻮﺟﺎﺩﺭﻭ ‪.Amedeo Avogadro‬‬

‫ﹰ‬ ‫ﺻﺎﳊﺎ ﻟﻌﺪ ﺍﳌﻜﻮﻧﺎﺕ‬ ‫ﻭﻣﻦ ﺍﻟﻮﺍﺿﺢ ﺃﻥ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻋﺪﺩ ﻫﺎﺋﻞ‪ ،‬ﻭﻫﺬﺍ ﻣﺎ ﳚﻌﻠﻪ‬ ‫ﺍﳌﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ‪ ،‬ﻣﺜﻞ ﺍﻟﺬﺭﺍﺕ‪ .‬ﻛﲈ ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﺘﺼﻮﺭ ﺃﻥ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻟﻦ‬ ‫ﻳﻜﻮﻥ ﻣﻨﺎﺳ ﹰﺒﺎ ﻟﻘﻴﺎﺱ ﻛﻤﻴﺔ ﻣﻦ ﻛﺮﺍﺕ ﺍﻟﻠﻌﺐ ﺍﻟﺰﺟﺎﺟﻴﺔ؛ ﻷﻥ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ‬ ‫ﻫﺬﻩ ﺍﻟﻜﺮﺍﺕ ﺳﻮﻑ ﻳﻐﻄﻲ ﺳﻄﺢ ﺍﻷﺭﺽ ﺇﱃ ﻋﻤﻖ ﻳﺘﺠﺎﻭﺯ ﺳﺘﺔ ﻛﻴﻠﻮﻣﱰﺍﺕ‪.‬‬ ‫ﻭﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﻟﺸﻜﻞ ‪ ،5-2‬ﻓﺈﻥ ﺍﺳﺘﺨﺪﺍﻡ ﺍﳌﻮﻝ ﻣﻨﺎﺳﺐ ﳊﺴﺎﺏ ﻛﻤﻴﺎﺕ ﻣﻦ‬ ‫ﺍﳌﻮﺍﺩ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻭﻳﺒﲔ ﺍﻟﺸﻜﻞ ﻛﻤﻴﺎﺕ ﻣﻘﺪﺍﺭﻫﺎ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﳌﺎﺀ‪ ،‬ﻭﺍﻟﻨﺤﺎﺱ‪،‬‬ ‫ﻭﺍﳌﻠﺢ‪ ،‬ﻭﻳﺘﻜﻮﻥ ﻛﻞ ﻣﻨﻬﺎ ﻣﻦ ﺟﺴﻴﲈﺕ ﳐﺘﻠﻔﺔ‪ .‬ﻓﺎﳉﺴﻴﲈﺕ ﺍﳌﻜﻮﻧﺔ ﳌﻮﻝ ﻣﻦ ﺍﳌﺎﺀ‬ ‫ﻫﻲ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ‪ .‬ﻭﺍﳌﻜﻮﻧﺔ ﳌﻮﻝ ﻣﻦ ﺍﻟﻨﺤﺎﺱ ﻫﻲ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﺍﳌﻜﻮﻧﺔ ﳌﻮﻝ‬ ‫ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻫﻲ ﻭﺣﺪﺍﺕ ﺻﻴﻐﺔ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪.‬‬ ‫ﺍﻟﺸﻜﻞ ‪ 5-2‬ﻛﻤﻴﺔ ﻛﻞ ﻣﺎﺩﺓ ﻣﺒﻴﻨﺔ ﻫﻲ ‪ ،6.02 10‬ﺃﻭ ‪ 1mol‬ﻣﻦ ﺍﳉﺴﻴﲈﺕ‬ ‫ﺍﳌﻜﻮﻧﺔ ﻟﻠﲈﺩﺓ‪ .‬ﺍﳉﺴﻴﲈﺕ ﺍﳌﻜﻮﻧﺔ ﻟﻜﻞ ﻣﺎﺩﺓ ﻣﻮﺿﺤﺔ ﺩﺍﺧﻞ ﺍﳌﺮﺑﻊ‪.‬‬ ‫‪23‬‬

‫‪ᨫ°üdG IóMh‬‬ ‫‪á«FÉ«ª«µdG‬‬

‫‪NaCl ΩƒjOƒ°üdG ójQƒ∏c‬‬

‫‪ÇõL‬‬

‫‪IQP‬‬

‫‪Cu ¢SÉëædG‬‬

‫‪H2O AÉŸG‬‬

‫‪43‬‬

‫الﺘحﻮﻳﻞ ﺑﻴﻦ المﻮﻻت ﻭالﺠسﻴمات‬ ‫‪Converting Between Moles and Particles‬‬

‫–‪ äɪ«°ùL ¤EG ä’ƒŸG πjƒ‬ﳊﺴﺎﺏ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﺍﻟﺴﻜﺮﻭﺯ ﰲ ‪ 3.5 mol‬ﻣﻨﻪ‪ ،‬ﻧﺴﺘﺨﺪﻡ‬ ‫ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ‪-‬ﺃﻱ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﻭﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ‪ -‬ﻛﻤﻌﺎﻣﻞ ﻟﻠﺘﺤﻮﻳﻞ‪.‬‬ ‫‪ 1 mol‬ﻣﻦ ﺍﳉﺴﻴﲈﺕ ﳛﺘﻮﻱ ﻋﲆ‬

‫‪.OQh 1 ¿RQO ¤EG IOQh 12‬‬

‫ﺍﻟﺸﻜﻞ ‪ 5-3‬ﻟﻜﻲ ﺗﺘﻤﻜﻦ ﻣﻦ ﲢﻠﻴﻞ‬ ‫ﺗﻌﺮﻑ‬ ‫ﺍﻟــﻮﺣــﺪﺍﺕ ﳚــﺐ ﹼ‬ ‫ﺍﻟــﻌــﻼﻗــﺔ ﺍﻟــﺮﻳــﺎﺿــﻴــﺔ‬ ‫ﺍﻟﺼﺤﻴﺤﺔ ﺑﲔ ﺍﻟﻮﺣﺪﺍﺕ‬ ‫ﺍﻟﺘﻲ ﺳﺘﺤﻮﳍﺎ‪ .‬ﻭﺍﻟﻌﻼﻗﺔ‬ ‫ﺍﳌﻮﺿﺤﺔ ﻫﻨﺎ ـ ‪ 12‬ﻭﺭﺩﺓ =‪1‬‬ ‫ﺩﺭﺯﻥ ﻭﺭﺩ ـ ﻳﻤﻜﻦ ﺍﺳﺘﻌﲈﳍﺎ‬ ‫ﻟﻜﺘﺎﺑﺔ ﻣﻌﺎﻣﲇ ﲢﻮﻳﻞ‪.‬‬

‫‪6.02 × 10 Particles‬‬ ‫‪23‬‬

‫ﺃﻱ ﺃﻥ‪ :‬ﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ )‪ =(Particles‬ﻋﺪﺩ ﺍﳌﻮﻻﺕ )‪6.02 × 10 Particles × (mol‬‬ ‫‪23‬‬

‫‪1 mol‬‬

‫ﺃﻱ ﺃﻥ‪ :‬ﺟﺰﻳﺌﺎﺕ ﺍﻟﺴﻜﺮﻭﺯ= ‪ 3.5 mol‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ ×‬

‫‪23‬‬ ‫‪ 6.02 × 10 molecules‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ‬

‫‪1 mol‬‬

‫= ‪ 2.11 × ×10 molecules‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ‬ ‫‪24‬‬

‫ﺃﻱ ﺃﻥ ﻫﻨﺎﻙ‬

‫‪molecules‬‬

‫‪ 2.11×10‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ ﰲ ‪ 3.5 mol‬ﻣﻨﻪ‪.‬‬

‫‪24‬‬

‫–‪ ä’ƒe ¤EG äɪ«°ù÷G πjƒ‬ﳊﺴﺎﺏ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﰲ ﻋﺪﺩ ﻣﻌﲔ ﻣﻦ ﺍﳉﺴﻴﲈﺕ‪،‬ﻳﻤﻜﻨﻚ‬ ‫ﺍﺳﺘﺨﺪﺍﻡ ﻣﻘﻠﻮﺏ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻛﻤﻌﺎﻣﻞ ﻟﻠﺘﺤﻮﻳﻞ‪.‬‬ ‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ )‪ = (mol‬ﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ )‪× (Particles‬‬

‫ﻭﳊﺴﺎﺏ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺴﻜﺮﻭﺯ ﰲ ﻋﻴﻨﺔ ﲢﺘﻮﻱ ﻋﲆ‬ ‫ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ‪ ،‬ﻓﻴﻜﻮﻥ‪:‬‬

‫‪1 mol‬‬ ‫‪Particles 6.02 × 10‬‬

‫‪23‬‬

‫‪ 2.11 × 10‬ﻣﻨﻪ‪ ،‬ﻧﺴﺘﻌﻤﻞ‬

‫‪24‬‬

‫المطﻮﻳات‬

‫ﺿﻤـﻦ ﻣﻌﻠﻮﻣﺎﺕ ﻣﻦ ﻫـﺬﺍ ﺍﻟﺒﻨﺪ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺴﻜﺮﻭﺯ‬ ‫ﹼ‬ ‫ﻓﻲ ﻣﻄﻮﻳﺘﻚ‪.‬‬ ‫‪24‬‬

‫=‬

‫‪ 2.11×10 molecules‬ﻣﻦ ﺳﻜﺮﻭﺯ×‬

‫‪1 mol‬‬ ‫‪molecules‬‬

‫ﺃﻱ ﺃﻥ ﻫﻨﺎﻙ ‪ 3.5 mol‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ ﰲ‬

‫‪ 6.02 × 10‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ‬

‫‪23‬‬

‫= ‪ 3.5 mol‬ﻣﻦ ﺍﻟﺴﻜﺮﻭﺯ‬

‫‪ 2.11 × 10‬ﻣﻨﻪ‪.‬‬

‫‪24‬‬

‫‪ ?äCGôb GPÉe‬ﺍﻛﺘﺐ ﻣﻌﺎﻣﲇ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻠﺬﻳﻦ ﻳﻤﻜﻦ ﺍﳊﺼﻮﻝ ﻋﻠﻴﻬﲈ ﻣﻦ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .1‬ﻳﺴﺘﺨﺪﻡ ﺍﳋﺎﺭﺻﲔ ‪ Zn‬ﻟﺘﻜﻮﻳﻦ ﻃﺒﻘﺔ ﻋﲆ ﺍﳊﺪﻳﺪ ﳊﲈﻳﺘﻪ ﻣﻦ ﺍﻟﺘﺂﻛﻞ‪ .‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺫﺭﺍﺕ ‪ Zn‬ﰲ ‪ 2.5 mol‬ﻣﻨﻪ‪.‬‬ ‫‪ .2‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﳉﺰﻳﺌﺎﺕ ﰲ ‪ 11.5 mol‬ﻣﻦ ﺍﳌﺎﺀ ‪.H2O‬‬ ‫‪ .3‬ﺗﺴﺘﺨﺪﻡ ﻧﱰﺍﺕ ﺍﻟﻔﻀﺔ ‪ AgNO3‬ﻟﺼﻨﺎﻋﺔ ﺃﻧﻮﺍﻉ ﻣﺘﻌﺪﺩﺓ ﻣﻦ ﻫﺎﻟﻴﺪﺍﺕ ﺍﻟﻔﻀﺔ ﺍﳌﺴﺘﺨﺪﻣﺔ ﰲ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺼﻮﻳﺮ‬ ‫ﺍﻟﻔﻮﺗﻮﺟﺮﺍﰲ‪ .‬ﻣﺎ ﻋﺪﺩ ﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻐﺔ ‪ AgNO3‬ﰲ ‪ 3.25 mol‬ﻣﻨﻬﺎ؟‬ ‫‪ .4‬ﲢﺪﹼ ‪ :‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﲔ ﰲ ‪ 5.0 mol‬ﻣﻦ ‪.O2‬‬

‫‪44‬‬

‫‪5-1 ∫Éãe‬‬

‫ﲢﻮﻳﻞ ﺍﳉﺴﻴﲈﺕ ﺇﱃ ﻣﻮﻻﺕ ﻳﺴﺘﺨﺪﻡ ﺍﻟﻨﺤﺎﺱ ‪ Cu‬ﰲ ﺻﻨﺎﻋﺔ ﺍﻷﺳﻼﻙ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ‪ .‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻨﺤﺎﺱ‬ ‫‪24‬‬ ‫ﺍﻟﺘﻲ ﲢﺘﻮﻱ ﻋﲆ ‪ 4.5 × 10 atoms‬ﻣﻨﻪ‪.‬‬

‫‪ádCÉ°ùŸG π«∏– 1‬‬

‫ﻟﺪﻳﻚ ﻋﺪﺩ ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﻋﻠﻴﻚ ﺃﻥ ﲢﺴﺐ ﻋﺪﺩ ﺍﳌﻮﻻﺕ‪ .‬ﻟﻮ ﻗﺎﺭﻧﺖ ‪ 4.5 × 1024 atoms‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ ‪Cu‬‬

‫ﻣﻊ ‪ ،6.02 × 10‬ﻭﻫﻮ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ﺍﳌﻮﻝ‪ ،‬ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﺘﻮﻗﻊ ﺃﻥ ﺍﻹﺟﺎﺑﺔ ﳚﺐ ﺃﻥ ﺗﻜﻮﻥ ﺃﻗﻞ ﻣﻦ ‪.10 mol‬‬ ‫‪23‬‬

‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫‪24‬‬ ‫ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ = ‪ 4.50 × 10 atoms‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬

‫‪ 1mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬ ‫‪2‬‬

‫ﻣﻮﻻﺕ ‪mol ? = Cu‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺳﺘﺨﺪﻡ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ )ﻣﻘﻠﻮﺏ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ( ﻭﺍﻟﺬﻱ ﻳﺮﺑﻂ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﺑﻌﺪﺩ ﺍﻟﺬﺭﺍﺕ‪.‬‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﹼ‬

‫‪3‬‬

‫‪23‬‬ ‫= ‪ 6.02 × 10 atoms‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬

‫ﺍﳌﻄﻠﻮﺏ‬

‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ )‪ = (mol‬ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ )‪× (atoms‬‬

‫‪1 mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬

‫‪atoms‬‬

‫‪ 6.02 × 10‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬

‫‪23‬‬

‫ﻋﻮﺽ ﻭﺍﴐﺏ ﺍﻷﺭﻗﺎﻡ‬ ‫‪1 mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬ ‫ﹼ‬ ‫‪24‬‬ ‫= ‪ 7.48 mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬ ‫×‬ ‫ﺍﻟﻨﺤﺎﺱ‬ ‫ﻣﻦ‬ ‫‪4.50‬‬ ‫×‬ ‫‪10‬‬ ‫‪atoms‬‬ ‫=‬ ‫ﻭﺍﻟﻮﺣﺪﺍﺕ ﻭﺍﻗﺴﻤﻬﺎ‬ ‫‪ 6.02 × 1023 atoms‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﺍﻹﺟﺎﺑﺔ ﻣﻜﺘﻮﺑﺔ ﺑﺸﻜﻞ ﺻﺤﻴﺢ ﻭﻫﻲ ﺃﻗﻞ ﻣﻦ ‪ 10 mol‬ﻣﻮﻻﺕ ﻛﲈ ﻫﻮ ﻣﺘﻮﻗﻊ‪ ،‬ﻛﲈ ﺃﻥ ﻭﺣﺪﺍﲥﺎ ﺻﺤﻴﺤﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .5‬ﻣﺎ ﻋﺪﺩ ﺍﳌﻮﻻﺕ)‪ (mol‬ﰲ ﻛﻞ ﻣﻦ‪:‬‬ ‫‪atoms .a‬‬

‫‪ 5.75 × 10‬ﻣﻦ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ‪A1‬‬

‫‪24‬‬

‫‪ 2.50 × 10 atoms .b‬ﻣﻦ ﺍﳊﺪﻳﺪ ‪Fe‬‬ ‫‪20‬‬

‫‪ .6‬ﲢﺪﹼ ‪ :‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﰲ ﻛﻞ ﻣﻦ‪:‬‬

‫‪ 3.75 × 1024 molecules .a‬ﻣﻦ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ‪CO2‬‬

‫‪ 3.58 × 10 molecules .b‬ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﳋﺎﺭﺻﲔ ‪ZnCl2 II‬‬ ‫‪23‬‬

‫‪45‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪5-1‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺍﻟﻤﻮﻝ ﻭﺣﺪﺓ ﺗﺴﺘﺨﺪﻡ ﻟﻌﺪ ﺟﺴﻴﻤﺎﺕ‬ ‫ﺍﻟﻤﺎﺩﺓ ﺑﺸـﻜﻞ ﻏﻴـﺮ ﻣﺒﺎﺷـﺮ‪ .‬ﺍﻟﻤﻮﻝ‬ ‫ﺍﻟﻮﺍﺣﺪ ﻣﻦ ﺍﻟﻤﺎﺩﺓ ﺍﻟﻨﻘﻴﺔ ﻳﺤﺘﻮﻱ ﻋﻠﻰ‬ ‫ﻋـﺪﺩ ﺃﻓﻮﺟـﺎﺩﺭﻭ ﻣﻦ ﺍﻟﺠﺴـﻴﻤﺎﺕ‪.‬‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ ﺗﺸﻤﻞ ﺍﻟــﺬﺭﺍﺕ‪،‬‬ ‫ﻭﺍﻷﻳﻮﻧــــﺎﺕ‪ ،‬ﻭﺍﻟﺠــﺰﻳﺌــﺎﺕ‪،‬‬ ‫ﻭﻭﺣــــﺪﺍﺕ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪،‬‬ ‫ﻭﺟﺴﻴﻤﺎﺕ ﺃﺧﺮ￯ ﻣﺸﺎﺑﻬﺔ‪.‬‬ ‫ﺍﻟﻤـــﻮﻝ ﺍﻟﻮﺍﺣـــﺪ ﻣـﻦ ﺫﺭﺍﺕ‬ ‫ﺍﻟﻜﺮﺑـﻮﻥ–‪ 12‬ﻟـﻪ ﻛﺘﻠـﺔ ﻣﻘﺪﺍﺭﻫﺎ‬ ‫‪ 12g‬ﺗﻤﺎ ﹰﻣﺎ‪.‬‬ ‫ﻳﻤﻜــﻦ ﺍﺳﺘﺨـــﺪﺍﻡ ﻣﻌـﺎﻣـﻼﺕ‬ ‫ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﻜﺘﻮﺑﺔ ﻣﻦ ﻋﻼﻗﺔ ﻋﺪﺩ‬ ‫ﺃﻓﻮﺟﺎﺩﺭﻭ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻤﻮﻻﺕ‬ ‫ﻭﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬‬

‫‪46‬‬

‫ﻓﺴﺮ ﻟﻤﺎﺫﺍ ﻳﺴﺘﺨﺪﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﺍﻟﻤﻮﻝ؟‬ ‫‪ .7‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﹼ‬ ‫‪ .8‬ﺍﺫﻛﺮ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺮﻳﺎﺿﻴﺔ ﺑﻴﻦ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻭﺍﻟﻤﻮﻝ‪.‬‬ ‫‪ .9‬ﻋﺪﹼ ﺩ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﺴﺘﺨﺪﻣﺔ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻭﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫ﻓﺴﺮ ﻭﺟﻪ ﺍﻟﺸﺒﻪ ﺑﻴﻦ ﺍﻟﻤﻮﻝ ﻭﺍﻟﺪﺭﺯﻥ‪.‬‬ ‫‪ .10‬ﹼ‬ ‫ﻃﺒﻖ ﻛﻴﻒ ﻳﻌﺪ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻓﻲ ﻋﺪﺩ ﻣﻌﻴﻦ ﻣﻦ ﻣﻮﻻﺕ ﺍﻟﻤﺎﺩﺓ‪.‬‬ ‫‪ .11‬ﹼ‬ ‫‪ .12‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺑﻮﺣﺪﺓ ﺍﻟﻜﺘﻞ ﺍﻟﺬﺭﻳﺔ ﻟـ ‪ 0.25 mol‬ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ– ‪.12‬‬ ‫‪ .13‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻓﻲ ﻛﻞ ﻣﻦ ﺍﻟﻤﻮﺍﺩ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪ 11.5 mol .a‬ﻣﻦ ﺍﻟﻔﻀﺔ ‪Ag‬‬

‫‪ 18.0 mol .b‬ﻣﻦ ﺍﻟﻤﺎﺀ ‪H2O‬‬ ‫‪ 0.15 mol .c‬ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪NaCl‬‬

‫‪ 1.35 ×10-2 mol .d‬ﻣﻦ ﺍﻟﻤﻴﺜﺎﻥ ‪CH4‬‬

‫‪ .14‬ﺭﺗـﺐ ﺍﻟﻌﻴﻨـﺎﺕ ﺍﻟﺜـﻼﺙ ﺍﻵﺗﻴـﺔ ﻣـﻦ ﺍﻷﺻﻐﺮ ﺇﻟـﻰ ﺍﻷﻛﺒﺮ ﺑﺤﺴـﺐ ﻋﺪﺩ‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ‪:‬‬ ‫‪25‬‬ ‫‪ 1.25 × 10 atoms‬ﻣﻦ ﺍﻟﺨﺎﺭﺻﻴﻦ‬

‫‪ 3.56 mol‬ﻣﻦ ﺍﻟﺤﺪﻳﺪ ‪Fe‬‬

‫‪ 6.78 × 10 molecules‬ﻣﻦ ﺍﻟﺠﻠﻮﻛﻮﺯ ‪C6H12O6‬‬ ‫‪22‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪5-2‬‬ ‫ا ﻫﺪاف‬

‫ﺗﺮﺑﻂ ﻛﺘﻠﺔ ﺍﻟــﺬﺭﺓ ﺑﻜﺘﻠﺔ ﻣﻮﻝ‬ ‫ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺬﺭﺍﺕ‪.‬‬ ‫ﲢﻮﻝ ﺑﲔ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻌﻨﴫ‬ ‫ﻭﻛﺘﻠﺘﻪ‪.‬‬ ‫ﲢﻮﻝ ﺑﲔ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻌﻨﴫ‬ ‫ﻭﻋﺪﺩ ﺫﺭﺍﺗﻪ‪.‬‬

‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﻣﻌﺎﻣـﻞ ﺍﻟﺘﺤﻮﻳـﻞ‪ :‬ﻧﺴـﺒﺔ ﺑـﲔ ﻗﻴﻢ‬ ‫ﻣﺘﻜﺎﻓﺌـﺔ‪ ،‬ﻳﺴـﺘﺨﺪﻡ ﻟﻠﺘﻌﺒـﲑ ﻋـﻦ‬ ‫ﺍﻟﻜﻤﻴﺔ ﻧﻔﺴﻬﺎ ﺑﻮﺣﺪﺍﺕ ﳐﺘﻠﻔﺔ‪.‬‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‬

‫اﻟﻜﺘﻠﺔ واﻟﻤﻮل‬ ‫‪Mass and the Mole‬‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﳛﺘﻮﻱ ﺍﳌـﻮﻝ ﹰ‬ ‫ﺩﺍﺋﲈ ﻋﲆ ﺍﻟﻌـﺪﺩ ﻧﻔﺴـﻪ ﻣـﻦ ﺍﳉﺴﻴﲈﺕ‪،‬‬ ‫ﻭﻣـﻊ ﺫﻟﻚ‪ ،‬ﻓﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﴏ ﺍﳌﺨﺘﻠﻔﺔ ﳍﺎ ﻛﺘﻞ ﳐﺘﻠﻔﺔ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻋﻨﺪ ﴍﺍﺀ ﺩﺭﺯﻥ ﻣﻦ ﺍﻟﺒﻴﺾ‪ ،‬ﺑﺈﻣﻜﺎﻧﻚ ﺍﺧﺘﻴﺎﺭ‬ ‫ﺃﺣﺠﺎﻡ ﳐﺘﻠﻔﺔ‪ :‬ﺻﻐﲑﺓ‪ ،‬ﻭﺳﻂ‪ ،‬ﻭﻛﺒﲑﺓ‪ .‬ﻻ ﻳﺆﺛﺮ ﺣﺠﻢ ﺍﻟﺒﻴﻀﺔ ﰲ ﻋﺪﺩ ﻣﺎ ﳛﺘﻮﻳﻪ‬ ‫ﺗﻜﻮﻥ ﺍﳌﻮﻝ‪.‬‬ ‫ﺍﻟﺼﻨﺪﻭﻕ‪ .‬ﻭﻫﺬﺍ ﻭﺿﻊ ﻣﺸﺎﺑﻪ ﳊﺠﻢ ﺍﻟﺬﺭﺍﺕ ﺍﻟﺘﻲ ﹼ‬

‫ﻛﺘلﺔ المﻮﻝ ‪The Mass of a Mole‬‬ ‫ﻟﻦ ﺗﺘﻮﻗﻊ ﺃﻥ ﻛﺘﻠﺔ ﺩﺭﺯﻥ ﻣﻦ ﺍﻟﻠﻴﻤﻮﻥ ﺗﺴﺎﻭﻱ ﻛﺘﻠﺔ ﺩﺭﺯﻥ ﻣﻦ ﺍﻟﺒﻴﺾ؛ ﻷﻥ‬ ‫ﺍﻟﺒﻴﺾ ﻭﺍﻟﻠﻴﻤﻮﻥ ﳜﺘﻠﻔﺎﻥ ﰲ ﺍﳊﺠﻢ ﻭﺍﻟﱰﻛﻴﺐ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪ ،‬ﻓﻤﻦ ﻏﲑ ﺍﳌﻔﺎﺟﺊ‬ ‫ﺇ ﹰﺫﺍ ﺃﻥ ﺗﻜﻮﻥ ﳍﲈ ﻛﺘﻞ ﳐﺘﻠﻔﺔ‪ ،‬ﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﻟﺸﻜﻞ ‪ .5-4‬ﻟﺬﻟﻚ ﻓﺈﻥ‬ ‫ﻛﻤﻴﺘﲔ ﻣﻘﺪﺍﺭ ﻛﻞ ﻣﻨﻬﲈ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﻣﺎﺩﺗﲔ ﳐﺘﻠﻔﺘﲔ ﳍﲈ ﻛﺘﻠﺘﺎﻥ ﳐﺘﻠﻔﺘﺎﻥ؛‬ ‫ﻛﻴﻤﻴﺎﺋﻴﺎ ﳐﺘﻠ ﹰﻔﺎ‪ .‬ﻓﻠﻮ ﻭﺿﻌﺖ ﻣﻮ ﹰ‬ ‫ﻷﻥ ﱟ‬ ‫ﻻ ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ﺍﻟﻜﺮﺑﻮﻥ‬ ‫ﻟﻜﻞ ﻣﻨﻬﲈ ﺗﺮﻛﻴ ﹰﺒﺎ‬ ‫ﹰﹼ‬ ‫ﻣﺜﻼ‪ ،‬ﻭﻣﻮ ﹰ‬ ‫ﹰ‬ ‫ﻻ ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ﺍﻟﻨﺤﺎﺱ ﰲ ﻣﻴﺰﺍﻧﲔ ﻓﺴﱰ￯ ﻓﺮ ﹰﻗﺎ ﰲ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﻛﺎﻟﺬﻱ‬

‫ﺗﺮﺍﻩ ﰲ ﻛﺘﻞ ﺍﻟﺒﻴﺾ‪ ،‬ﻭﺍﻟﻠﻴﻤﻮﻥ‪ .‬ﻭﻫﺬﺍ ﳛﺪﺙ ﻷﻥ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﲣﺘﻠﻒ‬ ‫‪23‬‬ ‫ﻋﻦ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﻟﺬﻟﻚ ﻓﺈﻥ ﻛﺘﻠﺔ ‪ 6.02 × 10 atoms‬ﻣﻦ ﺍﻟﻜﺮﺑﻮﻥ‬ ‫‪23‬‬ ‫ﻻ ﺗﺴﺎﻭﻱ ﻛﺘﻠﺔ ‪ 6.02 × 10 atoms‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‪.‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 5-4‬ﻛﺘﻠـﺔ ﺩﺭﺯﻥ ﻣـﻦ ﺍﻟﻠﻴﻤـﻮﻥ‬ ‫ﺍﻹﻏﺮﻳﻖ‬ ‫ﺗﺴﺎﻭﻱ ﻣﻦ‬ ‫ﺍﻟﺸﻜﻞ ‪ 1-3‬ﻛﺜﲑ‬ ‫ﻓﻼﺳﻔﺔﺩﺭﺯﻥ ﻣﻦ‬ ‫ﺿﻌﻒ ﻛﺘﻠﺔ‬ ‫ﻣﻦ‬ ‫ﻣﻜﻮﻧﺔ‬ ‫ﺍﻋﺘﻘﺪﻭﺍ‬ ‫ﺍﻟﻔـﺮﻕ‬ ‫ﺍﻟﺒﻴـﺾﺃﻥﺗﻘﺮﻳﺍﳌ ﹰﺒـﺎ‪.‬‬ ‫ـﺎﺩﺓﻭﻳﻌـﺪﹼ‬ ‫ﻭﺍﳌﺎﺀ‪،‬‬ ‫ﺑﻴـﻦ ﻋﻨﺎﴏ‪:‬‬ ‫ﺃﺭﺑﻌﺔ‬ ‫ﻣـﻨﻄـﻘﻴﺎ؛ ﻷﻥ‬ ‫ﺍﻟﻜﺘﻠﺘــﲔﺍﻟﱰﺍﺏ‪ ،‬ﹰﹼ‬ ‫ﺑﺮﺑﻂ‬ ‫ﺍﻟﻠﻴﻤــﻮﻥﻭﺍﻟﻨﺎﺭ‪.‬‬ ‫ﻭﺍﳍــﻮﺍﺀ‪،‬‬ ‫ﻭﻗﺎﻣﻮﺍﺍﻟﺒﻴﺾ‬ ‫ﳐﺘﻠﻒ ﻋـﻦ‬ ‫ﻭﺇﻥ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲﻋﻨﴫ‪.‬‬ ‫ﻣﻌﻴﻨﺔ ﻟﻜﻞ‬ ‫ﺧﻮﺍﺹ‬ ‫ﻭﺣﺠﻤﻪ‪.‬‬ ‫ﺗﺮﻛﻴﺒﻪ‬ ‫ﰲ‬ ‫ﻣﺰﺝ ﺍﳋــﻮﺍﺹ ﺍﳌﺘﻌﺎﻛﺴﺔ ﻣﺜﻞ‬ ‫ﺳﺎﺧﻦ ﻭﺑــﺎﺭﺩ‪ ،‬ﺭﻃﺐ ﻭﺟﺎﻑ‬ ‫ﻳﻌﻜﺲ ﺍﻟﺘﲈﺛﻞ ﺍﳌــﻼﺣــﻆ ﰲ‬ ‫ﺍﻟﻄﺒﻴﻌﺔ‪ .‬ﻫﺬﻩ ﺍﻷﻓﻜﺎﺭ ﺍﻷﻭﻟﻴﺔ‬ ‫ﱂ ﺗﻜﻦ ﺻﺤﻴﺤﺔ ﻭﻻ ﻋﻠﻤﻴﺔ‪.‬‬

‫‪47‬‬

‫ﺍﻟﺸﻜﻞ ‪ 5-5‬ﻣﻮﻝ ﻣﻦ ﺍﳊﺪﻳﺪ‪ ،‬ﳑﺜﻼ ﺑﻜﻴﺲ ﻣﻦ ﺍﳉﺴﻴﲈﺕ‪،‬‬ ‫ﳛﺘﻮﻱ ﻋﲆ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻭﻟﻪ‬ ‫ﻛﺘﻠﺔ ﻣﺴﺎﻭﻳﺔ ﻟﻜﺘﻠﺘﻪ ﺍﻟﺬﺭﻳﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ‪.‬‬ ‫ﻃﺒﻖ ﻣﺎ ﻛﺘﻠﺔ ﻣﻮﻝ ﻣﻦ ﺍﻟﻨﺤﺎﺱ؟‬

‫ﳌﺰﻳﺪ ﻣﻦ ﺍﳌﻌﻠﻮﻣﺎﺕ ﻋﻦ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ‪:‬‬ ‫‪www.obeikaneducation.com‬‬

‫= ‪ 6.02 × 10 atoms‬ﻣﻦ ﺍﻟﺤﺪﻳﺪ‬ ‫‪23‬‬

‫‪ 1mol‬ﻣﻦ ﺍﻟﺤﺪﻳﺪ‬ ‫‪C11-02C-828378-08‬‬

‫‪ á«dƒŸG á∏àµdG‬ﻛﻴﻒ ﺗﺮﺗﺒﻂ ﻛﺘﻠﺔ ﺫﺭﺓ ﻭﺍﺣﺪﺓ ﺑﻜﺘﻠﺔ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺗﻠﻚ ﺍﻟﺬﺭﺓ؟ ﺗﺬﻛﺮ‬ ‫ﺃﻥ ﺍﳌﻮﻝ ﻳﻌﺮﻑ ﻋﲆ ﺃﻧﻪ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ‪ 12-‬ﰲ ‪ 12 g‬ﻣﻨﻪ‪ .‬ﻭﻣﻦ ﺛﻢ ﻓﻜﺘﻠـﺔ‬ ‫‪1mol‬ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑـﻮﻥ ‪ 12-‬ﻫﻲ ‪ .12 g‬ﻭﺳـﻮﺍ ﹰﺀ ﻛﻨﺖ ﹰﹼ‬ ‫ﻣﻬﺘﲈ ﺑﺬﺭﺓ ﻭﺍﺣﺪﺓ ﺃﻭ ﺑﻌﺪﺩ‬ ‫ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﻟﺬﺭﺍﺕ ) ‪ (1 mol‬ﻓﺈﻥ ﻛﺘﻞ ﲨﻴﻊ ﺍﻟﺬﺭﺍﺕ ﺗﻢ ﺗﻌﻴﻴﻨﻬﺎ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻜﺘﻠﺔ ﺫﺭﺓ‬ ‫ﹴ‬ ‫ﻭﺍﺣﺪ ﻣﻦ ﺃﻱ ﻣﺎﺩﺓ ﻧﻘﻴﺔ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺮﺑﻮﻥ ‪ .12-‬ﻭﺗﺴﻤﻰ ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ ﳌﻮﻝ‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻷﻱ ﻋﻨﴫ ﺗﺴﺎﻭﻱ ﻋﺪﺩ ﹰﹼﻳﺎ ﻛﺘﻠﺘﻪ ﺍﻟﺬﺭﻳﺔ‪ ،‬ﻭﻭﺣﺪﲥﺎ ‪ .g/mol‬ﻭﻛﲈ ﻫﻮ ﻣﺒﲔ‬ ‫ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺪﻭﺭﻱ‪ ،‬ﻓﺈﻥ ﻛﺘﻠﺔ ﺫﺭﺓ ﺍﳊﺪﻳﺪ ﺍﻟﻮﺍﺣﺪﺓ ﻣﻘﺪﺍﺭﻫﺎ ‪ .55.845 amu‬ﻭﻋﻠﻴﻪ‪،‬‬ ‫ﻓﺎﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﺤﺪﻳﺪ ﺗﺴﺎﻭﻱ ‪ .55.845 g/mol‬ﻻﺣﻆ ﺃﻧﻪ ﺑﻘﻴﺎﺱ ‪ 55.845 g‬ﻣﻦ‬ ‫‪23‬‬ ‫ﺍﳊﺪﻳﺪ ﺗﻜﻮﻥ ﺑﻄﺮﻳﻘﺔ ﻏﲑ ﻣﺒﺎﴍﺓ ﻗﺪ ﻋﺪﺩﺕ ‪ 6.02 × 10 atoms‬ﻣﻨﻪ‪ .‬ﺍﻟﺸﻜﻞ ‪5-5‬‬ ‫ﻳﻮﺿﺢ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻭﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻌﻨﴫ‪.‬‬

‫ﻣﺨﺘﺒﺮ ﺣﻞ اﻟﻤﺸﻜﻼت‬ ‫ﺻﻴﺎﻏﺔ ﻧﻤﻮذج‬

‫‪ ?äCGôb GPÉe‬ﺍﺳـﺘﻨﺘﺞ ﻟﻤﺎﺫﺍ ﻛﺎﻥ ﻣﻦ ﺍﻟﺼﻌـﺐ ﻋﻠﻰ ﺩﻳﻤﻘﺮﻳﻄﺲ ﺃﻥ ﻳﺪﺍﻓﻊ‬ ‫ﺍﻟﻤﻮﻟﻴـﺔ‪ ،‬ﻭﻋـﺪﺩ ﺃﻓﻮﺟـﺎﺩﺭﻭ ﻭﺍﻟﻜﺘﻠـﺔ‬ ‫ﻛﻴـﻒ ﺗﺮﺗﺒـﻂ ﺍﻟﻜﺘﻠـﺔ‬ ‫ﻋﻦ ﺃﻓﻜﺎﺭﻩ؟‬

‫ﺍﻟﺬﺭﻳـﺔ؟ ﻳﻤﻜـﻦ ﺃﻥ ﻳﻮﻓـﺮ ﺍﻟﻨﻤـﻮﺫﺝ ﺍﻟﻨـﻮﻭﻱ ﻟﻠـﺬﺭﺓ ﺻـﻮﺭﺓ‬ ‫ﻣﺒﺴـﻄﺔ ﻟﻼﺭﺗﺒﺎﻃـﺎﺕ ﺑﻴـﻦ ﺍﻟﻤـﻮﻝ‪ ،‬ﻭﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻭﻋﺪﺩ‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬‬ ‫اﻟﺘﺤﻠﻴﻞ‬

‫ﻳﻈﻬـﺮ ﺍﻟﺮﺳـﻢ ﻋـﻦ ﺍﻟﻴﺴـﺎﺭ ﻧﻤـﺎﺫﺝ ﺃﻧﻮﻳـﺔ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪1-‬‬ ‫ﻭﺍﻟﻬﻴﻠﻴـﻮﻡ ‪ .4-‬ﺗﺤﺘﻮﻱ ﻧـﻮﺍﺓ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ 1-‬ﻋﻠﻰ ﺑﺮﻭﺗﻮﻥ‬ ‫ﻭﺍﺣـﺪ ﺑﻜﺘﻠـﺔ ﻣﻘﺪﺭﺍﻫـﺎ ‪ ،1.007 amu‬ﻭﻗـﺪ ﻗـﺪﺭﺕ ﻛﺘﻠﺔ‬ ‫ﺍﻟﺒﺮﻭﺗـﻮﻥ ﺑﺎﻟﺠﺮﺍﻣـﺎﺕ ﻋﻠـﻰ ﺃﻧﻬـﺎ ‪.1.672 × 10-24 g‬‬ ‫ﺗﺤﺘـﻮﻱ ﻧﻮﺍﺓ ﺍﻟﻬﻴﻠﻴﻮﻡ– ‪ 4‬ﻋﻠـﻰ ﺑﺮﻭﺗﻮﻧﻴﻦ ﻭﻧﻴﻮﺗﺮﻭﻧﻴﻦ‪ ،‬ﻭﻟﻬﺎ‬ ‫ﻛﺘﻠﺔ ﻣﻘﺪﺭﺍﻫﺎ ‪.4 amu‬‬ ‫اﻟﺘﻔﻜﻴﺮ اﻟﻨﺎﻗﺪ‬

‫ﻃﺒـﻖ‪ :‬ﻣﺎ ﻛﺘﻠﺔ ﺫﺭﺓ ﺍﻟﻬﻴﻠﻴﻮﻡ ﺍﻟﻮﺍﺣﺪﺓ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ؟ )ﻛﺘﻠﺔ‬ ‫‪ .1‬ﹼ‬ ‫ﺍﻟﻨﻴﻮﺗﺮﻭﻥ ﻣﺴﺎﻭﻳﺔ ﺗﻘﺮﻳ ﹰﺒﺎ ﻟﻜﺘﻠﺔ ﺍﻟﺒﺮﻭﺗﻮﻥ( ‪.‬‬

‫‪48‬‬

‫‪Helium - 4‬‬ ‫ﺍﻟﻬﻴﻠﻴﻮﻡ‪4-‬‬

‫‪Hydrogen - 1‬‬ ‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪1-‬‬

‫‪C11-12C-828378-08‬ﻋﻠﻰ ﺳـﺘﺔ ﺑﺮﻭﺗﻮﻧﺎﺕ‬ ‫‪ .2‬ﺍﺭﺳـﻢ ﺍﻟﻜﺮﺑﻮﻥ ‪ 12-‬ﻳﺤﺘﻮﻱ‬ ‫ﻭﺳـﺘﺔ ﻧﻴﻮﺗﺮﻭﻧـﺎﺕ‪ .‬ﺍﺭﺳـﻢ ﻧـﻮﺍﺓ ﺍﻟﻜﺮﺑـﻮﻥ ‪،12-‬‬ ‫ﻭﺍﺣﺴﺐ ﻛﺘﻠﺔ ﺍﻟﺬﺭﺓ ﺍﻟﻮﺍﺣﺪﺓ ﺑﻮﺣﺪﺗﻲ ‪.g ، amu‬‬ ‫ﻃﺒﻖ‪ :‬ﻣﺎ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪ 1-‬ﻓﻲ ﻋﻴﻨﺔ ﻛﺘﻠﺘﻬﺎ‬ ‫‪ .3‬ﹼ‬ ‫‪1.007g‬؟ ﺗﺬﻛـﺮ ﺃﻥ ‪ 1.007 amu‬ﻫـﻲ ﻛﺘﻠﺔ ﺫﺭﺓ‬ ‫ﻗـﺮﺏ ﺇﺟﺎﺑﺘﻚ ﺇﻟﻰ‬ ‫ﻭﺍﺣﺪﺓ ﻣـﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴـﻦ‪ .1-‬ﹼ‬ ‫ﺃﻗﺮﺏ ﺟﺰﺀ ﻣﻦ ﻣﺎﺋﺔ‪.‬‬ ‫ﻃﺒﻖ‪ :‬ﻟﻮ ﻛﺎﻧﺖ ﻟﺪﻳﻚ ﻋﻴﻨﺘﺎﻥ ﻣﻦ ﺍﻟﻬﻴﻠﻴﻮﻡ ﻭﺍﻟﻜﺮﺑﻮﻥ‬ ‫‪ .4‬ﹼ‬ ‫ﺗﺤﺘﻮﻳـﺎﻥ ﻋﻠﻰ ﻋـﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﻟـﺬﺭﺍﺕ‪ ،‬ﻓﻜﻢ‬ ‫ﺗﻜﻮﻥ ﻛﺘﻠﺔ ﻛﻞ ﻋﻴﻨﺔ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ؟‬ ‫‪ .5‬ﺍﺳـﺘﻨﺘﺞ ﻣـﺎﺫﺍ ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﺴـﺘﻨﺘﺞ ﻋـﻦ ﺍﻟﻌﻼﻗﺔ ﺑﻴﻦ‬ ‫ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﻭﻛﺘﻠﺔ ﻛﻞ ﻋﻴﻨﺔ؟‬

‫اﺳﺘﺨﺪاﻡ الﻜﺘلﺔ المﻮلﻴﺔ ‪Using Molar Mass‬‬ ‫–‪ á∏àc ¤EG ä’ƒŸG πjƒ‬ﺍﻓﺮﺽ ﺃﻧﻪ ﺧﻼﻝ ﻋﻤﻠﻚ ﰲ ﳐﺘﱪ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﺣﺘﺠﺖ‬ ‫ﺇﱃ ‪ 3.00mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ ‪ Cu‬ﻟﺘﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ‪ ،‬ﻓﻜﻴﻒ ﺗﻘﻴﺲ ﻫﺬﻩ ﺍﻟﻜﻤﻴﺔ؟‬ ‫ﻳﻤﻜﻦ ﲢﻮﻳﻞ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻨﺤﺎﺱ ﺇﱃ ﻛﺘﻠﺔ ﻣﻜﺎﻓﺌﺔ ﺗﻘﺎﺱ ﺑﺎﳌﻴﺰﺍﻥ‪ .‬ﻭﳊﺴﺎﺏ‬ ‫ﻛﺘﻠﺔ ﻋﺪﺩ ﻣﻌﲔ ﻣﻦ ﺍﳌﻮﻻﺕ ﺍﴐﺏ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﰲ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪:‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ )‪ = (g‬ﻋﺪﺩ ﺍﳌﻮﻻﺕ )‪× (mol‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ )‪(g‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﺭﺟﻊ ﺇﱃ ﺍﳌﻮﻗﻊ ﺍﻹﻟﻜﺘﺮﻭﻧﻲ‪:‬‬ ‫‪www.obeikaneducation.com‬‬

‫‪1 mol‬‬

‫ﺇﺫﺍ ﺗﻔﺤﺼـﺖ ﺍﳉـﺪﻭﻝ ﺍﻟـﺪﻭﺭﻱ ﻟﻠﻌﻨـﺎﴏ ﻓﺴـﺘﺠﺪ ﺃﻥ ﺍﻟﻨﺤـﺎﺱ ‪ ،29-‬ﻟـﻪ‬ ‫ﻛﺘﻠـﺔ ﺫﺭﻳـﺔ ﻣﻘﺪﺍﺭﻫـﺎ ‪ ،63.45 amu‬ﻭﺃﻧـﺖ ﺗﻌﻠـﻢ ﺃﻥ ﺍﻟﻜﺘﻠـﺔ ﺍﳌﻮﻟﻴـﺔ ﻟﻠﻌﻨـﴫ‬ ‫)‪ (g/mol‬ﺗﺴﺎﻭﻱ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ )ﻣﻌﱪ ﻋﻨﻬﺎ ﺑﻮﺣﺪﺓ ‪ ،(amu‬ﻟﺬﻟﻚ‪،‬‬ ‫ﻓﻜﺘﻠﺔ ﺍﻟﻨﺤﺎﺱ ﺍﳌﻮﻟﻴﺔ ﻫﻲ ‪ ،63.546 g/mol‬ﻭﺑﺎﺳﺘﺨﺪﺍﻣﻬﺎ‪ ،‬ﻳﻤﻜﻨﻚ‬ ‫ﲢﻮﻳﻞ ‪ 3.00 mol‬ﻧﺤﺎﺱ ﺇﱃ ﺟﺮﺍﻣﺎﺕ ﻧﺤﺎﺱ‪.‬‬ ‫‪ 63.546g‬ﻣﻦ ‪Cu‬‬ ‫ﻛﺘﻠﺔ ﺍﻟﻨﺤﺎﺱ ﺑﺎﳉﺮﺍﻣﺎﺕ ) ‪ 3.00 mol = (g‬ﻣﻦ ‪× Cu‬‬ ‫‪1mol‬ﻣﻦ ‪Cu‬‬

‫= ‪191g‬ﻣﻦ ‪Cu‬‬

‫ﻟﺬﻟﻚ‪ ،‬ﻭﻛﲈ ﻫﻮ ﻣﻮﺿﺢ ﰲ ﺍﻟﺸﻜﻞ ‪ ،5-6‬ﻳﻤﻜﻨﻚ ﻗﻴﺎﺱ ‪ 3.00 mol‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‬ ‫ﺍﻟﻼﺯﻣﺔ ﻟﻠﺘﻔﺎﻋﻞ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻣﻴﺰﺍﻥ ﻟﺘﻌﻴﲔ ‪ 191g‬ﻣﻦ ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﺍﻟﻌﻜﴘ )ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﺍﳌﻮﻻﺕ( ﻳﺘﻀﻤﻦ ﹰ‬ ‫ﺃﻳﻀﺎ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺑﻮﺻﻔﻬﺎ ﻣﻌﺎﻣﻞ‬ ‫ﲢﻮﻳﻞ‪ ،‬ﻭﻟﻜﻦ ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻫﻮ ﺍﳌﺴﺘﺨﺪﻡ‪ .‬ﻓﻬﻞ ﺑﺈﻣﻜﺎﻧﻚ ﺗﻔﺴﲑ‬ ‫ﺍﻟﺴﺒﺐ؟‬

‫المطﻮﻳات‬

‫ﺿﻤـﻦ ﻣﻌﻠﻮﻣﺎﺕ ﻣﻦ ﻫـﺬﺍ ﺍﻟﺒﻨﺪ ﻓﻲ‬ ‫ﹼ‬ ‫ﻣﻄﻮﻳﺘﻚ‪.‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 5-6‬ﻟﻘﻴﺎﺱ ‪3.00 mol‬‬ ‫ﻣـﻦ ﺍﻟﻨﺤـﺎﺱ‪ ،‬ﺿـﻊ‬

‫ﻭﺭﻗـﺔ ﻭﺯﻥ ﻋﲆ ﺍﳌﻴﺰﺍﻥ‪،‬‬ ‫ﻭﺻ ﱢﻔـــﺮﻩ‪ ،‬ﺛـﻢ ﺿــﻊ‬ ‫‪ 191 g‬ﻣـﻦ ﺍﻟﻨﺤﺎﺱ‪.‬‬

‫‪49‬‬

‫‪ AÉ«MC’G º∏Y ™e §HôdG‬ﻳﻜﺘﺸﻒ ﻋﻠﲈﺀ ﺍﳋﻠﻴﺔ ﺑﺮﻭﺗﻴﻨﺎﺕ ﺟﺪﻳﺪﺓ ﺑﺎﺳﺘﻤﺮﺍﺭ‪ .‬ﻭﺑﻌﺪ‬ ‫ﺍﻛﺘﺸﺎﻑ ﺟﺰﻱﺀ ﺣﻴﻮﻱ ﺟﺪﻳﺪ ﻳﻘﻮﻡ ﻋﺎﱂ ﺍﻷﺣﻴﺎﺀ ﺑﺘﻌﻴﲔ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ‬ ‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺗﻘﻨﻴﺔ ﻣﻄﻴﺎﻑ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﺍﻟﺬﻱ ﻳﻮﻓﺮ ـ ﺑﺎﻹﺿﺎﻓﺔ ﺇﱃ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ـ‬ ‫ﻣﻌﻠﻮﻣﺎﺕ ﺇﺿﺎﻓﻴﺔ ﺗﺴﺎﻋﺪ ﻋﲆ ﺍﻟﻜﺸﻒ ﻋﻦ ﺍﻟﱰﻛﻴﺐ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﻠﻤﺮﻛﺐ‪.‬‬

‫‪5-2 ∫Éãe‬‬ ‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﺍﳌﻮﻝ ﺇﱃ ﺍﻟﻜﺘﻠﺔ ‪ :‬ﺍﻟﻜﺮﻭﻡ ‪ Cr‬ﻋﻨﴫ ﺍﻧﺘﻘﺎﱄ‪ ،‬ﻳﺴﺘﺨﺪﻡ ﰲ ﻃﻼﺀ ﺍﳊﺪﻳﺪ ﻭﺍﻟﻔﻠﺰﺍﺕ ﳊﲈﻳﺘﻬﺎ ﻣﻦ ﺍﻟﺘﺂﻛﻞ‪ .‬ﺍﺣﺴﺐ‬ ‫ﻛﺘﻠﺔ ‪ 0.0450 mol‬ﻣﻦ ﺍﻟﻜﺮﻭﻡ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬ ‫ﻟﺪﻳﻚ ﻋﺪﺩ ﻣﻦ ﻣﻮﻻﺕ ﺍﻟﻜﺮﻭﻡ ﺍﻟﺘﻲ ﳚﺐ ﺣﺴﺎﺏ ﻛﺘﻠﺘﻬﺎ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻜﺮﻭﻡ ﻣﻦ ﺍﳉﺪﻭﻝ ﺍﻟﺪﻭﺭﻱ ﻟﻠﻌﻨﺎﴏ‪ .‬ﻭﺑﲈ ﺃﻥ‬ ‫ﺍﻟﻌﻴﻨﺔ ﺃﻗﻞ ﻣﻦ ‪ ،0.1 mol‬ﻓﺎﻹﺟﺎﺑﺔ ﳚﺐ ﺃﻥ ﺗﻜﻮﻥ ﺃﻗﻞ ﻣﻦ ‪ 0.1‬ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪.‬‬

‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ = ‪0.0450 mol‬‬

‫ﺍﳌﻄﻠﻮﺏ‬

‫ﻛﺘﻠﺔ ‪ = Cr‬؟‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻜﺮﻭﻡ = ‪52.00 g/mol‬‬ ‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬ ‫ﻋﻮﺽ ﺑﺎﻟﻘﻴﻢ ﺍﳌﻌﻠﻮﻣﺔ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﻭﺣﻠﻬﺎ‪.‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ )ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺟﺮﺍﻣﺎﺕ ﺍﻟﻜﺮﻭﻡ ﺑﻤﻮﻻﺗﻪ‪ ،‬ﺛﻢ ﹼ‬

‫ﻃﺒــــﻖ ﻣـــﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳــﻞ ﻛﺘﻠﺔ ﺍﻟﻜﺮﻭﻡ )‪ = (g‬ﻣﻮﻻﺕ ﺍﻟﻜﺮﻭﻡ )‪× (mol‬‬ ‫ﻋﻮﺽ ﺑﺎﻟﻤﻌﻄﻴﺎﺕ ﻭﺃﻭﺟﺪ ﺍﻟﺤﻞ = ‪ 0.0450 mol‬ﻣﻦ ‪× Cr‬‬ ‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫= ‪ 20.34g‬ﻣﻦ ‪Cr‬‬

‫ﺟﺮﺍﻣﺎﺕ ﺍﻟﻜﺮﻭﻡ )‪(g‬‬ ‫‪ 1mol‬ﻣﻦ ﺍﻟﻜﺮﻭﻡ‬

‫‪ 52.00g‬ﻣﻦ ‪Cr‬‬ ‫‪ 1mol‬ﻣﻦ ‪Cr‬‬

‫ﹸﺃﻋﻄﻲ ﺍﳉﻮﺍﺏ ﺑﺎﻟﻮﺣﺪﺍﺕ ﺍﻟﺼﺤﻴﺤﺔ )‪ ،(g‬ﻭﻫﻮ ﺃﻗﻞ ﻣﻦ ‪ 0.1 mol‬ﻛﲈ ﻫﻮ ﻣﺘﻮﻗﻊ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪ .15‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ )‪ (g‬ﻟﻜﻞ ﳑﺎ ﻳﲇ‪:‬‬ ‫‪ 3.57 mol .a‬ﻣﻦ ‪Al‬‬ ‫‪ 42.6 mol .b‬ﻣﻦ ‪Si‬‬

‫‪ .16‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ ﻟﻜﻞ ﳑﺎ ﻳﲇ ‪:‬‬ ‫‪ 3.54 × 10 mol .a‬ﻣﻦ ‪Co‬‬ ‫‪2‬‬

‫‪ 2.45 × 10-2 mol .b‬ﻣﻦ ‪Zn‬‬

‫‪50‬‬

‫‪5-3 ∫Éãe‬‬

‫ﺗﻮﺍﻓﺮﺍ ﰲ ﺍﻷﺭﺽ‪ ،‬ﻭﻳﻮﺟﺪ ﹰ‬ ‫ﺩﺍﺋﲈ ﻣﺘﺤﺪﹰ ﺍ ﻣﻊ ﻋﻨﺎﴏ‬ ‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﺍﳌﻮﻝ‪ :‬ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca‬ﻣﻦ ﺃﻛﺜﺮ ﺍﻟﻌﻨﺎﴏ‬ ‫ﹰ‬ ‫ﺃﺧﺮ￯ ﺑﺴﺒﺐ ﻧﺸﺎﻃﻪ ﺍﻟﻌﺎﱄ‪ .‬ﻣﺎ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﰲ ‪ 525 g‬ﻣﻨﻪ؟‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻋﻠﻴﻚ ﺃﻥ ﲢﻮﻝ ﻛﺘﻠﺔ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺇﱃ ﻣﻮﻻﺕ ﻛﺎﻟﺴﻴﻮﻡ‪ ،‬ﻓﻜﺘﻠﺔ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﻫﻨﺎ ﺃﻛﱪ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺑﺄﻛﺜﺮ ﻣﻦ ﻋﴩ‬ ‫ﻣﺮﺍﺕ‪ ،‬ﻟﺬﻟﻚ ﳚﺐ ﺃﻥ ﺗﻜﻮﻥ ﺍﻹﺟﺎﺑﺔ ﺃﻛﱪ ﻣﻦ ‪.10mol‬‬ ‫‪܃∏£ŸG‬‬

‫‪äÉ«£©ŸG‬‬

‫ﻋﺪﺩ ﻣﻮﻻﺕ ‪ = Ca‬؟‬

‫ﺍﻟﻜﺘﻠﺔ = ‪ 525 g‬ﻣﻦ ‪Ca‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟـ ‪40.08 g/mol = Ca‬‬ ‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺳﺘﺨﺪﻡ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ )ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﻣﻮﻻﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺑﺠﺮﺍﻣﺎﺗﻪ‪ ،‬ﻭﻋﻮﺽ ﺍﻟﻘﻴﻢ ﺍﳌﻌﻠﻮﻣﺔ‪،‬‬ ‫ﻭﺣﻞ‪:‬‬ ‫‪ 1 mol‬ﻣﻦ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﻣﻮﻻﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ )‪ = (mol‬ﻛﺘﻠﺔ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ )‪× (g‬‬ ‫ﺟﺮﺍﻣﺎﺕ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ )‪(g‬‬ ‫ﻋﻮﺽ ﺑﺎﻟﻤﻌﻄﻴﺎﺕ ﻭﺃﻭﺟﺪ ﺍﻟﺤﻞ‬ ‫‪3‬‬

‫= ‪× 525 g Ca‬‬

‫‪ÜGƒ÷G ˃≤J‬‬

‫‪ 1 mol‬ﻣﻦ ‪Ca‬‬ ‫‪ 40.08 g‬ﻣﻦ ‪Ca‬‬

‫=‬

‫‪ 13.1 mol‬ﻣﻦ ‪Ca‬‬

‫ﺍﳉﻮﺍﺏ ﺃﻛﱪ ﻣﻦ ‪ 10mol‬ﻣﻮﻻﺕ ﻛﲈ ﻫﻮ ﻣﺘﻮﻗﻊ‪ ،‬ﻭﻟﻪ ﻭﺣﺪﺓ ﺍﻟﻘﻴﺎﺱ ﺍﳌﻄﻠﻮﺑﺔ‪ ،‬ﺍﳌﻮﻝ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .17‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﳌﻮﻻﺕ)‪ (mol‬ﰲ ﱟ‬ ‫ﻛﻞ ﳑﺎ ﻳﲇ‪:‬‬ ‫‪ 25.5 g .a‬ﻣﻦ ‪Ag‬‬ ‫‪ 300.0 g .b‬ﻣﻦ ‪S‬‬

‫ﺣﻮﻝ ﹰﹼ‬ ‫ﻛﻼ ﻣﻦ ﺍﻟﻜﺘﻞ ﺍﻟﺘﺎﻟﻴﺔ ﺇﱃ ﻣﻮﻻﺕ‪:‬‬ ‫‪ .18‬ﲢﺪﹼ ‪ :‬ﹼ‬ ‫‪ 1.25 × 103 g .a‬ﻣﻦ ‪Zn‬‬ ‫‪ 1.00 kg .b‬ﻣﻦ ‪Fe‬‬

‫‪51‬‬

‫‪äGQòdGh á∏àµdG ÚH πjƒëàdG‬‬ ‫ﺇﻧﻚ ﻻ ﺗﺴﺘﻄﻴﻊ ﺃﻥ ﺗﻘﻮﻡ ﺑﺘﺤﻮﻳﻞ ﻣﺒﺎﴍ ﻣﻦ ﻛﺘﻠﺔ ﺍﳌﺎﺩﺓ ﺇﱃ ﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ ﺍﳌﻜﻮﻧﺔ‬ ‫ﳍﺎ‪ ،‬ﺇﺫ ﻻ ﺑﺪ ﺃﻥ ﲢﻮﻝ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﰲ ﺍﻟﺒﺪﺍﻳﺔ‪ ،‬ﻭﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﺍﳌﻜﻮﻧﺔ ﻣﻦ‬ ‫ﺧﻄﻮﺗﲔ ﻣﻮﺿﺤﺔ ﰲ ﺍﳌﺜﺎﻝ ‪.5-4‬‬

‫‪5-4 ∫Éãe‬‬

‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﺍﻟﺬﺭﺍﺕ‪ :‬ﺍﻟﺬﻫﺐ ‪ Au‬ﻫﻮ ﺃﺣﺪ ﻓﻠﺰﺍﺕ ﺍﻟﻌﻤﻠﺔ )ﺍﻟﻨﺤﺎﺱ‪ ،‬ﻭﺍﻟﻔﻀﺔ‪ ،‬ﻭﺍﻟﺬﻫﺐ(‪ .‬ﻣﺎ ﻋﺪﺩ ﺫﺭﺍﺕ‬ ‫ﺍﻟﺬﻫﺐ ﰲ ﻋﻤﻠﺔ ﺫﻫﺒﻴﺔ ﻛﺘﻠﺘﻬﺎ ‪31.1 g‬؟‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﻋﻠﻴﻚ ﺃﻥ ﲢﺴﺐ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ﻛﺘﻠﺔ ﻣﻌﻴﻨﺔ ﻣﻦ ﺍﻟﺬﻫﺐ‪ ،‬ﻭﺑﲈ ﺃﻧﻚ ﻻ ﺗﺴﺘﻄﻴﻊ ﺍﻟﺘﺤﻮﻳﻞ ﻣﺒﺎﴍﺓ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﻋﺪﺩ‬ ‫ﺍﻟﺬﺭﺍﺕ‪ ،‬ﻓﻌﻠﻴﻚ ﺃﻭ ﹰ‬ ‫ﻻ ﺃﻥ ﲢﻮﻝ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﻣﻮﻻﺕ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪ ،‬ﺛﻢ ﲢﻮﻝ ﺍﳌﻮﻻﺕ ﺇﱃ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ‬ ‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‪.‬‬ ‫ﺑﲈ ﺃﻥ ﻛﺘﻠﺔ ﺍﻟﺬﻫﺐ ﺍﳌﻌﻄﺎﺓ ﻫﻲ ﹸﺳﺪﺱ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﺬﻫﺐ )‪ .(196.97 g/mol‬ﻟﺬﺍ‪ ،‬ﻓﻌﺪﺩ ﺫﺭﺍﺕ ﺍﻟﺬﻫﺐ ﳚﺐ‬ ‫ﺃﻥ ﺗﻜﻮﻥ ﹸﺳﺪﹾ ﺱ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﺗﻘﺮﻳ ﹰﺒﺎ‪.‬‬ ‫ﺍﳌﻄﻠﻮﺏ‬ ‫ﺍﳌﻌﻄﻴﺎﺕ‬ ‫ﻋﺪﺩ ﺫﺭﺍﺕ ‪ = Au‬؟‬ ‫ﺍﻟﻜﺘﻠﺔ = ‪ 31.1 g‬ﻣﻦ ‪Au‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ = ‪196.97 g/mol‬‬

‫ﺍﺳﺘﺨﺪﻡ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ)ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﻣﻮﻻﺕ ﺍﻟﺬﻫﺐ ﺑﺠﺮﺍﻣﺎﺗﻪ‪.‬‬

‫ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺬﻫﺐ )‪ = (mol‬ﻛﺘﻠﺔ ﺍﻟﺬﻫﺐ )‪ 1 mol ×(Au‬ﻣﻦ ﺍﻟﺬﻫﺐ‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﺟﺮﺍﻣﺎﺕ ﺍﻟﺬﻫﺐ )‪(g‬‬ ‫‪1 mol‬ﻣﻦ ‪Au‬‬ ‫= ‪0.158 mol‬ﻣﻦ ‪Au‬‬ ‫ﻋﻮﺽ ﺑﺎﻟﻤﻌﻄﻴﺎﺕ‪ ،‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ = ‪31.1 g‬ﻣﻦ ‪× Au‬‬ ‫‪196.97 g‬ﻣﻦ ‪Au‬‬

‫ﻟﺘﺤﻮﻳﻞ ﺍﳌﻮﻻﺕ ﺇﱃ ﻋﺪﺩ ﺫﺭﺍﺕ‪ ،‬ﺍﴐﺏ ﰲ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‬

‫‪ 6.02 × 1023 atoms‬ﻣﻦ ﺍﻟﺬﻫﺐ‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺬﻫﺐ )‪ = (atom‬ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺬﻫﺐ )‪× (mol‬‬ ‫‪ 1 mol‬ﻣﻦ ﺍﻟﺬﻫﺐ‬ ‫ﻋﻮﺽ ﺑﺎﻟﻤﻌﻄﻴﺎﺕ‪ ،‬ﻭﺃﻭﺟﺪ ﺍﻟﺤﻞ‬

‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫= ‪0.158 mol‬ﻣﻦ ‪× Au‬‬

‫‪6.02 × 1023 atom‬ﻣﻦ ‪Au‬‬

‫= ‪ 9.51 × 1022 atoms‬ﻣﻦ ‪Au‬‬

‫‪ 1mol‬ﻣﻦ ‪Au‬‬

‫ﻭﻋ ﹼﱪ ﻋﻨﻪ ﺑﺎﻟﻮﺣﺪﺓ ﺍﻟﺼﺤﻴﺤﺔ )‪.(atoms‬‬ ‫ﹸﻋ ﹼﱪ ﻋﻦ ﺍﳉﻮﺍﺏ ﺑﺸﻜﻞ ﺻﺤﻴﺢ‪ ،‬ﺇﺫ ﺑﻠﻎ ﺳﺪﺱ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﺗﻘﺮﻳ ﹰﺒﺎ‪ ،‬ﻛﲈ ﻫﻮ ﻣﺘﻮﻗﻊ‪ ،‬ﹸ‬

‫‪52‬‬

‫‪5-5 ∫Éãe‬‬

‫ﲢﻮﻳﻞ ﺍﻟﺬﺭﺍﺕ ﺇﱃ ﻛﺘﻠﺔ‪ :‬ﺍﳍﻴﻠﻴﻮﻡ ‪ He‬ﻏﺎﺯ ﻧﺒﻴﻞ‪ ،‬ﻓﺈﺫﺍ ﺍﺣﺘﻮ￯ ﺑﺎﻟﻮﻥ ﻋﲆ ‪ 5.50 × 10 atoms‬ﻣﻦ ﺍﳍﻴﻠﻴﻮﻡ‪ ،‬ﻓﺎﺣﺴﺐ ﻛﺘﻠﺔ‬ ‫ﺍﳍﻴﻠﻴﻮﻡ ﻓﻴﻪ‪.‬‬ ‫‪22‬‬

‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﺣﻮﻝ ﺃﻭ ﹰﻻ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﺇﱃ ﻣﻮﻻﺕ‪ ،‬ﺛﻢ ﺣﻮﻝ ﺍﳌﻮﻻﺕ ﺇﱃ‬ ‫ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﳍﻴﻠﻴﻮﻡ ﻣﻌﻠﻮﻣﺔ ﻟﺪﻳﻚ‪ ،‬ﻭﻋﻠﻴﻚ ﺇﳚﺎﺩ ﻛﺘﻠﺔ ﺍﻟﻐﺎﺯ‪ .‬ﹼ‬ ‫ﺟﺮﺍﻣﺎﺕ‪.‬‬ ‫‪äÉ«£©ŸG‬‬

‫‪22‬‬

‫ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﳍﻴﻠﻴﻮﻡ = ‪ 5.50 × 10 atoms‬ﻣﻦ ‪He‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻬﻴﻠﻴﻮﻡ= ‪ 4.00 g/mol‬ﻣﻦ ‪He‬‬

‫‪2‬‬

‫‪܃∏£ŸG‬‬ ‫ﺍﻟﻜﺘﻠﺔ = ‪ g‬؟ ﻣﻦ ‪He‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺳﺘﺨﺪﻡ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ )ﻣﻘﻠﻮﺏ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﳌﻮﻻﺕ ﺑﻌﺪﺩ ﺍﻟﺬﺭﺍﺕ‬

‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﳍﻴﻠﻴﻮﻡ )‪ = (mol‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﳍﻴﻠﻴﻮﻡ )‪× (atoms‬‬ ‫ﻋﻮﺽ ‪22 atoms‬‬

‫‪ 5.50 ×10‬ﻣﻦ ‪He‬‬

‫ﻟﺘﺤﻮﻳﻞ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﺇﱃ ﻛﺘﻠﺔ‪ ،‬ﺍﴐﺏ ﰲ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‬

‫‪ 6.02 × 10‬ﻣﻦ ﺍﳍﻴﻠﻴﻮﻡ‬

‫‪ 1 mol He‬ﻣﻦ ‪He‬‬

‫‪ 6.02 × 1023 atoms‬ﻣﻦ ‪He‬‬

‫ﻛﺘﻠﺔ ﺍﳍﻴﻠﻴﻮﻡ ﺑﺎﳉﺮﺍﻣﺎﺕ )‪ = (g‬ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﳍﻠﻴﻮﻡ )‪× (mol‬‬

‫ﻋﻮﺽ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ 0.0914 mol = He‬ﺍﻟﻜﺘﻠﺔ‬ ‫ﺍﳌﻮﻟﻴﺔ ‪ ،4.00g/mol = He‬ﻭﺃﻭﺟﺪ ﺍﳊﻞ‬ ‫‪3‬‬

‫= ‪ 5.50 × 1022 atom‬ﻣﻦ ‪× He‬‬ ‫= ‪ 0.0914 mol‬ﻣﻦ ‪He‬‬

‫ﺿﺮﺏ ﻭﻗﺴﻤﺔ ﺍﻷﺭﻗﺎﻡ ﻭﺍﻟﻮﺣﺪﺍﺕ‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬

‫‪23 atoms‬‬

‫‪ 1 mol‬ﻣﻦ ﺍﳍﻴﻠﻴﻮﻡ‬

‫ﺟﺮﺍﻣﺎﺕ ﺍﳍﻴﻠﻴﻮﻡ )‪(g‬‬ ‫‪ 1mol‬ﻣﻦ ﺍﳍﻴﻠﻴﻮﻡ‬

‫‪ 4.00g‬ﻣﻦ ‪He‬‬ ‫= ‪ 0.0914 mol‬ﻣﻦ ‪× He‬‬ ‫‪ 1mol‬ﻣﻦ ‪He‬‬

‫= ‪ 0.366g‬ﻣﻦ ‪He‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﹸﻋﱪ ﻋﻦ ﺍﳉﻮﺍﺏ ﺑﺎﻟﻮﺣﺪﺓ ﺍﻟﺼﺤﻴﺤﺔ )‪.(g‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪ .19‬ﻣﺎ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﰲ ‪ 11.5g‬ﻣﻦ ﺍﻟﺰﺋﺒﻖ؟‬

‫‪ .20‬ﻣﺎ ﻛﺘﻠﺔ ‪ 1.50 × 1015 atoms‬ﻣﻦ ‪N‬؟‬ ‫‪ .21‬ﺃﺣﺴﺐ ﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ ﰲ ﻛﻞ ﳑﺎ ﻳﲇ‪:‬‬

‫‪ 4.56 × 103 g .a‬ﻣﻦ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ‪Si‬‬

‫‪ 0.120kg .b‬ﻣﻦ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ ‪Ti‬‬

‫‪53‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 5-7‬ﻳﻌﺪ ﺍﳌﻮﻝ ﺃﺳـﺎﺱ ﺍﻟﺘﺤﻮﻳﻞ ﻣﺎ ﺑﲔ ﺍﻟﻜﺘﻠﺔ ﻭﺍﳉﺴـﻴﲈﺕ)ﺍﻟﺬﺭﺍﺕ‪،‬ﺍﻷﻳﻮﻧﺎﺕ‪،‬ﺍﳉﺰﻳﺌﺎﺕ(‪.‬ﰲ ﺍﻟﺸـﻜﻞ ﲤﺜﻞ ﺍﻟﻜﺘﻠﺔ ﰲ ﺍﳌﻴﺰﺍﻥ ﻭﺍﳌﻮﻻﺕ ﰲ ﺣﻘﻴﺒﺔ‬ ‫ﲢﺘﻮﻱ ﻋﲆ ﺍﳉﺴﻴﲈﺕ‪ ،‬ﻭﺍﳉﺴﻴﲈﺕ ﺗﻨﺘﴩ ﻣﻦ ﺍﳊﻘﻴﺒﺔ‪ .‬ﲢﺘﺎﺝ ﺇﱃ ﺧﻄﻮﺗﲔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﺍﳉﺴﻴﲈﺕ ﺃﻭ ﺍﻟﻌﻜﺲ‪.‬‬

‫ﹰ‬ ‫ﲢﻮﻳﻼ ﺑﲔ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﻭﺍﳌﻮﻻﺕ‪ ،‬ﻭﺍﳉﺴﻴﲈﺕ‪ ،‬ﺃﻧﺖ ﺗﺪﺭﻙ ﺃﻥ ﺍﳌﻮﻝ‬ ‫ﺍﻵﻥ ﺑﻌﺪ ﺃﻥ ﺃﺟﺮﻳﺖ‬ ‫ﺃﺳﺎﺱ ﺍﳊﺴﺎﺑﺎﺕ‪ .‬ﻓﺎﻟﻜﺘﻠﺔ ﹰ‬ ‫ﺩﺍﺋﲈ ﲢﻮﻝ ﺇﱃ ﻣﻮﻻﺕ ﻗﺒﻞ ﲢﻮﻳﻠﻬﺎ ﺇﱃ ﺫﺭﺍﺕ‪ ،‬ﻭﺍﻟﺬﺭﺍﺕ‬ ‫ﲢﻮﻝ ﺇﱃ ﻣﻮﻻﺕ ﻗﺒﻞ ﺃﻥ ﲢﺴﺐ ﻛﺘﻠﺘﻬﺎ‪.‬‬

‫ﺍﻟﺸﻜﻞ‪ 5-7‬ﻳﺒﲔ ﺧﻄﻮﺍﺕ ﺍﻟﺘﺤﻮﻳﻞ‪ .‬ﰲ ﺍﻷﻣﺜﻠﺔ ﺍﳊﺴﺎﺑﻴﺔ ﺍﻟﺘﻲ ﻣﺮﺕ ﺑﻚ‪ ،‬ﺍﺳﺘﻌﻤﻠﺖ‬ ‫ﺧﻄﻮﺗﲔ ﰲ ﺍﻟﺘﺤﻮﻳﻞ‪ ،‬ﻓﺈﻣﺎ ﲢﻮﻳﻞ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﻣﻮﻻﺕ ﺛﻢ ﺇﱃ ﺫﺭﺍﺕ‪ ،‬ﺃﻭ ﲢﻮﻳﻞ ﺍﻟﺬﺭﺍﺕ‬ ‫ﺇﱃ ﻣﻮﻻﺕ ﺛﻢ ﺇﱃ ﻛﺘﻠﺔ‪ .‬ﻭﻳﻤﻜﻨﻚ ﺩﻣﺞ ﺍﳋﻄﻮﺗﲔ ﰲ ﺧﻄﻮﺓ ﻭﺍﺣﺪﺓ‪ .‬ﺍﻓﱰﺽ ﺃﻧﻚ ﺗﺮﻳﺪ‬ ‫ﻣﻌﺮﻓﺔ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﲔ ﰲ ‪ 1.00g‬ﻣﻨﻪ‪ .‬ﺇﻥ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺤﻮﻳﻞ ﻫﺬﻩ ﺗﺘﻄﻠﺐ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﻣﻦ ﻛﺘﻠﺔ ﺇﱃ ﻣﻮﻻﺕ ﻭﻣﻦ ﻣﻮﻻﺕ ﺇﱃ ﺫﺭﺍﺕ‪ ،‬ﻭﻳﻤﻜﻦ ﺃﻥ ﲤﺜﻞ ﺫﻟﻚ ﰲ ﺍﳌﻌﺎﺩﻟﺔ‪.‬‬ ‫ﻣﻦ‬

‫‪8‬‬

‫ﻣﻦ‬ ‫ﻣﻦ‬

‫ﻣﻦ‬

‫ﻣﻦ‬

‫‪88‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪5 - 2‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺗﺴﻤﻰ ﺍﻟﻜﺘﻠﺔ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ ﻟﻮﺍﺣﺪ ﻣﻮﻝ ﻣﻦ‬ ‫ﺃﻱ ﻣﺎﺩﺓ ﻧﻘﻴﺔ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻷﻱ ﻋﻨﺼﺮ ﺗﺴـﺎﻭﻱ ﻋﺪﺩ ﹰﹼﻳﺎ‬ ‫ﻛﺘﻠﺘﻪ ﺍﻟﺬﺭﻳﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻷﻱ ﻣﺎﺩﺓ ﻫﻲ ﻛﺘﻠﺔ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‬ ‫ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻟﻬﺬﻩ ﺍﻟﻤﺎﺩﺓ‪.‬‬ ‫ﺗﺴـﺘﺨﺪﻡ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻤﻮﻻﺕ‬ ‫ﺇﻟـﻰ ﻛﺘﻠﺔ‪ ،‬ﻭﻳﺴـﺘﺨﺪﻡ ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴﺔ‬ ‫ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﻟﻰ ﻣﻮﻻﺕ‪.‬‬

‫‪54‬‬

‫‪ .22‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻟﺨـﺺ ﺍﻟﻔـﺮﻕ ﺑﻴـﻦ ﻛﻤﻴـﺎﺕ ﻣـﻮﻝ ﻭﺍﺣـﺪ ﻣـﻦ ﻣﺎﺩﺗﻴـﻦ‬ ‫ﻣﺨﺘﻠﻔﺘﻴﻦ ﺃﺣﺎﺩﻳﺘﻲ ﺍﻟﺬﺭﺍﺕ ﻣﻦ ﺣﻴﺚ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻭﺍﻟﻜﺘﻠﺔ؟‬ ‫‪ .23‬ﺍﺫﻛﺮ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻼﺯﻡ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻜﺘﻠﺔ ﻭﺍﻟﻤﻮﻻﺕ ﻟﺬﺭﺓ ﺍﻟﻔﻠﻮﺭ‪.‬‬ ‫‪ .24‬ﺍﺷﺮﺡ ﻛﻴﻒ ﺗﺮﺑﻂ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻛﺘﻠﺔ ﺍﻟﺬﺭﺓ ﺑﻜﺘﻠﺔ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺬﺭﺍﺕ‪.‬‬ ‫‪ .25‬ﺻﻒ ﺍﻟﺨﻄﻮﺍﺕ ﺍﻟﻼﺯﻣﺔ ﻟﺘﺤﻮﻳﻞ ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ ﺇﻟﻰ ﻋﺪﺩ ﺫﺭﺍﺗﻪ‪.‬‬ ‫‪ .26‬ﺭﺗﹼﺐ ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ ﺍﻷﺻﻐﺮ ﺇﻟﻰ ﺍﻷﻛﺒﺮ ﺑﺤﺴـﺐ ﺍﻟﻜﺘﻠﺔ‪ 1.0 mol :‬ﻣﻦ‬ ‫‪3.0 × 1024 atoms ، Ar‬ﻣﻦ ‪20 g ،Ne‬ﻣﻦ ‪. Kr‬‬ ‫‪ .27‬ﺣﺪﱢ ﺩ ﺍﻟﻜﻤﻴﺔ ﺍﻟﺘﻲ ﺗﺤﺴﺐ ﺑﻘﺴﻤﺔ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻌﻨﺼﺮ ﻋﻠﻰ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‪.‬‬ ‫ﺻﻤـﻢ ﺧﺮﻳﻄـﺔ ﻣﻔﺎﻫﻴﻤﻴـﺔ ﺗﻮﺿﺢ ﺍﻟﻤﻌﺎﻣـﻼﺕ ﺍﻟﻼﺯﻣﺔ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴـﻦ ﺍﻟﻜﺘﻠﺔ‪،‬‬ ‫‪ .28‬ﹼ‬ ‫ﻭﺍﻟﻤﻮﻻﺕ‪ ،‬ﻭﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪5-3‬‬ ‫ا ﻫﺪاف‬

‫ﺗﺘﻌﺮﻑ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﺘﻲ ﺗﺮﺑﻂ‬ ‫ﺍﳌﻮﻝ ﺑﺎﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫ﲢﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﳌﺮﻛﺐ‪.‬‬ ‫ﺗﻄﺒﻖ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﻟﺘﺤﺪﻳﺪ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﺃﻭ‬ ‫ﺍﻷﻳﻮﻧﺎﺕ ﰲ ﻛﺘﻠﺔ ﻣﻌﺮﻭﻓﺔ ﻣﻦ‬ ‫ﻣﺮﻛﺐ‪.‬‬

‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﳉﺴﻴﻢ‪ :‬ﺫﺭﺓ ﺃﻭ ﺟﺰﻱﺀ ﺃﻭ ﻭﺣﺪﺓ‬ ‫ﺻﻴﻐﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺃﻭﺃﻳﻮﻥ‪.‬‬

‫ﻣﻮﻻت اﻟﻤﺮﻛﺒﺎت‬ ‫‪Moles of Compounds‬‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﻤﻜﻦ ﺣﺴﺎﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻣﻦ ﺧﻼﻝ ﺻﻴﻐﺘﻪ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻛﲈ ﻳﻤﻜﻦ ﺍﺳﺘﻌﲈﻝ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﺎ ﺑﲔ ﺍﻟﻜﺘﻠﺔ ﻭﺍﳌﻮﻻﺕ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﲣﻴﻞ ﺣﻘﻴﺒﺘﲔ ﹸﻓﺤﺼﺘﺎ ﰲ ﺍﳌﻄﺎﺭ ‪ ،‬ﻭﺗﺒﲔ ﺃﻥ ﺇﺣﺪﺍﳘﺎ ﻗﺪ‬ ‫ﲡﺎﻭﺯﺕ ﺣﺪ ﺍﻟﻮﺯﻥ ﺍﳌﺴﻤﻮﺡ ﺑﻪ‪ .‬ﻭﺑﲈ ﺃﻥ ﻭﺯﻥ ﻛﻞ ﺣﻘﻴﺒﺔ ﻳﻌﺘﻤﺪ ﻋﲆ ﳎﻤﻮﻉ ﺍﻷﺷﻴﺎﺀ‬ ‫ﺍﳌﻮﺟﻮﺩﺓ ﺑﺪﺍﺧﻠﻬﺎ‪ ،‬ﻓﺈﻥ ﺗﻐﻴﲑﻫﺬﻩ ﺍﻷﺷﻴﺎﺀ ﻳﻐﲑ ﻭﺯﻥ ﻛﻞ ﻣﻨﻬﲈ‪.‬‬

‫الصﻴﻎ الﻜﻴمﻴاﺋﻴﺔ ﻭالمﻮﻝ‬ ‫‪Chemical Formulas and the Mole‬‬ ‫ﺗﻌﻠﻤﺖ ﺃﻥ ﺍﻷﻧﻮﺍﻉ ﺍﳌﺨﺘﻠﻔﺔ ﻣﻦ ﺍﳉﺴﻴﲈﺕ ﹸﺗﻌﺪ ﺑﺎﺳﺘﻌﲈﻝ ﺍﳌﻮﻝ‪ ،‬ﻭﻛﺬﻟﻚ ﺗﻌﻠﻤﺖ‬ ‫ﺃﻥ ﺍﻟﻜﺘﻞ ﺍﳌﻮﻟﻴﺔ ﺗﺴﺘﻌﻤﻞ ﻟﻠﺘﺤﻮﻳﻞ ﺑﲔ ﺍﳌﻮﻻﺕ ﻭﺍﻟﻜﺘﻠﺔ‪ ،‬ﻭﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ‬ ‫ﻟﻠﻌﻨﴫ‪ .‬ﻭﻟﻠﻘﻴﺎﻡ ﺑﺘﺤﻮﻳﻼﺕ ﻣﺸﺎﲠﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ﻭﺍﻷﻳﻮﻧﺎﺕ ﻓﺈﻧﻚ ﲢﺘﺎﺝ ﺇﱃ‬ ‫ﻣﻌﺮﻓﺔ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﳍﺎ‪.‬‬

‫ﺗﺬﻛﺮ ﺃﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺗﻌﱪﻋﻦ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﻭﺃﻧﻮﺍﻋﻬﺎ ﺍﳌﻮﺟﻮﺩﺓ‬ ‫ﰲ ﻭﺣﺪﺓ ﺻﻴﻐﺔ ﻭﺍﺣﺪﺓ ﻣﻨﻪ‪ .‬ﺧﺬ ﺑﻌﲔ ﺍﻻﻋﺘﺒﺎﺭﺍﳌﺮﻛﺐ ﺛﻨﺎﺋﻲ ﻛﻠﻮﺭﻭ ﺛﻨﺎﺋﻲ ﻓﻠﻮﺭﻭ‬ ‫ﻣﻴﺜﺎﻥ‪ ،‬ﻭﺻﻴﻐﺘﻪ ‪ CCl2F2‬ﺣﻴﺚ ﺗﺪﻝ ﺍﻷﺭﻗﺎﻡ ﺍﻟﺴﻔﻠﻴﺔ ﰲ ﺍﳌﺮﻛﺐ ﻋﲆ ﺃﻥ ﺟﺰﻳﺌﹰﺎ‬ ‫ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ‪ CCl2F2‬ﻳﺘﻜﻮﻥ ﻣﻦ ﺫﺭﺓ ﻛﺮﺑﻮﻥ )‪ ،(C‬ﻭﺫﺭﰐ ﻛﻠﻮﺭ)‪ (Cl‬ﻭﺫﺭﰐ‬ ‫ﻓﻠﻮﺭ)‪ .(F‬ﻭﻫﺬﻩ ﺍﻟﺬﺭﺍﺕ ﻣﺮﺗﺒﻄﺔ ﺑﻌﻀﻬﺎ ﺑﺒﻌﺾ ﻛﻴﻤﻴﺎﺋ ﹰﻴﺎ ‪ ،‬ﺑﻨﺴﺒﺔ ‪F: Cl: C‬‬ ‫ﻫﻲ ‪.2:2:1‬‬

‫ﺍﻟﺸـﻜـﻞ ‪ 5-8‬ﻳﻮﺿــﺢ ﺩﺭﺯﻥ ﻣـﻦ‬ ‫ﺟﺰﻳﺌـﺎﺕ ‪ CCl2F2‬ﲢﺘـﻮﻱ ﻋـﲆ ﺩﺭﺯﻥ‬ ‫ﻣـﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑـﻮﻥ‪ ،‬ﻭﺩﺭﺯﻧﲔ ﻣـﻦ ﺫﺭﺍﺕ‬ ‫ﺍﻟﻜﻠـﻮﺭ‪ ،‬ﻭﺩﺭﺯﻧـﲔ ﻣـﻦ ﺫﺭﺍﺕ ﺍﻟﻔﻠـﻮﺭ‪.‬‬

‫‪ è``àæà°SG‬ﻛـﻢ ﺫﺭﺓ ﻣـﻦ ﺍﻟﻜﺮﺑـﻮﻥ‪ ،‬ﻭﺍﻟﻜﻠﻮﺭ‪،‬‬ ‫ﻭﺍﻟﻔﻠﻮﺭ ﺗﻮﺟﺪ ﰲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ‪.CCl2F2‬‬

‫ﻭﺍﻵﻥ‪ ،‬ﺍﻓﺮﺽ ﺃﻥ ﻟﺪﻳﻚ ﻣﻮ ﹰ‬ ‫ﻻ ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ‪ CCl2F2‬ﻓﻬﺬﺍ ﻳﻌﻨﻲ ﺃﻧﻪ ﳛﺘﻮﻱ ﻋﲆ‬ ‫ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﳉﺴﻴﲈﺕ ﻭﺍﻟﺘﻲ ﲤﺜﻞ ﺑﺎﳉﺰﻳﺌﺎﺕ‪.‬ﻭﺳﺘﺒﻘﻰ ﺍﻟﻨﺴﺒﺔ ‪2 :2 :1‬‬ ‫ﺑﲔ ﺫﺭﺍﺕ ‪ F:Cl:C‬ﰲ ﹴ‬ ‫ﻣﻮﻝ ﻣﻦ ﺍﳌﺮﻛﺐ ﻛﲈ ﻫﻲ ﰲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻨﻪ‪.‬‬

‫ﻭﺍﻟﺸﻜﻞ ‪ 5-8‬ﻳﻮﺿﺢ ﺩﺭﺯﻥ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ‪ ، CCl2F2‬ﺇﺫ ﲢﺘﻮﻱ ﻋﲆ ﺩﺭﺯﻥ ﻭﺍﺣﺪ‬ ‫ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ‪ ،‬ﻭﺩﺭﺯﻧﲔ ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﻠﻮﺭ‪ ،‬ﻭﺩﺭﺯﻧﲔ ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻔﻠﻮﺭ‪.‬‬ ‫ﻓﺎﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ‪CCl2F2‬ﻻ ﲤﺜﻞ ﺟﺰﻳﺌﹰﺎ ﻣﻨﻔﺮﺩ ﹰﺍ ﻣﻦ ‪ CCl2F2‬ﻓﻘﻂ‪ ،‬ﺑﻞ ﲤﺜﻞ‬ ‫ﺃﻳﻀﺎ ﻣﻮ ﹰ‬ ‫ﹰ‬ ‫ﻻ ﻣﻦ ﺍﳌﺮﻛﺐ‪.‬‬

‫‪55‬‬

‫ﻗﺪ ﲢﺘﺎﺝ ﰲ ﺑﻌﺾ ﺍﳊﺴﺎﺑﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺇﱃ ﺍﻟﺘﺤﻮﻳﻞ ﺑﲔ ﻣﻮﻻﺕ ﺍﳌﺮﻛﺐ ﻭﻣﻮﻻﺕ ﺍﺣﺪ￯ ﺍﻟﺬﺭﺍﺕ ﺍﳌﻜﻮﻧﻪ ﻟﻪ‪ .‬ﻓﺎﻟﻨﺴﺐ‬ ‫ﺃﻭ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺘﺎﻟﻴﺔ‪ ،‬ﻳﻤﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻻﺳﺘﻌﲈﳍﺎ ﰲ ﺍﳊﺴﺎﺑﺎﺕ ﳉﺰﻱﺀ ‪.CCl2F2‬‬ ‫‪ 2 mol‬ﻣﻦ‬ ‫‪ 1 mol‬ﻣﻦ ‪CCl 2F 2‬‬

‫‪ 1 mol‬ﻣﻦ‬ ‫‪ 1 mol‬ﻣﻦ ‪CCl 2F 2‬‬

‫‪C atoms‬‬

‫‪ 2 mol‬ﻣﻦ‬ ‫‪ 1 mol‬ﻣﻦ ‪CCl 2F 2‬‬

‫‪F atoms‬‬

‫‪Cl atoms‬‬

‫ﻹﳚﺎﺩ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺕ ﺍﻟﻔﻠﻮﺭ ﰲ ‪ 5.50 moles‬ﻣﻦ ﺍﻟﻔﺮﻳﻮﻥ ‪ CCl2F2‬ﺍﴐﺏ ﻣﻮﻻﺕ ﺍﻟﻔﺮﻳﻮﻥ ﰲ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺬﻱ‬ ‫ﻳﺮﺑﻂ ﺑﲔ ﻣﻮﻻﺕ ﺫﺭﺍﺕ ﺍﻟﻔﻠﻮﺭ ﻭﻣﻮﻻﺕ ﺍﻟﻔﺮﻳﻮﻥ‪.‬‬ ‫= ‪ moles‬ﻣﻦ ‪× CCl 2F 2‬‬

‫‪ moles‬ﻣﻦ‬

‫‪ 1mol‬ﻣﻦ‬

‫= ‪ 5.50 mol‬ﻣﻦ ‪× CCl 2F 2‬‬

‫‪F atoms‬‬ ‫‪CCl 2F 2‬‬

‫‪ 2 mol‬ﻣﻦ‬

‫‪ 2 mol‬ﻣﻦ‬

‫= ‪ moles‬ﻣﻦ‬

‫‪C atoms‬‬ ‫‪CCl 2F 2‬‬

‫‪F atoms‬‬

‫= ‪ 11.0 mol‬ﻣﻦ‬

‫‪F atoms‬‬

‫ﻳﻤﻜﻦ ﺍﺳﺘﻌﲈﻝ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺬﻱ ﺍﺳﺘﻌﻤﻞ ﻟﻠﻔﻠﻮﺭ ﰲ ﻛﺘﺎﺑﺔ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﻟﺴﺎﺋﺮﺍﻟﻌﻨﺎﴏ ﰲ ﺍﳌﺮﻛﺐ ‪.‬ﻭﻋﺪﺩ ﻣﻮﻻﺕ‬ ‫ﺍﻟﻌﻨﴫﺍﻟﺘﻲ ﺗﻮﺿﻊ ﰲ ﺍﻟﺒﺴﻂ ﻫﻲ ﺍﻟﺮﻗﻢ ﺍﻟﺴﻔﲇ ﻟﻠﻌﻨﴫ ﰲ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪5-6 ∫Éãe‬‬

‫ﻋﻼﻗﺎﺕ ﺍﳌﻮﻝ ﺍﳌﺮﺗﺒﻄﺔ ﺑﺎﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺃﻛﺴﻴﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ )‪ (Al2O3‬ﺍﻟﺬﻱ ﻏﺎﻟ ﹰﺒﺎ ﻣﺎ ﻳﺴﻤﻰ ﺃﻟﻮﻣﻴﻨﺎ‪ ،‬ﻫﻮ ﺍﳌﺎﺩﺓ ﺍﳋﺎﻡ‬ ‫ﺍﻷﺳﺎﺳﻴﺔ ﻹﻧﺘﺎﺝ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ )‪ ،(Al‬ﺍﻟﺬﻱ ﻳﻮﺟﺪ ﰲ ﻣﻌﺪﻥ ﺍﻟﻜﻮﺭﻧﺪﻳﻮﻡ ﻭﺍﻟﺒﻮﻛﺴﻴﺖ‪.‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ‬ ‫‪3+‬‬

‫ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ) ‪ (Al‬ﰲ ‪ 1.25 mol‬ﻣﻦ ‪.Al2O3‬‬

‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘﺪ ﺃﻋﻄﻴﺖ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ ،Al2O3‬ﻭﻋﻠﻴﻚ ﺃﻥ ﲢﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ .Al3+‬ﻣﺴﺘﻌﻤ ﹰ‬ ‫ﻼ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﳌﺒﻨﻲ ﻋﲆ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺑﲔ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ Al3+‬ﻭﻣﻮﻻﺕ ‪ .Al2O3‬ﻛﻞ ﻣﻮﻝ ﻣﻦ ‪ Al2O3‬ﳛﺘﻮﻱ ﻋﲆ ﻣﻮﻝ ﻣﻦ‬ ‫‪3+‬‬ ‫ﺃﻳﻮﻧﺎﺕ ‪ ،Al‬ﻭﻋﻠﻴﻪ‪ ،‬ﻓﺎﻹﺟﺎﺑﺔ ﳚﺐ ﺃﻥ ﺗﻜﻮﻥ ﺿﻌﻒ ﻣﻮﻻﺕ ‪.Al2O3‬‬

‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫ﻋﺪﺩ ﻣﻮﻻﺕ= ‪1.25 mol‬ﻣﻦ ‪Al2O3‬‬ ‫‪2‬‬

‫ﺍﳌﻄﻠﻮﺏ‬

‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ = ؟ ‪ mol‬ﻣﻦ ﺃﻳﻮﻥ ‪Al+3‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬ ‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﻌﻼﻗﺔ ‪ 1mol‬ﻣﻦ ‪ ، Al2O3‬ﳛﺘﻮﻱ ﻋﲆ ‪ Al3+‬ﻟﻜﺘﺎﺑﺔ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‪.‬‬ ‫ﻋﻴﻦ ﻣﻌﺎﻣﻞ ﺗﺤﻮﻳﻞ ﻳﺮﺑﻂ ﺑﻴﻦ ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ Al 3+‬ﺑﻤﻮﻻﺕ ‪Al2O3‬‬

‫‪3‬‬

‫‪ 2 mol‬ﻣﻦ‬

‫‪3+‬‬

‫‪Al‬‬

‫‪ 1 mol‬ﻣﻦ‬ ‫ﻟﺘﺤﻮﻳﻞ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ Al2O3‬ﺍﳌﻌﺮﻭﻓﺔ ﺇﱃ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ Al3+‬ﺍﴐﺏ ﰲ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﳌﻜﺘﻮﺏ‪.‬‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫‪ 2 mol‬ﻣﻦ ﺃﻳﻮﻥ ‪Al3+‬‬ ‫= ‪ moles‬ﻣﻦ ﺃﻳﻮﻥ ‪Al3+‬‬ ‫= ‪ moles‬ﻣﻦ ‪× Al 2O 3‬‬ ‫‪ 1 mol‬ﻣﻦ ‪Al 2O 3‬‬ ‫ﻋـﻮﺽ ﻣﺴـﺘﻌﻴﻨﹰﺎ ﺑﺎﻟﻤﻌﻄﻴـﺎﺕ‪ ،‬ﻭﺟـﺪ‬ ‫‪ 2 mol‬ﻣﻦ ﺃﻳﻮﻥ ‪ 2.50mol Al3+‬ﻣﻦ ﺃﻳﻮﻥ ‪Al3+‬‬ ‫ﺍﻟﺤﻞ‬ ‫= ‪ 1.25 mol‬ﻣﻦ ‪× Al 2O 3‬‬ ‫‪ 1 mol‬ﻣﻦ ‪= Al O‬‬ ‫‪2 3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫‪3+‬‬

‫ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ Al‬ﺿﻌﻒ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ Al2O3‬ﻛﲈ ﻫﻮ ﻣﺘﻮﻗﻊ‪.‬‬

‫‪56‬‬

‫‪Al 2O 3‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ (ZnCl‬ﺑﻮﺻﻔﻪ ﺳﺒﻴﻜﺔ ﻟﺤﺎﻡ ﻟﺮﺑﻂ ﻓﻠﺰﻳﻦ ﺑﻌﻀﻬﻤﺎ‬ ‫‪ .29‬ﻳﺴﺘﻌﻤﻞ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ )‪2‬‬ ‫‬‫ﺑﺒﻌﺾ‪ ،‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ‪ Cl‬ﻓﻲ ‪ 2.50mol‬ﻣﻦ ‪.ZnCl2‬‬ ‫ﻣﺼﺪﺭﺍ ﻟﻠﻄﺎﻗﺔ‪،‬‬ ‫‪ .30‬ﺗﻌﺘﻤﺪ ﺍﻟﻨﺒﺎﺗﺎﺕ ﻭﺍﻟﺤﻴﻮﺍﻧﺎﺕ ﻋﻠﻰ ﺳﻜﺮﺍﻟﺠﻠﻮﻛﻮﺯ )‪ (C6H12O6‬ﺑﻮﺻﻔﻪ‬ ‫ﹰ‬ ‫ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﻛﻞ ﻋﻨﺼﺮ ﻓﻲ ‪ 1.25mol‬ﻣﻦ ‪.C6H12O6‬‬ ‫‪ .31‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻜﺒﺮﻳﺘﺎﺕ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ‪ 3.00mol‬ﻣﻦ ‪.Fe2 (SO4)3‬‬ ‫‪ .32‬ﻣﺎ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﻴﻦ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ‪ 5.00mol‬ﻣﻦ ‪P2O5‬؟‬ ‫‪ .33‬ﺗﺤﺪﱟ ‪ :‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ‪ 1.15 × 101 mol‬ﻣﻦ ﺍﻟﻤﺎﺀ‪.‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 5-9‬ﺑـﲈ ﺃﻥ ﻛﻞ ﻣـﺎﺩﺓ ﲢﺘﻮﻱ‬ ‫ﻋﲆ ﺃﻋﺪﺍﺩ ﻭﺃﻧﻮﺍﻉ ﳐﺘﻠﻔﺔ‬ ‫ﻣﻦ ﺍﻟـﺬﺭﺍﺕ‪ ،‬ﻓﺈﻥ ﻛﺘﻠﻬﺎ‬ ‫ﺍﳌﻮﻟﻴـﺔ ﳐﺘﻠﻔـﺔ‪ .‬ﻓﺎﻟﻜﺘﻠﺔ‬ ‫ﺍﳌﻮﻟﻴﺔ ﻟـﻜﻞ ﻣﺮﻛﺐ ﻫﻲ‬ ‫ﺣﺎﺻﻞ ﳎﻤﻮﻉ ﻛﺘﻞ ﲨﻴﻊ‬ ‫ﺍﻟﻌﻨـﺎﴏ ﺍﳌﻜﻮﻧـﻪ ﻟـﻪ‪.‬‬

‫ﹼ‬ ‫للمرﻛﺒات‬ ‫الﻜﺘلﺔ المﻮلﻴﺔ‬ ‫‪The Molar Mass of Compounds‬‬ ‫ﻛﺘﻠﺔ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﳌﺮﻛﺐ ﺗﺴﺎﻭﻱ ﳎﻤﻮﻉ ﻛﺘﻞ ﺍﳉﺴﻴﲈﺕ ﺍﻟﺘﻲ ﻳﺘﻜﻮﻥ ﻣﻨﻬﺎ ﺍﳌﺮﻛﺐ‪.‬‬ ‫ﺍﻓﺮﺽ ﺃﻧﻚ ﺗﺮﻏﺐ ﰲ ﺗﻌﻴﲔ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﳌﺮﻛﺐ ﻛﺮﻭﻣﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‬ ‫)‪ ،(K2CrO4‬ﺍﺑﺪﺃ ﺑﺎﻟﺒﺤﺚ ﻋﻦ ﺍﻟﻜﺘﻞ ﺍﳌﻮﻟﻴﺔ ﻟﻜﻞ ﻋﻨﴫ ﰲ ‪ ،K2CrO4‬ﺛﻢ ﺍﴐﺏ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻜﻞ ﻋﻨﴫ ﰲ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻌﻨﴫ ﺍﳌﻤﺜﻠﺔ ﰲ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪،‬‬ ‫ﺛﻢ ﺍﲨﻊ ﻛﺘﻞ ﺍﻟﻌﻨﺎﴏ ﻛﺎﻓﺔ ﻟﺘﺤﺼﻞ ﻋﲆ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪.K2CrO4‬‬

‫ﻛﺮﻭﻣﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ )‪( K2CrO4‬‬

‫ﻣﻦ‬ ‫=‪ 2 mol‬ﻣﻦ ‪78.20 g = K 39.10 g × K‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪K‬‬

‫‪ 52.0 g‬ﻣﻦ‬ ‫‪52.0 g = Cr‬‬ ‫=‪ 1 mol‬ﻣﻦ ‪× Cr‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪Cr‬‬

‫‪ 16.0 g‬ﻣﻦ‬ ‫‪64.0 g = O‬‬ ‫=‪ 4 mol‬ﻣﻦ ‪× O‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪O‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﹺ‬ ‫ﻟـ ‪194.20g = K2CrO4‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺗﻮﺿﺢ ﻗﺎﻧﻮﻥ ﺣﻔﻆ ﺍﻟﻜﺘﻠﺔ؛ ﻓﺎﻟﻜﺘﻠﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻠﻤﺘﻔﺎﻋﻼﺕ‬ ‫ﺗﺴﺎﻭﻱ ﻛﺘﻠﺔ ﺍﳌﺮﻛﺐ ﺍﳌﺘﻜﻮﻥ‪ .‬ﺍﻟﺸﻜﻞ ‪ 5-9‬ﻳﻮﺿﺢ ﻛﺘ ﹰ‬ ‫ﻼ ﻣﺘﻜﺎﻓﺌﺔ ﳌﻮﻝ ﻭﺍﺣﺪ ﻣﻦ‬ ‫ﻛﺮﻭﻣﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‪ ،‬ﻭﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‪ ،‬ﻭﺍﻟﺴﻜﺮﻭﺯ‪.‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪(NaCl‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪ .34‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻜﻞ ﻣﺮﻛﺐ ﺃﻳﻮﻧﻲ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪CaCl2.b‬‬ ‫‪NaOH.a‬‬ ‫‪ .35‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻜﻞ ﻣﺮﻛﺐ ﺃﻳﻮﻧﻲ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪CCl4.c‬‬ ‫‪HCN.b‬‬ ‫‪C2H5OH.a‬‬

‫‪KC2H3O2.c‬‬

‫‪ .36‬ﺗﺤﺪﱟ ‪ :‬ﹼﺻﻨﻒ ﻛ ﹰ‬ ‫ﻼ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺑﻮﺻﻔﻪ ﻣﺮﻛ ﹰﺒﺎ ﺟﺰﻳﺌ ﹰﻴﺎ ﺃﻭﺃﻳﻮﻧ ﹰﻴﺎ‪ ،‬ﺛﻢ ﺍﺣﺴﺐ ﻛﺘﻠﺘﻪ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ‪:‬‬ ‫‪C12H22O11.c‬‬ ‫‪(NH4)3PO4.b‬‬ ‫‪Sr(NO3)2.a‬‬

‫ﺍﻟﺴﻜﺮﻭﺯ )‪(C12H22O11‬‬

‫‪57‬‬

‫ﺗحﻮﻳﻞ مﻮﻻت المرﻛﺐ ﺇلﻰ ﻛﺘلﺔ‬ ‫‪Converting Moles of a Compound to Mass‬‬

‫ﹴ‬ ‫ﲡﺮﺑﺔ ﻣﺎ‪ ،‬ﻓﻌﻠﻴﻚ ﺃﻭ ﹰ‬ ‫ﻻ ﺃﻥ ﲢﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻄﻠﻮﺑﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ ﻣﻦ ﺧﻼﻝ ﻋﺪﺩ‬ ‫ﺇﺫﺍ ﺃﺭﺩﺕ ﺇﳚﺎﺩ ﻋﺪﺩ ﻣﻮﻻﺕ ﻣﺮﻛﺐ ﻟﻌﻤﻞ‬ ‫ﺍﳌﻮﻻﺕ ‪ ،‬ﺛﻢ ﻳﻤﻜﻨﻚ ﻗﻴﺎﺱ ﻫﺬﻩ ﺍﻟﻜﺘﻠﺔ ﺑﺎﳌﻴﺰﺍﻥ‪ .‬ﻓﻔﻲ ﺍﳌﺜﺎﻝ ‪ ،5-2‬ﺗﻌﻠﻤﺖ ﻛﻴﻔﻴﺔ ﲢﻮﻳﻞ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻌﻨﺎﴏ ﺇﱃ ﻛﺘﻠﺔ‬ ‫ﺑﺎﺳﺘﻌﲈﻝ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺑﻮﺻﻔﻬﺎ ﻣﻌﺎﻣﻞ ﲢﻮﻳﻞ‪ ،‬ﻭﺗﺴﺘﻌﻤﻞ ﺍﻟﻄﺮﻳﻘﺔ ﻧﻔﺴﻬﺎ ﻣﻊ ﺍﳌﺮﻛﺒﺎﺕ ﻏﲑ ﺃﻧﻪ ﻳﺘﻌﲔ ﻋﻠﻴﻚ ﺣﺴﺎﺏ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ‪.‬‬

‫‪5-7 ∫Éãe‬‬

‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﻣﻮﻝ ﺇﱃ ﻛﺘﻠﺔ ﰲ ﺍﳌﺮﻛﺒﺎﺕ ﺗﻌﻮﺩ ﺍﻟﺮﺍﺋﺤﺔ ﺍﳌﻤﻴﺰﺓ ﻟﻠﺜﻮﻡ ﺇﱃ ﻭﺟﻮﺩﺍﳌﺮﻛﺐ ‪ . (C3H5)2S‬ﻓﲈ ﻛﺘﻠﺔ‬ ‫‪ 2.50 mol‬ﻣﻦ ]‪ [(C3H5)2S‬؟‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘﺪ ﺃﻋﻄﻴﺖ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ ، (C3H5)2S‬ﻭﻋﻠﻴﻚ ﺃﻥ ﲢﻮﻝ ﺍﳌﻮﻻﺕ ﺇﱃ ﻛﺘﻠﺔ ﺑﺎﺳﺘﻌﲈﻝ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺑﻮﺻﻔﻬﺎ ﻣﻌﺎﻣﻞ‬ ‫ﲢﻮﻳﻞ‪ .‬ﻭﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻫﻲ ﺣﺎﺻﻞ ﳎﻤﻮﻉ ﺍﻟﻜﺘﻞ ﺍﳌﻮﻟﻴﺔ ﻟﻜﻞ ﺍﻟﻌﻨﺎﴏ ﰲ ‪.(C3H5)2S‬‬ ‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ = ‪ 2.50mol‬ﻣﻦ ‪(C3H5)2S‬‬ ‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟـ ‪.(C3H5)2S‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ S‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪S‬‬

‫ﺍﳌﻄﻠﻮﺏ‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ = ؟ ‪ g/mol‬ﻣﻦ ‪(C3H5)2S‬‬

‫ﺍﻟﻜﺘﻠﺔ = ؟ ‪ g‬ﻣﻦ ‪(C3H5)2S‬‬

‫‪ 32.07 g‬ﻣﻦ‬ ‫‪ 1 mol‬ﻣﻦ ‪× S‬‬ ‫‪ 32.07 g= S‬ﻣﻦ ‪S‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪S‬‬

‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ C‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪C‬‬

‫ﻣﻦ‬ ‫‪ 6 mol‬ﻣﻦ ‪ 72.06 g= C 12.01 g × C‬ﻣﻦ ‪C‬‬

‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ H‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪H‬‬

‫‪ 1.008 g‬ﻣﻦ‬ ‫‪ 10 mol‬ﻣﻦ ‪× H‬‬ ‫‪ 10.08 g= H‬ﻣﻦ ‪H‬‬

‫‪ 1 mol‬ﻣﻦ‬

‫‪C‬‬

‫‪ 1 mol‬ﻣﻦ‬

‫‪H‬‬

‫ﺣﺎﺻﻞ ﺟﻤﻊ ﺍﻟﻜﺘﻞ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ = )‪ 114.21 g/ mol = (10.08g + 72.06g + 32.07g‬ﻣﻦ ‪(C3H5)2S‬‬

‫ﺍﺳﺘﻌﻤﻞ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ) ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﳉﺮﺍﻣﺎﺕ ﺑﺎﳌﻮﻻﺕ‪.‬‬ ‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﻋــﻮﺽ ﻣـﺴـﺘـﻌـﻴـﻨﹰﺎ‬ ‫ﻭﺣﻞ‬ ‫ﺑﺎﻟﻤﻌﻄﻴﺎﺕ‪ ،‬ﹸ‬

‫‪ g‬ﻣﻦ‬ ‫= ‪ moles‬ﻣﻦ ‪× (C3H5)2S‬‬ ‫‪ 1 mol‬ﻣﻦ ‪(C3H5)2S‬‬ ‫‪(C3H5)2S‬‬

‫= ‪ g‬ﻣﻦ ‪(C3H5)2S‬‬

‫‪ 114.21 g‬ﻣﻦ‬ ‫= ‪ 2.50 mol‬ﻣﻦ ‪× (C3H5)2S‬‬ ‫‪ 286 g = (C3H5)2S‬ﻣﻦ ‪(C3H5)2S‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪(C3H5)2S‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .37‬ﻣﺎ ﻛﺘﻠﺔ ‪ 3.25mol‬ﻣﻦ ﺣﻤﺾ ﺍﻟﻜﺒﺮﻳﺘﻴﻚ ‪H2SO4‬؟‬ ‫‪ .38‬ﻣﺎ ﻛﺘﻠﺔ ‪ 4.35 ×10-2 mol‬ﻣﻦ ﻛﻠﻮﺭﻳﺪﺍﻟﺨﺎﺭﺻﻴﻦ ‪ ZnCl2‬؟‬ ‫‪ .39‬ﺗﺤﺪﱟ ‪ :‬ﺍﻛﺘﺐ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﺒﺮﻣﻨﺠﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‪ ،‬ﺛﻢ ﺍﺣﺴﺐ ﻛﺘﻠﺔ ‪ 2.55mol‬ﻣﻦ ﻫﺬﺍ ﺍﻟﻤﺮﻛﺐ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪.‬‬

‫‪58‬‬

‫ﺗحﻮﻳﻞ ﻛﺘلﺔ المرﻛﺐ ﺇلﻰ مﻮﻻت‬ ‫‪Converting the Mass of a Compound to Moles‬‬

‫ﺇﺫﺍ ﻧﺘﺞ ﻋﻦ ﺇﺣﺪ￯ ﺍﻟﺘﺠﺎﺭﺏ ﺍﻟﺘﻲ ﺃﺟﺮﻳﺘﻬﺎ ﰲ ﺍﳌﺨﺘﱪ‪ 5.55g‬ﻣﻦ ﻣﺮﻛﺐ ﻣﺎ‪ ،‬ﻓﲈ ﻋﺪﺩ ﺍﳌﻮﻻﺕ ﰲ ﻫﺬﻩ ﺍﻟﻜﺘﻠﺔ؟ ﻭﻟﺘﺤﺪﻳﺪ‬ ‫ﺫﻟﻚ‪ ،‬ﺍﻓﱰﺽ ﺃﻧﻚ ﺣﺴﺒﺖ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻭﻭﺟﺪﲥﺎ ‪ ،185.0g/mol‬ﻭﺑﲈ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺗﺮﺑﻂ ﺍﳉﺮﺍﻣﺎﺕ‬ ‫ﺑﺎﳌﻮﻻﺕ‪ ،‬ﻓﺈﻧﻚ ﲢﺘﺎﺝ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪ ،‬ﺇﱃ ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﺑﻮﺻﻔﻪ ﻣﻌﺎﻣﻞ ﲢﻮﻳﻞ‪.‬‬ ‫‪ 1 mol‬ﻣﻦ ﺍﳌﺮﻛﺐ‬ ‫‪ 5.50 g‬ﻣﻦ ﺍﳌﺮﻛﺐ ×‬ ‫‪ 185g‬ﻣﻦ ﺍﳌﺮﻛﺐ‬

‫‪5-8 ∫Éãe‬‬

‫=‪ 0.0297 mol‬ﻣﻦ ﺍﳌﺮﻛﺐ‬

‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﻣﻮﻻﺕ ﰲ ﺍﳌﺮﻛﺒﺎﺕ ﻳﺴﺘﻌﻤﻞ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca(OH)2‬ﻹﺯﺍﻟﺔ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﱪﻳﺖ‬ ‫ﻣﻦ ﻏﺎﺯﺍﺕ ﺍﻟﻌﺎﺩﻡ ﺍﳌﻨﺒﻌﺜﺔ ﻣﻦ ﳏﻄﺎﺕ ﺍﻟﻄﺎﻗﺔ‪ ،‬ﻭﰲ ﻣﻌﺎﳉﺔ ﻋﴪ ﺍﳌﺎﺀ ﻹﺯﺍﻟﺔ ﺃﻳﻮﻧﺎﺕ ‪ Ca2+‬ﻭ ‪ .Mg2+‬ﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ‬ ‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﰲ ‪ 325g‬ﻣﻦ ﺍﳌﺮﻛﺐ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﺪﻳﻚ ‪ 325g‬ﻣﻦ ‪ Ca(OH)2‬ﻭﺍﳌﻄﻠﻮﺏ ﺇﳚﺎﺩ ﻋﺪﺩ ﻣﻮﻻﺕ ‪.Ca(OH)2‬ﺍﺣﺴﺐ ﺃﻭ ﹰ‬ ‫ﻻ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻝ ‪.Ca(OH)2‬‬ ‫ﺍﳌﻄﻠﻮﺏ‬

‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫ﺍﻟﻜﺘﻠﺔ = ‪ 325g‬ﻣﻦ ‪Ca(OH)2‬‬ ‫‪2‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ = ؟ ‪ g/mol‬ﻣﻦ ‪Ca(OH)2‬‬ ‫ﻋﺪﺩ ﺍﳌﻮﻻﺕ = ؟ ‪ mol‬ﻣﻦ ‪Ca(OH)2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪.Ca(OH)2‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ Ca‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪Ca‬‬

‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ O‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪O‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ H‬ﻓﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪H‬‬

‫‪ 40.08 g‬ﻣﻦ‬ ‫‪40.08 g= Ca‬‬

‫‪ 1 mol‬ﻣﻦ ‪× Ca‬‬ ‫‪ 1 mol‬ﻣﻦ‬ ‫‪ 16.0 g‬ﻣﻦ‬ ‫‪32.0 g= O‬‬ ‫‪ 2 mol‬ﻣﻦ ‪× O‬‬ ‫‪ 1 mol‬ﻣﻦ ‪O‬‬ ‫‪Ca‬‬

‫‪ 1.008 g‬ﻣﻦ‬ ‫‪2.016 g= H‬‬ ‫‪ 2 mol‬ﻣﻦ ‪× H‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪H‬‬

‫ﺣﺎﺻﻞ ﺟﻤﻊ ﺍﻟﻜﺘﻞ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ = )‪ 74.10 g/ mol = (2.016g + 32.00g + 40.08g‬ﻣﻦ ‪Ca(OH)2‬‬ ‫ﺍﺳﺘﻌﻤﻞ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ )ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﳌﻮﻻﺕ ﺑﺎﳉﺮﺍﻣﺎﺕ‪.‬‬ ‫‪ 1 mol‬ﻣﻦ ‪Ca(OH)2‬‬

‫ﻋــﻮﺽ ﻣـﺴـﺘـﻌـﻴـﻨﹰﺎ‬ ‫=‪ 325 g‬ﻣﻦ ‪× Ca(OH)2‬‬ ‫ﻭﺣﻞ‬ ‫ﺑﺎﻟﻤﻌﻄﻴﺎﺕ‪ ،‬ﹸ‬

‫‪3‬‬

‫‪ 74.10 g‬ﻣﻦ ‪Ca(OH)2‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫= ‪ 4.39 mol‬ﻣﻦ ‪Ca(OH)2‬‬

‫ﻗﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﹺﻟـ ‪ Ca(OH)2‬ﺇﱃ ‪ ، 75g/mol‬ﻭﻛﺬﻟﻚ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻌﻄﺎﺓ ﻣﻦ ‪Ca(OH)2‬‬ ‫ﻟﻠﺘﺤﻘﻖ ﻣﻦ ﺻﺤﺔ ﺍﳉﻮﺍﺏ‪ ،‬ﹼ‬ ‫ﺇﱃ ‪ .300g‬ﻭﺑﲈ ﺃﻥ ﺍﻟﻌﺪﺩ ‪ 300‬ﺃﺭﺑﻌﺔ ﺃﺿﻌﺎﻑ ﺍﻟﻌﺪﺩ ‪ .75‬ﻟﺬﺍ ﻓﺎﳉﻮﺍﺏ ﻣﻘﺒﻮﻝ‪. ،‬ﻛﲈ ﺃﻥ ﻭﺣﺪﺓ ﺍﳌﻮﻝ ﺻﺤﻴﺤﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪ .40‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﻟﻜﻞ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻵﺗﻴﺔ‪:‬‬ ‫‪ 22.6 g .a‬ﻣﻦ ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ‪AgNO3‬‬

‫‪ 6.5g .b‬ﻣﻦ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﺨﺎﺭﺻﻴﻦ ‪ZnSO4‬‬

‫‪ 2.50kg .a‬ﻣﻦ ‪Fe2O3‬‬

‫‪ 25.4mg .b‬ﻣﻦ ‪PbCl4‬‬

‫‪ .41‬ﺗﺤﺪﱟ ﺻﻨﻒ ﻛ ﹰ‬ ‫ﻼ ﻣﻦ ﺍﻟﻤﺮﻛﺒﻴﻦ ﺍﻟﺘﺎﻟﻴﻴﻦ ﺇﻟﻰ ﺃﻳﻮﻧﻲ ﺃﻭ ﺟﺰﻳﺌﻲ‪،‬ﺛﻢ ﺣﻮﻝ ﺍﻟﻜﺘﻞ ﺍﻟﻤﻌﻄﺎﺓ ﺇﻟﻰ ﻣﻮﻻﺕ‪:‬‬

‫‪59‬‬

‫ﺗحﻮﻳﻞ ﻛﺘلﺔ مرﻛﺐ ﺇلﻰ ﻋﺪد ﺟسﻴمات‬ ‫‪Converting the Mass of a Compound to Number of particles‬‬

‫‪‬‬

‫ﺿﻤﻦ ﻣﻌﻠﻮﻣﺎﺕ ﻫﺬﺍ ﺍﻟﺠﺰﺀ‬ ‫ﻓﻲ ﻣﻄﻮﻳﺘﻚ‪.‬‬

‫ﺗﻌﺮﻓﺖ ﻛﻴﻔﻴﺔ ﺇﻳﺠﺎﺩ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﻓﻲ ﻛﺘﻠﺔ ﻣﻌﻴﻨﺔ ﻣﻦ ﺍﻟﻤﺮﻛﺐ‪.‬ﺍﻵﻥ ﺳﻮﻑ ﺗﺘﻌﻠﻢ ﻛﻴﻔﻴﺔ‬ ‫ﺣﺴﺎﺏ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ – ﺍﻟﺠﺰﻳﺌﺎﺕ ﺃﻭﺍﻷﻳﻮﻧﺎﺕ ﺃﻭ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻐﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ -‬ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﻛﺘﻠﺔ ﻣﻌﻴﻨﺔ‪.‬‬ ‫ﺍﻟﻤﻜﻮﻧﺔ ﻟﻬﺎ‪ ،‬ﺇﺫ‬ ‫ﺗﺬﻛﺮ ﺑﺄﻧﻪ ﻻ ﻳﻤﻜﻦ ﺍﻟﺘﺤﻮﻳﻞ ﻣﺒﺎﺷﺮﺓ ﻣﻦ ﻛﺘﻠﺔ ﺍﻟﻤﺎﺩﺓ ﺇﻟﻰ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ‬ ‫ﹼ‬ ‫ﻻ ﺑﺪ ﻣﻦ ﺃﻥ ﺗﺤﻮﻝ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻌﻄﺎﺓ ﺇﻟﻰ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﻓﻲ ﺍﻟﺒﺪﺍﻳﺔ‪ ،‬ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﻀﺮﺏ ﻓﻲ‬ ‫ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﺛﻢ ﻳﻤﻜﻨﻚ ﺑﻌﺪ ﺫﻟﻚ ﺗﺤﻮﻳﻞ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﺇﻟﻰ ﻋﺪﺩ ﺟﺴﻴﻤﺎﺕ‬ ‫ﻣﻦ ﺧﻼﻝ ﺍﻟﻀﺮﺏ ﻓﻲ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‪ .‬ﻭﻟﺘﺤﺪﻳﺪ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪،‬‬ ‫ﺳﻮﻑ ﺗﺤﺘﺎﺝ ﺇﻟﻰ ﻣﻌﺎﻣﻼﺕ ﺗﺤﻮﻳﻞ ﺗﻌﻄﻲ ﻧﺴﺒﺔ ﺃﻋﺪﺍﺩ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‬ ‫ﺇﻟﻰ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻨﻪ ‪ ،‬ﻭﻫﻲ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺍﻟﻤﺜﺎﻝ ‪ 5-9‬ﻳﺒﻴﻦ ﻛﻴﻔﻴﺔ ﺣﻞ‬ ‫ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺍﻟﻤﺴﺎﺋﻞ‪.‬‬

‫‪5-9 ∫Éãe‬‬ ‫ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﻛﺘﻠﺔ ﺇﻟﻰ ﻣﻮﻻﺕ ﺛﻢ ﺇﻟﻰ ﺟﺴﻴﻤﺎﺕ ﻳﺴﺘﻌﻤﻞ ﻛﻠﻮﺭﻳﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ )‪ (AlCl3‬ﻟﺘﻜﺮﻳﺮ ﺍﻟﺒﺘﺮﻭﻝ ﻭﺻﻨﺎﻋﺔ ﺍﻟﻤﻄﺎﻁ‬ ‫ﻭﺍﻟﺸﺤﻮﻡ‪ .‬ﻓﺈﺫﺍ ﻛﺎﻥ ﻟﺪﻳﻚ ﻋﻴﻨﺔ ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﻛﺘﻠﺘﻬﺎ ‪ 35.6g‬ﻓﺠﺪ‪:‬‬ ‫‪ .a‬ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ .b‬ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻜﻠﻮﺭ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ .c‬ﺍﻟﻜﺘﻠﺔ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ ﻟﻮﺣﺪﺓ ﺻﻴﻐﺔ ﻭﺍﺣﺪﺓ ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻷﻟﻮﻣﻨﻴﻮﻡ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﺪﻳﻚ ‪ 35.6g‬ﻣﻦ ‪ AlCl3‬ﻭﻋﻠﻴﻚ ﺃﻥ ﺗﺤﺴﺐ ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ﻛﻞ ﻣﻦ ‪ ،Al3+‬ﻭ‪ Cl-‬ﻭﻛﺘﻠﺔ ﺻﻴﻐﺔ ﻭﺣﺪﺓ ﻭﺍﺣﺪﺓ ﻣﻦ ‪AlCl3‬‬ ‫ﻋﻠﻤﺎ ﺑﺄﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻭﺍﻟﻨﺴﺐ ﻣﻦ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻫﻲ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﻄﻠﻮﺑﺔ‪،‬‬ ‫ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪ .‬ﹰ‬ ‫‪3+‬‬ ‫‬‫ﻭﺑﻤﺎ ﺃﻥ ﻧﺴﺒﺔ ﺃﻳﻮﻧﺎﺕ ‪ Al‬ﺇﻟﻰ ﺃﻳﻮﻧﺎﺕ ‪ Cl‬ﻓﻲ ﺍﻟﺼﻴﻐﺔ ﻫﻲ ‪ ،3:1‬ﻟﺬﺍ ﻓﺈﻥ ﻋﺪﺩ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﻤﺤﺴﻮﺑﺔ ﻳﺠﺐ ﺃﻥ ﺗﻜﻮﻥ‬ ‫ﺑﺎﻟﻨﺴﺒﺔ ﻧﻔﺴﻬﺎ‪.‬‬ ‫ﺍﳌﻌﻄﻴﺎﺕ‬

‫ﺍﻟﻜﺘﻠﺔ = ‪ 35.6g‬ﻣﻦ ‪AlCl3‬‬

‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪.AlCl3‬‬ ‫ﺍﺿﺮﺏ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ Al‬ﻓﻲ ﻛﺘﻠﺘﻪ ﺍﻟﻤﻮﻟﻴﺔ‬

‫ﺍﺿﺮﺏ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ Cl‬ﻓﻲ ﻛﺘﻠﺘﻪ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬

‫ﺍﳌﻄﻠﻮﺏ‬ ‫ﻋﺪﺩ ﺍﻷﻳﻮﻧﺎﺕ = ؟ ‪Al ions‬‬ ‫ﻋﺪﺩ ﺍﻷﻳﻮﻧﺎﺕ = ؟ ‪Cl- ions‬‬ ‫ﺍﻟﻜﺘﻠﺔ = ؟ ‪ g/unit‬ﻣﻦ ‪AlCl3‬‬ ‫‪3+‬‬

‫ﻣﻦ‬ ‫=‪ 1 mol‬ﻣﻦ ‪ 26.98 g= Al 26.98 g × Al‬ﻣﻦ ‪Al‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪Al‬‬

‫ﻣﻦ‬ ‫=‪ 3 mol‬ﻣﻦ ‪ 106.35 g= Cl 35.45 g × Cl‬ﻣﻦ ‪Cl‬‬ ‫‪ 1 mol‬ﻣﻦ‬

‫‪Cl‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ= )‪ 133.33 g/mol = (106.35g + 26.98g‬ﻣﻦ ‪AlCl3‬‬

‫ﺣﺎﺻﻞ ﺟﻤﻊ ﺍﻟﻜﺘﻞ‬ ‫ﺍﺳﺘﻌﻤﻞ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ) ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﻟﻤﻮﻻﺕ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪.‬‬

‫‪60‬‬

‫ﻃﺒﻖ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ‬ ‫ﻋـﻮﺽ ﻛﺘﻠـﺔ ‪ ،AlCl3‬ﻭﻣﻘﻠـﻮﺏ ﺍﻟﻜﺘﻠﺔ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ ‪ ،‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫ﺍﺿﺮﺏ ﺍﻷﻋﺪﺍﺩ ﻭﺍﻟﻮﺣﺪﺍﺕ ﻭﺍﻗﺴﻤﻬﺎ‪.‬‬

‫‪ 1 mol‬ﻣﻦ ‪AlCl3‬‬ ‫= ‪ g‬ﻣﻦ ‪× AlCl3‬‬ ‫‪ g‬ﻣﻦ ‪AlCl3‬‬ ‫= ‪ 35.6 g‬ﻣﻦ ‪× AlCl3‬‬

‫= ‪ moles‬ﻣﻦ ‪AlCl3‬‬

‫‪ 1 mol‬ﻣﻦ ‪AlCl3‬‬

‫‪ 133.33 g‬ﻣﻦ ‪AlCl3‬‬

‫= ‪ 0.276 mol‬ﻣﻦ ‪× AlCl3‬‬ ‫= ‪1.61×1023 Formula unit‬‬

‫‪6.02×1023 Formula unit‬‬ ‫‪ 1 mol‬ﻣﻦ ‪AlCl3‬‬

‫ﺍﺿﺮﺏ ﺍﻷﻋﺪﺍﺩ ﻭﺍﻟﻮﺣﺪﺍﺕ ﻭﺍﻗﺴﻤﻬﺎ‪.‬‬

‫= ‪ 1.61 ×1023 Formula unit‬ﻣﻦ ‪× AlCl3‬‬

‫ﺍﺿﺮﺏ ﺍﻷﻋﺪﺍﺩ ﻭﺍﻟﻮﺣﺪﺍﺕ ﻭﺍﻗﺴﻤﻬﺎ‪.‬‬

‫= ‪ 1.6 ×1023 Formula unit‬ﻣﻦ ‪× AlCl3‬‬

‫= ‪ 1.61 ×1023 ions‬ﻣﻦ ‪Al3+‬‬

‫= ‪ 4.83 ×1023 ions‬ﻣﻦ ‪Cl-‬‬

‫ﻋﻮﺽ ‪ 133.33g‬ﻣﻦ ‪ ،AlCl3‬ﺛﻢ ﹸﺣﻞ‪.‬‬

‫= ‪ 0.267mol‬ﻣﻦ ‪AlCl3‬‬

‫‪ 1 ions‬ﻣﻦ ‪Al3+‬‬

‫‪ 1 Formula unit‬ﻣﻦ ‪AlCl3‬‬ ‫‪ 3 ions‬ﻣﻦ ‪Cl-‬‬

‫‪ 1 Formula unit‬ﻣﻦ ‪AlCl3‬‬

‫ﺍﺣﺴﺐ ﻛﺘﻠﺔ ‪ AlCl3‬ﺑﺎﺳﺘﻌﻤﺎﻝ ﻣﻘﻠﻮﺏ ﻋﺪﺩ ﺍﻓﻮﺟﺎﺩﺭﻭ‬

‫ﻣﻦ‬ ‫= ‪× AlCl3 133.33 g‬‬ ‫‪23‬‬ ‫‪6.02 ×10 Formula unit‬‬ ‫‪1 mol‬‬ ‫‪1 mol‬‬

‫= ‪ 1.61 ×1023 g/formula unit‬ﻣﻦ ‪Al3+‬‬ ‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ‪ Cl-‬ﻳﺴﺎﻭﻱ ﺛﻼﺛﺔ ﺃﺿﻌﺎﻑ ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ‪ Al‬ﻛﻤﺎ ﻫﻮ ﻣﺘﻮﻗﻊ‪ ،‬ﻳﻤﻜﻦ ﺣﺴﺎﺏ ﻛﺘﻠﺔ ﻭﺣﺪﺓ ﺻﻴﻐﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻣﻦ ‪ AlCl3‬ﺑﻄﺮﻳﻘﺔ ﻣﺨﺘﻠﻔﺔ‪ .‬ﺍﻗﺴﻢ ﻛﺘﻠﺔ )‪ (35.6g‬ﻣﻦ ‪ AlCl3‬ﻋﻠﻰ ﻋﺪﺩ ﻭﺣﺪﺍﺕ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﻜﺘﻠﺔ‬ ‫)‪ (1.61 ×1023‬ﻟﺤﺴﺎﺏ ﻛﺘﻠﺔ ﻭﺣﺪﺓ ﺻﻴﻐﺔ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻭﺍﺣﺪﺓ‪ .‬ﻛﻠﺘﺎ ﺍﻹﺟﺎﺑﺘﻴﻦ ﻣﺘﻄﺎﺑﻘﺘﺎﻥ‪.‬‬ ‫‪3+‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫ﻣﺼﺪﺭﺍ ﻟﻠﻮﻗﻮﺩ‪ ،‬ﻭﳜﻠﻂ ﺃﺣﻴﺎ ﹰﻧﺎ ﻣﻊ ﺍﳉﺎﺯﻭﻟﲔ‪ .‬ﺇﺫﺍ ﻛﺎﻥ ﻟﺪﻳﻚ ﻋﻴﻨﺔ ﻣﻦ ﺍﻹﻳﺜﺎﻧﻮﻝ‬ ‫‪ .42‬ﻳﺴﺘﻌﻤﻞ ﺍﻹﻳﺜﺎﻧﻮﻝ ) ‪(C2H5OH‬‬ ‫ﹰ‬ ‫) ‪ (C2H5OH‬ﻛﺘﻠﺘﻬﺎ ‪ 45.1g‬ﺟﺪ‪:‬‬ ‫‪ .b‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﳍﻴﺪﺭﻭﺟﲔ ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ .a‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ .c‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﲔ ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ .43‬ﻋﻴﻨﺔ ﻣﻦ ﻛﱪﻳﺘﻴﺖ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪ (Na2SO3‬ﻛﺘﻠﺘﻬﺎ ‪ ،2.25g‬ﺟﺪ‪:‬‬ ‫‪2‬‬‫‪ . b‬ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ‪ SO3‬ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ . a‬ﻋﺪﺩ ﺃﻳﻮﻧﺎﺕ ‪ Na+‬ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ‪.‬‬ ‫‪ . c‬ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ ﻟﻮﺣﺪﺓ ﺻﻴﻐﺔ ﻭﺍﺣﺪﺓ‪ ،‬ﻣﻦ ‪ Na2SO3‬ﰲ ﺍﻟﻌﻴﻨﺔ‪.‬‬ ‫‪ .44‬ﻋﻴﻨﺔ ﻣﻦ ﺛﺎﲏ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ )‪ (CO2‬ﻛﺘﻠﺘﻬﺎ ‪.52.0g‬ﺟﺪ‪:‬‬ ‫‪ .b‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﲔ ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ ‪.‬‬ ‫‪ .a‬ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﺍﳌﻮﺟﻮﺩﺓ ﻓﻴﻬﺎ ‪.‬‬ ‫‪ .c‬ﻛﺘﻠﺔ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ ‪ CO2‬ﺑﺎﳉﺮﺍﻣﺎﺕ‪.‬‬ ‫‪24‬‬ ‫‪ .45‬ﻣﺎ ﻛﺘﻠﺔ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ )‪ (NaCl‬ﺍﻟﺘﻲ ﲢﺘﻮﻱ ﻋﲆ ‪4.59 × 10 Formula unit‬؟‬ ‫‪ .46‬ﲢﺪﱟ ‪ :‬ﻋﻴﻨﺔ ﻣﻦ ﻛﺮﻭﻣﺎﺕ ﺍﻟﻔﻀﺔ ﻛﺘﻠﺘﻬﺎ ‪25.8g‬‬ ‫‪ .b‬ﻋﺪﺩ ﺍﻷﻳﻮﻧﺎﺕ ﺍﳌﻮﺟﺒﺔ ﻓﻴﻬﺎ ‪.‬‬ ‫‪ .a‬ﺍﻛﺘﺐ ﺻﻴﻐﺔ ﻛﺮﻭﻣﺎﺕ ﺍﻟﻔﻀﺔ‪.‬‬ ‫‪ .d‬ﻣﻘﺪﺍﺭ ﺍﻟﻜﺘﻠﺔ ﺑﺎﳉﺮﺍﻣﺎﺕ ﻟﻮﺣﺪﺓ ﺻﻴﻐﺔ ﻭﺍﺣﺪﺓ ﻣﻨﻬﺎ‪.‬‬ ‫‪ .c‬ﻋﺪﺩ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻴﻬﺎ‪.‬‬

‫‪61‬‬

‫ﺍﻟﺸﻜﻞ‪ 5-10‬ﻻﺣﻆ ﺍﻟﻤﻮﻗﻊ ﺍﻟﻤﺮﻛﺰﻱ‬ ‫ﻟﻠﻤﻮﻝ‪ ،‬ﺇﺫﺍ ﺗﺤﺮﻛﺖ ﻣﻦ ﻳﻤﻴﻦ ﺍﻟﺸﻜﻞ ﺃﻭ‬ ‫ﻳﺴﺎﺭﻩ ﺃﻭ ﺃﻋﻼﻩ ﺃﻭ ﻷﻱ ﻣﻜﺎﻥ ﺁﺧﺮ ﻳﺠﺐ‬ ‫ﺃﻥ ﺗﻤﺮ ﻣﻦ ﺧﻼﻝ ﺍﻟﻤﻮﻝ‪ .‬ﻭﺗﺰﻭﺩﻧﺎ‬ ‫ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﻜﺘﻮﺑﺔ ﻋﻠﻰ ﺍﻷﺳﻬﻢ‬ ‫ﺑﻤﻌﻠﻮﻣﺎﺕ ﻋﻦ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺤﻮﻳﻞ‪.‬‬

‫ﻣﻮﻝ ﻣﻦ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ‬ ‫‪ 1 mol‬ﻣﻦ ﺍﳌﺮﻛﺐ‬ ‫ﻣﻮﻝ ﻣﻦ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ‬

‫ﻣﻮﻝ ﻣﻦ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ‬ ‫‪1 mol compound‬‬

‫‪ 6.02 × 1023‬ﻣﻦ ﺍﳉﺴﻴﲈﺕ‬

‫‪1 mol‬‬ ‫ﻋﺪﺩ ﺍﳉﺮﺍﻣﺎﺕ‬

‫‪1 mol‬‬

‫ﻋﺪﺩ ﺍﳉﺮﺍﻣﺎﺕ‬

‫‪1 mol‬‬

‫‪ 6.02 × 1023‬ﻣﻦ ﺍﳉﺴﻴﲈﺕ‬ ‫ﺍﳉﺴﻴﲈﺕ ﺍﳌﻤﺜﻠﺔ‬

‫‪1 mol‬‬

‫ﻣﻮﻝ ﻣﻦ ﺍﳌﺮﻛﺐ‬

‫ﻛﺘﻠﺔ ﺍﳌﺮﻛﺐ‬

‫‪C11-06C-828378-08‬ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﻭﺍﻟﻤﻮﻻﺕ‪ ،‬ﻭﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬‬ ‫ﻣﻠﺨﺼﺎ‬ ‫ﻳﺘﻀﻤﻦ ﺍﻟﺸﻜﻞ ‪5-10‬‬ ‫ﹰ‬ ‫ﻻﺣﻆ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻭﻣﻘﻠﻮﺑﻬﺎ ﻫﻤﺎ ﻣﻌﺎﻣﻼ ﺍﻟﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻜﺘﻠﺔ ﻭﻋﺪﺩ‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬ﻭﺃﻥ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻭﻣﻘﻠﻮﺑﻪ ﻫﻤﺎ ﻣﻌﺎﻣﻼ ﺍﻟﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻤﻮﻻﺕ ﻭﻋﺪﺩ‬ ‫ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬ﻭﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﺍﻟﻤﻮﻻﺕ ﻭﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ‬ ‫ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪ ،‬ﺍﺳﺘﻌﻤﻞ ﻧﺴﺐ ﻣﻮﻻﺕ ﺍﻟﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ ﺇﻟﻰ ﻣﻮﻝ‬ ‫ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ﺃﻭ ﻣﻘﻠﻮﺑﻪ‪ ،‬ﻛﻤﺎ ﻫﻮ ﻣﺒﻴﻦ ﻋﻠﻰ ﺍﻷﺳﻬﻢ ﺍﻟﻤﺘﺠﻬﺔ ﺇﻟﻰ ﺍﻷﻋﻠﻰ ﺃﻭ‬ ‫ﺍﻷﺳﻔﻞ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ ،5-10‬ﻭﻫﺬﻩ ﺍﻟﻨﺴﺐ ﺗﺸﺘﻖ ﻣﻦ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪5 - 3‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺗـﺪﻝ ﺍﻷﺭﻗـﺎﻡ ﻓـﻲ ﺍﻟﺼﻴـﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻋﻠﻰ‬ ‫ﻋـﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻌﻨﺼﺮ ﻓﻲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ‬ ‫ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫ﺗﹸﺤﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻣﻦ ﺍﻟﻜﺘﻞ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ ﻟﺠﻤﻴﻊ ﺍﻟﻌﻨﺎﺻﺮﻓﻴﻪ‪.‬‬ ‫ﺗﺴـﺘﻌﻤﻞ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳـﻞ ﺍﻟﻤﺒﻨﻴﺔ ﻋﻠﻰ‬ ‫ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﻠﻤﺮﻛـﺐ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ‬ ‫ﻣﻮﻻﺕ ﺍﻟﻤﺮﻛﺐ ﻭﻛﺘﻠﺘﻪ‪.‬‬

‫‪ .47‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺻﻒ ﻛﻴﻒ ﺗﺤﺪﺩ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ‪.‬‬ ‫‪ .48‬ﺣﺪﺩ ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﻄﻠﻮﺑﺔ ﻟﻠﺘﺤﻮﻳﻞ ﺑﻴﻦ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻤﺮﻛﺐ ﻭﻛﺘﻠﺘﻪ‪.‬‬ ‫‪ .49‬ﻭﺿـﺢ ﻛﻴـﻒ ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﺤـﺪﺩ ﻋﺪﺩ ﺍﻟـﺬﺭﺍﺕ ﺃﻭ ﺍﻷﻳﻮﻧﺎﺕ ﻓـﻲ ﻛﺘﻠﺔ ﻣﻌﻴﻨﺔ ﻣﻦ‬ ‫ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫‪ .50‬ﻃﺒﻖ‪ :‬ﻣﺎ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺕ ﻛﻞ ﻣﻦ ‪ O ، C ،K‬ﻓﻲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ‪K2C2O4‬؟‬ ‫‪ .51‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺒﺮﻭﻣﻴﺪ ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ ‪.MgBr2‬‬ ‫‪ .52‬ﺍﺣﺴﺐ‪ :‬ﻣﺎ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ Ca2+‬ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ‪ 1000 mg‬ﻣﻦ ‪CaCO3‬؟‬ ‫ﺭﺳﻤﺎ ﺑﻴﺎﻧ ﹰﻴﺎ ﺑﺎﻷﻋﻤﺪﺓ ﻳﻈﻬﺮ ﻋﺪﺩ ﻣﻮﻻﺕ ﻛﻞ ﻋﻨﺼﺮ ﻣﻮﺟﻮﺩ ﻓﻲ ‪ 500 g‬ﻣﻦ‬ ‫‪ .53‬ﺻﻤﻢ ﹰ‬ ‫ﺍﻟﺴﻤﻴﺔ‪.‬‬ ‫ﺍﻟﺪﺍﻳﻮﻛﺴﻴﻦ )‪ ،(C12H4Cl4O2‬ﺍﻟﺸﺪﻳﺪ‬ ‫ﹼ‬

‫ﻟﻤﺰﻳﺪ ﻣﻦ ﺍﻻﺧﺘﺒﺎﺭﺍﺕ ﺍﻟﻘﺼﻴﺮﺓ ﺍﺭﺟﻊ ﺇﻟﻰ ﺍﻟﻤﻮﻗﻊ‪www.obeikaneducation.com :‬‬

‫‪62‬‬

‫‪5-4‬‬ ‫ا ﻫﺪاف‬

‫ﺗﻔﴪ ﻣﺎ ﺍﳌﻘﺼﻮﺩ ﺑﺎﻟﱰﻛﻴﺐ‬ ‫ﺍﻟﻨﺴﺒﻲ ﺍﳌﺌﻮﻱ ﻟﻠﻤﺮﻛﺐ‪.‬‬ ‫ﲢﺪﺩ ﺍﻟﺼﻴﻐﺘﲔ ﺍﻷﻭﻟﻴﺔ ﻭﺍﳉﺰﻳﺌﻴﺔ‬ ‫ﻟﻠﻤﺮﻛﺐ ﻣﻦ ﺧﻼﻝ ﺍﻟﱰﻛﻴﺐ‬ ‫ﺍﻟﻨﺴﺒﻲ ﺍﳌﺌﻮﻱ ﻭﺍﻟﻜﺘﻞ ﺍﳊﻘﻴﻘﻴﺔ‬ ‫ﻟﻠﻤﺮﻛﺐ‪.‬‬ ‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﻟﻨﺴﺒﺔ ﺍﳌﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ‪ :‬ﻧﺴﺒﺔ ﻛﺘﻠﺔ‬ ‫ﻛﻞ ﻋﻨﴫ ﺇﱃ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻜﻠﻴﺔ‬ ‫ﻟﻠﻤﺮﻛﺐ‪.‬‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﻟﱰﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﳌﺌﻮﻱ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ‬

‫اﻟﺼﻴﻐﺔ ا وﻟﻴﺔ واﻟﺼﻴﻐﺔ اﻟﺠﺰﻳﺌﻴﺔ‬ ‫‪Empirical and Molecular Formulas‬‬ ‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻟﺼﻴﻐـﺔ ﺍﳉﺰﻳﺌﻴﺔ ﳌﺮﻛـﺐ ﻣﺎ ﻫﻲ ﻧﺎﺗـﺞ ﴐﺏ ﺻﻴﻐﺘﻪ ﺍﻷﻭﻟﻴﺔ ﰲ‬ ‫ﻋﺪﺩ ﺻﺤﻴﺢ‪ ،‬ﻭﺗﻀﻢ ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ ﻓﻘﻂ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﻟﻌﻠﻚ ﻻﺣﻈﺖ ﺃﻥ ﺑﻌﺾ ﻋﺒﻮﺍﺕ ﺍﳌﴩﻭﺑﺎﺕ ﺃﻭ ﻭﺟﺒﺎﺕ‬ ‫ﺍﻟﻄﻌـﺎﻡ ﲢﺪﺩ ﻛﻤﻴﺔﺍﻟﺴـﻌﺮﺍﺕ ﺍﳊﺮﺍﺭﻳﺔ ﰲ ﺟﺰﺀ ﻣﻨﻬﺎ)ﻗﻄﻌـﺔ‪ ،‬ﻣﻌﻠﻘﺔ‪(...،g، ml ،‬‬ ‫ﻓﻜﻴﻒ ﻳﻤﻜﻨﻚ ﲢﺪﻳﺪ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻜﻠﻴﺔ ﻟﻠﺴﻌﺮﺍﺕ ﺍﳊﺮﺍﺭﻳﺔ ﰲ ﺍﻟﻌﺒﻮﺓ ﺃﻭ ﺍﻟﻮﺟﺒﺔ؟‬

‫الﺘرﻛﻴﺐ الﻨسﺒﻲ المﺌﻮﻱ‬ ‫‪Percent Composiotion‬‬

‫ﻏﺎﻟﺒـ ﹰﺎ ﻣـﺎ ﻳﻨﺸـﻐﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﻮﻥ ﰲ ﺗﻄﻮﻳـﺮ ﺍﳌﺮﻛﺒـﺎﺕ ﻟﻼﺳـﺘﻌﲈﻻﺕ ﺍﻟﺼﻨﺎﻋﻴـﺔ‪،‬‬ ‫ﻭﺍﻟﺪﻭﺍﺋﻴـﺔ‪ ،‬ﻭﺍﳌﻨﺰﻟﻴـﺔ‪ ،‬ﻛﲈ ﰲ ﺍﻟﺸـﻜﻞ ‪ ،5-11‬ﻓﺒﻌـﺪ ﺃﻥ ﻳﻘـﻮﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺼﻨﺎﻋﻲ‬ ‫)ﺍﻟـﺬﻱ ﳛﴬ ﻣﺮﻛﺒﺎﺕ ﺟﺪﻳﺪﺓ( ﺑﺘﺤﻀﲑﻣﺮﻛﺐ ﺟﺪﻳﺪ ﻳﻘـﻮﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺘﺤﻠﻴﲇ‬ ‫ﺑﺘﺤﻠﻴﻞ ﺍﳌﺮﻛﺐ ﻟﻴﻘﺪﻡ ﺩﻟﻴ ﹰ‬ ‫ﻼ ﻋﻤﻠ ﹰﻴﺎ ﻋﲆ ﺗﺮﻛﻴﺒﻪ ﻭﺻﻴﻐﺘﻪ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬ ‫ﺇﻥ ﻣﻬﻤـﺔ ﺍﻟﻜﻴﻤﻴﺎﺋـﻲ ﺍﻟﺘﺤﻠﻴـﲇ ﻫﻲ ﲢﺪﻳـﺪ ﺍﻟﻌﻨﺎﴏﺍﻟﺘﻲ ﳛﻮﳞـﺎ ﺍﳌﺮﻛﺐ‪ ،‬ﻭﲢﺪﻳﺪ‬ ‫ﻧﺴـﺒﻬﺎ ﺍﳌﺌﻮﻳـﺔ ﺑﺎﻟﻜﺘﻠﺔ‪ .‬ﻓﺎﻟﺘﺤﺎﻟﻴـﻞ ﺍﻟﻮﺯﻧﻴﺔ ﻭﺍﳊﺠﻤﻴﺔ ﻫﻲ ﺇﺟـﺮﺍﺀﺍﺕ ﻋﻤﻠﻴﺔ ﻣﺒﻨﻴﺔ‬ ‫ﻋﲆ ﻗﻴﺎﺱ ﻛﺘﻞ ﺍﳌﻮﺍﺩ ﺍﻟﺼﻠﺒﺔ ﻭﺣﺠﻮﻡ ﺍﻟﺴﻮﺍﺋﻞ‪.‬‬ ‫ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻣﻦ ﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﻟﻌﻤﻠﻴﺔ ﻓﻌﻠﻰ ﺳـﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‪،‬ﺇﺫﺍ ﺃﺧﺬﺕ ﻋﻴﻨﺔ‬

‫ﻛﺘﻠﺘﻬـﺎ ‪ 100g‬ﻣـﻦ ﻣﺮﻛـﺐ ﻳﺤﺘـﻮﻱ ﻋﻠـﻰ ‪ 55g‬ﻣﻦ ﻋﻨﺼـﺮ ‪ X‬ﻭ‪ 45g‬ﻣـﻦ ﻋﻨﺼﺮ ‪.Y‬‬ ‫ﻓﺎﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻷﻱ ﻋﻨﺼﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ﻳﻤﻜﻦ ﺣﺴـﺎﺑﻬﺎ ﺑﻘﺴـﻤﺔ ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ‬ ‫ﻋﻠﻰ ﻛﺘﻠﺔ ﺍﻟﻤﺮﻛﺐ ﻭﺍﻟﻀﺮﺏ ﻓﻲ ﻣﺌﺔ‪.‬‬

‫ﺍﻟﻨﺴﺒﺔ ﺍﳌﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ )ﻟﻠﻌﻨﴫ( =‬

‫ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼــــــــﺮ‬ ‫ﻛﺘﻠﺔ ﺍﳌﺮﻛــــــــﺐ‬

‫× ‪100‬‬

‫ﺍﻟﺸﻜﻞ ‪ 5-11‬ﺗﺤﻀﺮ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺠﺪﻳﺪﺓ‬ ‫ﺑﻜﻤﻴﺎﺕ ﺻﻐﻴﺮﺓ ﻣﻦ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺼﻨﺎﻋﻲ‬ ‫ﻛﻤﺎ ﻓﻲ ﺍﻟﺼﻮﺭﺓ ﺍﻟﻴﻤﻨﻰ‪ ،‬ﺛﻢ ﻳﻘﻮﻡ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫ﺍﻟﺘﺤﻠﻴﻠﻲ ﻛﻤﺎ ﻓﻲ ﺍﻟﺼﻮﺭﺓ ﺍﻟﻴﺴﺮ￯‪،‬‬ ‫ﺑﺘﺤﻠﻴﻞ ﺍﻟﻤﺮﻛﺐ ﻟﻴﺆﻛﺪ ﺻﺤﺔ ﺗﺮﻛﻴﺒﻪ‬ ‫ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻭﺻﻴﻐﺘﻪ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪.‬‬

‫‪63‬‬

‫ﻭﺑﻤﺎ ﺃﻥ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺗﻌﻨﻲ ﺍﻷﺟﺰﺍﺀ ﻓﻲ ﻣﺌﺔ ‪ ،‬ﻓﺈﻥ ﻣﺠﻤﻮﻉ ﺍﻟﻨﺴـﺐ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ﻳﺠﺐ‬ ‫ﺃﻥ ﻳﻜﻮﻥ ‪.100‬‬ ‫‪ 55g‬ﻣﻦ ﺍﻟﻌﻨﺼﺮ ‪x‬‬ ‫× ‪ 55% =100‬ﻣﻦ ‪x‬‬ ‫‪ 100g‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ‬ ‫‪ 45g‬ﻣﻦ ﺍﻟﻌﻨﺼﺮ ‪y‬‬ ‫‪ 100g‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ‬

‫× ‪ 45% =100‬ﻣﻦ ‪y‬‬

‫ﻭﻟﻬﺬﺍ‪ ،‬ﻓﺈﻥ ﺍﻟﻤﺮﻛﺐ ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 55%‬ﻣﻦ ‪ X‬ﻭ‪ 45%‬ﻣﻦ ‪ .Y‬ﻭ ﹸﺗﺴﻤﻰ ﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻠﻤﺮﻛﺐ‪.‬‬ ‫ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻣﻦ ﺧﻼﻝ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻳﻤﻜﻦ ﺗﺤﺪﻳﺪ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻤﺮﻛﺐ ﹰ‬ ‫ﺃﻳﻀﺎ ﻣﻦ ﺧﻼﻝ‬ ‫ﺍﻟﺼﻴﻐـﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻭﻟﻌﻤﻞ ﺫﻟﻚ‪ ،‬ﺍﻓﺘﺮﺽ ﺃﻥ ﻟﺪﻳﻚ ﻣﻮ ﹰ‬ ‫ﻻ ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ﻭﺍﺳـﺘﻌﻤﻞ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﺤﺴـﺎﺏ‬ ‫ﻭﺃﺧﻴﺮﺍ ﺍﺳـﺘﻌﻤﻞ ﺍﻟﻌﻼﻗﺔ ﺃﺩﻧﺎﻩ‬ ‫ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﻠﻤﺮﻛـﺐ‪ ،‬ﺛﻢ ﺍﺣﺴـﺐ ﻛﺘﻠﺔ ﻛﻞ ﻋﻨﺼﺮ ﻓـﻲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣـﻦ ﺍﻟﻤﺮﻛﺐ‪،‬‬ ‫ﹰ‬ ‫ﻟﺤﺴﺎﺏ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﻋﻨﺼﺮ‪.‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻣﻦ ﺧﻼﻝ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ ﻓﻲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻤﺮﻛﺐ‬ ‫× ‪100‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ =‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ‬

‫ﺗـﺠــﺮﺑــﺔ‬ ‫ْ‬ ‫اﻟﻌﻠﻜﺔ‬ ‫ّﺣﻠﻞ‬

‫ﻫﻞ ﺍ ﹸﳌﺤ ﹼﻠﻴﺎﺕ ﻭﺍﻟﻨﻜﻬﺎﺕ ﺗﻀﺎﻑ ﺇﱃ ﺍﻟﻄﺒﻘﺔ ﺍﳋﺎﺭﺟﻴﺔ ﻟﻠﻌﻠﻜﺔ ﺃﻡ ﺗﻜﻮﻥ‬

‫ﳐﻠﻮﻃﺔ ﲠﺎ؟‬

‫ﻣﻌﺎ ‪.‬‬ ‫ﺗﺪﻉ ﺍﻟﻘﻄﻊ ﺗﺘﺠﻤﻊ ﹰ‬

‫ﺣﺬﺭﺍ ﻋﻨﺪ ﺍﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﻘﺺ‪.‬‬ ‫ﺗﺤﺬﻳﺮ‪:‬ﻛﻦ ﹰ‬ ‫‪ .6‬ﺍﺳـﺘﻌﻤﻞ ﹸﻣﻨﺨﻞ ﻗﻴـﺎﺱ )‪ (10cm×10cm‬ﻟﺘﺼﻔﻴـﺔ ﺍﻟﻤﺎﺀ ﻣﻦ‬ ‫ﻗﻄﻊ ﺍﻟﻌﻠﻜﺔ‪ .‬ﻭﺟﻔﻔﻬﺎ ﺑﻤﻨﺎﺷﻒ ﻭﺭﻗﻴﺔ‪ ،‬ﺛﻢ ﻗﺲ ﻛﺘﻠﺘﻬﺎ ﻭﺳﺠﻠﻬﺎ‪.‬‬

‫اﻟﺨﻄﻮات‬ ‫اﻟﺘﺤﻠﻴﻞ‬ ‫‪AK FHBL GCIM H‬‬ ‫‪F K‬‬ ‫ﻧﻤﻮﺫﺝ‪G‬‬ ‫‪I N‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪D AE BF ACG BDH CE I DFJ EG‬‬ ‫ﺍﻟﺴﻼﻣﺔ ﻓﻲ‪O‬‬ ‫ﺍﻟﻤﺨﺘﺒﺮ‪.‬‬ ‫‪N‬‬ ‫‪JDN K‬‬ ‫‪IE‬‬ ‫‪JLP‬‬ ‫‪M‬‬ ‫‪M‬ﺍﻗﺮﺃ ‪LH‬‬ ‫‪O.1‬‬ ‫‪PJ OK P‬‬ ‫ﺍﻟﻤﺤ ﹼﻠﻴﺎﺕ ﻭﺍﻟﻨﻜﻬﺎﺕ ﺍﻟﻤﺬﺍﺑﺔ ﻟﻘﻄﻌﺔ ﺍﻟﻌﻠﻜﺔ ﺍﻟﺘﻲ‬ ‫‪ .1‬ﺍﺣﺴـﺐ ﻛﺘﻠﺔ ﹸ‬ ‫‪ .2‬ﺃﺯﻝ ﺍﻟﻐـﻼﻑ ﻋـﻦ ﻗﻄﻌﺘﻲ ﻋﻠﻜﺔ‪ ،‬ﺛﻢ ﻗﺲ ﻛﺘﻠـﺔ ﻛﻞ ﻣﻨﻬﻤﺎ‬ ‫ﻟﻢ ﺗﻘﻄﻊ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺴـﺎﻭﻱ ﺍﻟﻔﺮﻕ ﺑﻴﻦ ﻛﺘﻠﺔ ﺍﻟﻌﻠﻜﺔ ﺍﻷﺻﻠﻴﺔ ﻭﻛﺘﻠﺔ‬ ‫ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﻴﺰﺍﻥ ﻭﺳﺠﻠﻬﺎ‪.‬‬ ‫ﺍﻟﻌﻠﻜﺔ ﺍﻟﺠﺎﻓﺔ‪.‬‬ ‫‪ .3‬ﺃﺿـﻒ ‪ 150 ml‬ﻣـﻦ ﻣـﺎﺀ ﺍﻟﺼﻨﺒـﻮﺭ ﺍﻟﺒـﺎﺭﺩ ﺇﻟـﻰ ﻛﺄﺱ‬ ‫ﺍﻟﻤﺤ ﹼﻠﻴﺎﺕ ﻭﺍﻟﻨﻜﻬﺎﺕ ﺍﻟﻤﺬﺍﺑﺔ ﻟﻘﻄﻌﺔ ﺍﻟﻌﻠﻜﺔ ﺍﻟﺘﻲ‬ ‫‪ .2‬ﺍﺣﺴﺐ ﻛﺘﻠﺔ ﹸ‬ ‫ﺳـﻌﺔ ‪ .250 ml‬ﻭﺿﻊ ﺇﺣﺪ￯ ﻗﻄﻌﺘﻲ ﺍﻟﻌﻠﻜﺔ ﻓﻲ ﺍﻟﻜﺄﺱ‬ ‫ﻗﻄﻌﺎ ﺻﻐﻴﺮﺓ‪.‬‬ ‫ﻗﻄﻌﺖ ﹰ‬ ‫ﻭﺣﺮﻛﻬﺎ ﺑﻘﻀﻴﺐ ﺗﺤﺮﻳﻚ ﻣﺪﺓ ﺩﻗﻴﻘﺘﻴﻦ‪.‬‬ ‫‪ .3‬ﻃﺒـﻖ‪ :‬ﺍﺣﺴـﺐ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻟﻜﺘﻠـﺔ ﺍﻟﻌﻠﻜﺔ )ﺃﻱ ﺑﻌـﺪ ﺫﻭﺑﺎﻥ‬ ‫ﻭﺍﻟﻤﺤ ﹼﻠﻴﺎﺕ(‪.‬‬ ‫‪ .4‬ﺃﺧﺮﺝ ﻗﻄﻌﺔ ﺍﻟﻌﻠﻜﺔ ﻭﺟﻔﻔﻬﺎ ﺑﺎﺳﺘﻌﻤﺎﻝ ﻣﻨﺎﺷﻒ ﻭﺭﻗﻴﺔ‪ ،‬ﺛﻢ‬ ‫ﺍﻟﻨﻜﻬﺎﺕ ﹸ‬ ‫‪ .4‬ﺍﺳـﺘﻨﺘﺞ‪ :‬ﻣﺎﺫﺍ ﻳﻤﻜﻦ ﺃﻥ ﺗﺴـﺘﻨﺘﺞ ﻣﻦ ﺍﻟﻨﺴـﺒﺘﻴﻦ ﺍﻟﻤﺌﻮﻳﺘﻴﻦ؟ ﻫﻞ‬ ‫ﻗﺲ ﻛﺘﻠﺘﻬﺎ ﻭﺳﺠﻠﻬﺎ‪.‬‬ ‫ﺍﻟﻤﺤ ﹼﻠﻴﺎﺕ ﻭﺍﻟﻨﻜﻬـﺎﺕ ﻣﺨﻠﻮﻃﺔ‬ ‫ﺍﻟﻌﻠﻜـﺔ ﻣﻐﻄـﺎﺓ ﺑﺎﻟﺴـﻜﺮ ﺃﻡ ﺃﻥ ﹸ‬ ‫ﻣﻘﺼـﺎ ﻟﺘﻘﻄﻴـﻊ ﻗﻄﻌـﺔ ﺍﻟﻌﻠﻜـﺔ ﺍﻟﺜﺎﻧﻴـﺔ ﺇﻟﻰ ﻗﻄﻊ‬ ‫‪ .5‬ﺍﺳـﺘﻌﻤﻞ ﹰ‬ ‫ﺑﺎﻟﻌﻠﻜﺔ؟‬ ‫ﺻﻐﻴـﺮﺓ‪ .‬ﻭﻛﺮﺭ ﺍﻟﺨﻄﻮﺓ ﺍﻟﺜﺎﻟﺜﺔ ﻣﺴـﺘﻌﻤ ﹰ‬ ‫ﻼ ﻣـﺎ ﹰﺀ ﺟﺪﻳﺪﹰ ﺍ‪ .‬ﻭﻻ‬

‫‪64‬‬

‫‪5-10 ∫Éãe‬‬ ‫ﺣﺴﺎﺏ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﺣﺪﺩ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﺜﺎﻧﻲ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ ‪.CO2‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘﺪ ﺃﻋﻄﻴﺖ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻓﻘﻂ‪ .‬ﻭﻟﻬﺬﺍ ﺍﻓﺘﺮﺽ ﺃﻥ ﻟﺪﻳﻚ ﻣﻮ ﹰ‬ ‫ﻻ ﻭﺍﺣﺪﹰ ﺍ ﻣﻦ ‪ ،CO2‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ‬ ‫ﻭﻛﺘﻠﺔ ﻛﻞ ﻋﻨﺼﺮﻓﻲ ﺍﻟﻤﻮﻝ ﺍﻟﻮﺍﺣﺪ ﻟﺘﺤﺪﻳﺪ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﻋﻨﺼﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬

‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬ ‫ﺍﻟﺼﻴﻐﺔ = ‪CO2‬‬ ‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺍﻟﻤﻄﻠﻮﺏ‬ ‫ﻧﺴﺒﺔ ‪ = C‬؟‬ ‫ﻧﺴﺒﺔ ‪ = O‬؟‬

‫ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻭﻧﺴﺒﺔ ﻛﻞ ﻋﻨﺼﺮ ﻓﻴﻪ‪.‬‬

‫‪ 12.01g‬ﻣﻦ ‪C‬‬ ‫ﺍﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻜﺮﺑﻮﻥ ﻓﻲ ﻋﺪﺩ ﺫﺭﺍﺗﻪ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪ 1mol .‬ﻣﻦ ‪× C‬‬ ‫‪ 1mol‬ﻣﻦ ‪C‬‬ ‫‪ 16.00g‬ﻣﻦ ‪O‬‬ ‫ﺍﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻓﻲ ﻋﺪﺩ ﺫﺭﺍﺗﻪ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫= ‪32.00g‬ﻣﻦ ‪O‬‬ ‫‪ 2mol‬ﻣﻦ ‪× O‬‬ ‫‪ 1mol‬ﻣﻦ ‪O‬‬ ‫ﺍﺟﻤﻊ ﻛﺘﻞ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ = ‪ 44.01g/mol = 32.00g + 12.01g‬ﻣﻦ ‪CO2‬‬ ‫= ‪12.01g‬ﻣﻦ ‪C‬‬

‫ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﻋﻨﺼﺮ‬

‫ﻋﻮﺽ ﻛﺘﻠﺔ ﺍﻟﻜﺮﺑﻮﻥ ﻓﻲ ‪ 1 mol‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ= ‪12.01g/mol‬‬ ‫‪12.01g‬‬ ‫ﹺ‬ ‫× ‪27.29% = 100‬‬ ‫ﻭﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟـ ‪ ،44.01g/mol = CO2‬ﻭﺍﺣﺴـﺐ ﻧﺴـﺒﺔ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟـ ‪= C‬‬ ‫‪44.1g‬‬ ‫ﺍﻟﻜﺮﺑﻮﻥ‪.‬‬

‫‪32.00g‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ﺍﻷﻛﺴﺠﻴﻦ ﻓﻲ ‪ 1 mol‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ = ‪32.00g/mol‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪= O‬‬ ‫ﻭﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟـ ‪ ،44.01g/mol = CO2‬ﻭﺍﺣﺴـﺐ ﻧﺴـﺒﺔ‬ ‫‪44.1g‬‬ ‫ﺍﻷﻛﺴﺠﻴﻦ‪.‬‬ ‫‪ CO2‬ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 27.29%‬ﻣﻦ ‪ C‬ﻭ‪ 72.71%‬ﻣﻦ ‪.O‬‬

‫× ‪72.71% = 100‬‬

‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﺑﻤﺎ ﺃﻥ ﺟﻤﻴﻊ ﺍﻟﻜﺘﻞ ﻭﺍﻟﻜﺘﻞ ﺍﻟﻤﻮﻟﻴﺔ ﻓﻴﻬﺎ ﺃﺭﺑﻌﺔ ﺃﺭﻗﺎﻡ ﻣﻌﻨﻮﻳﺔ‪ ،‬ﻟﺬﺍ ﻓﺈﻥ ﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ ﻣﻌﻄﺎﺓ ﺑﺼﻮﺭﺓ ﺻﺤﻴﺤﺔ ‪ .‬ﻭﻟﻮﺃﺧﺬﻧﺎ‬ ‫ﺑﻌﻴﻦ ﺍﻻﻋﺘﺒﺎﺭ ﺣﺪﻭﺙ ﺧﻄﺄ ﻓﻲ ﺗﺪﻭﻳﺮ ﺍﻟﻤﻨﺎﺯﻝ‪ ،‬ﻓﺈﻥ ﻣﺠﻤﻮﻉ ﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻳﺴﺎﻭﻱ ‪ 100%‬ﻛﻤﺎ ﻫﻮ ﻣﻄﻠﻮﺏ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .54‬ﻣـﺎ ﺍﻟﺘﺮﻛﻴـﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌـﻮﻱ ﻟﺤﻤـﺾ ﺍﻟﻔﺴـﻔﻮﺭﻳﻚ‬ ‫)‪(H3PO4‬؟‬ ‫‪ .55‬ﺃﻱ ﺍﻟﻤﺮﻛﺒﻴـﻦ ﺍﻟﺘﺎﻟﻴﻴـﻦ ﺗﻜﻮﻥ ﻓﻴﻪ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ‬ ‫ﻟﻠﻜﺒﺮﻳﺖ ﺃﻋﻠﻰ‪ H2SO3 :‬ﺃﻡ ‪H2SO4‬؟‬ ‫‪ .56‬ﻳﺴـﺘﻌﻤﻞ ﻛﻠﻮﺭﻳـﺪ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ )‪ (CaCl2‬ﻟﻤﻨﻊ ﺍﻟﺘﺠﻤﺪ‪.‬‬ ‫ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﻋﻨﺼﺮ ﻓﻲ ‪.CaCl2‬‬

‫‪ .57‬ﹺ‬ ‫ﺗﺤـﺪﱟ ‪ :‬ﺗﺴـﺘﻌﻤﻞ ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﺼﻮﺩﻳـﻮﻡ ﻓـﻲ ﺻﻨﺎﻋـﺔ‬ ‫ﺍﻟﻤﻨﻈﻔﺎﺕ‪.‬‬ ‫ﺍﻟﻤﻜﻮﻧـﺔ ﻟﻜﺒﺮﻳﺘـﺎﺕ ﺍﻟﺼﻮﺩﻳـﻮﻡ‪ ،‬ﺛﻢ‬ ‫‪ .a‬ﺣـﺪﺩ ﺍﻟﻌﻨﺎﺻـﺮ‬ ‫ﹼ‬ ‫ﺍﻛﺘﺐ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻬﺬﺍ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫‪ .b‬ﺍﺣﺴـﺐ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳـﺔ ﺑﺎﻟﻜﺘﻠـﺔ ﻟـﻜﻞ ﻋﻨﺼـﺮ ﻓـﻲ‬ ‫ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ‪.‬‬

‫‪65‬‬

‫الصﻴﻐﺔ اﻷﻭلﻴﺔ ‪Empirical Formula‬‬

‫ﺍﻟﺸـﻜﻞ ‪ 5-12‬ﹼ‬ ‫ﺗﺬﻛـﺮ ﻫﺬﺍ ﺍﻟﺸـﻜﻞ‬ ‫ﻋﻨـﺪ ﺣـﻞ ﺍﻟﻤﺴـﺎﺋﻞ ﺍﻟﻤﺘﻌﻠﻘـﺔ‬ ‫ﺑﺎﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ‪ .‬ﻳﻤﻜﻨﻚ‬ ‫ﺩﺍﺋﻤـﺎ ﺑﺄﻥ ﻟﺪﻳـﻚ ﻋﻴﻨﺔ‬ ‫ﺍﻻﻓﺘـﺮﺍﺽ ﹰ‬ ‫ﻛﺘﻠﺘﻬـﺎ ‪ 100g‬ﻣـﻦ ﺍﻟﻤﺮﻛـﺐ‪،‬‬ ‫ﻭﺍﺳﺘﻌﻤﻞ ﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ ﻟﻠﻌﻨﺎﺻﺮ‬ ‫ﺑﻮﺻﻔﻬﺎ ﻛﺘ ﹰ‬ ‫ﻼ‪.‬‬ ‫‪100.00% SO3‬‬

‫‪40.05% S‬‬

‫‪59.95% O‬‬

‫= ‪ 59.95g‬ﻣﻦ ‪× O‬‬

‫‪100.00 g SO3‬‬

‫‪66‬‬ ‫‪C11-15C-828378-08‬‬

‫‪ 32.07g‬ﻣﻦ‬

‫‪ 1 mol‬ﻣﻦ‬

‫‪ 16.00g‬ﻣﻦ‬

‫‪S‬‬ ‫‪S‬‬

‫‪O‬‬ ‫‪O‬‬

‫= ‪ 1.249 mol‬ﻣﻦ‬ ‫= ‪ 3.747 mol‬ﻣﻦ‬

‫‪S‬‬

‫‪O‬‬

‫ﻭﻟﺬﺍ ﻓﺈﻥ ﻧﺴﺒﺔ ﺫﺭﺍﺕ ‪ ،S‬ﺇﻟﻰ ﺫﺭﺍﺕ ‪ O‬ﻓﻲ ﺍﻟﻤﺮﻛﺐ ﻫﻲ ‪ .3.747 :1.249‬ﻭﻋﻨﺪﻣﺎ‬ ‫ﻻ ﺗﻜﻮﻥ ﺍﻟﻘﻴﻢ ﻓﻲ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ ﻓﻼ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻟﻬﺎ ﻓﻲ ﺍﻟﺼﻴﻐﺔ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ‪ ،‬ﻭﻟـﺬﺍ ﻳﺠـﺐ ﺗﺤﻮﻳﻠﻬـﺎ ﺇﻟﻰ ﺃﻋـﺪﺍﺩ ﺻﺤﻴﺤـﺔ‪ ،‬ﻭﻟﺠﻌﻞ ﺍﻟﻘﻴﻤـﺔ ﺍﻟﻤﻮﻟﻴﺔ‬ ‫ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ‪ ،‬ﺍﻗﺴـﻢ ﺍﻟﻘﻴﻤﺘﻴﻦ ﺍﻟﻤﻮﻟﻴﺘﻴـﻦ ﻋﻠﻰ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻣﻮﻟﻴﺔ ﻭﻫﻲ ﻟﻠﻜﺒﺮﻳﺖ‬ ‫)‪ ،(1.249‬ﻭﻫـﺬﺍ ﻻ ﻳﻐﻴﺮ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺑﻴﻦ ﺍﻟﻌﻨﺼﺮﻳﻦ ﻷﻥ ﻛﻠﻴﻬﻤﺎ ﺳﻴﻘﺴـﻢ ﻋﻠﻰ‬ ‫ﺍﻟﺮﻗﻢ ﻧﻔﺴﻪ‪.‬‬ ‫‪1.249‬‬

‫‪59.95 g O‬‬

‫‪ 1 mol‬ﻣﻦ‬

‫= ‪ 40.05g‬ﻣﻦ ‪× S‬‬

‫‪ 1.249 mol‬ﻣﻦ‬

‫ﺗﺘﺤﻮﻝ ﺇﱃ‬ ‫ﱠ‬

‫‪40.05 g S‬‬

‫ﻋﻨﺪﻣـﺎ ﹸﻳﻌـﺮﻑ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻤﺮﻛﺐ ﻣـﺎ‪ ،‬ﻓﺈﻧﻪ ﻳﻤﻜﻦ ﺣﺴـﺎﺏ ﺻﻴﻐﺘﻪ‪،‬‬ ‫ﻭﺫﻟﻚ ﺑﺘﺤﺪﻳﺪ ﺃﺻﻐﺮ ﻧﺴـﺒﺔ ﻣـﻦ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻤـﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻴﻪ‪ .‬ﻭﺗﻤﺜﻞ‬ ‫ﻫـﺬﻩ ﺍﻟﻨﺴـﺒﺔ ﻓﻲ ﺻـﻮﺭﺓ ﺃﺭﻗﺎﻡ ﻓﻲ ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴـﺔ‪ .‬ﻓﺎﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴـﺔ ﻟﻤﺮﻛﺐ ﻫﻲ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺘﻲ ﺗﺒﻴﻦ ﺃﺻﻐﺮ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪ .‬ﻭﻗﺪ‬ ‫ﺗﻜـﻮﻥ ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴﺔ ﻫﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻧﻔﺴـﻬﺎ ﺃﻭ ﻣﺨﺘﻠﻔﺔ ﻋﻨﻬﺎ‪ .‬ﻭﺇﺫﺍ ﺍﺧﺘﻠﻔﺖ‬ ‫ﹰ‬ ‫ﺑﺴـﻴﻄﺎ ﻟﻠﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﺩﺍﺋﻤـﺎ ﻣﻀﺎﻋ ﹰﻔﺎ‬ ‫ﺍﻟﺼﻴﻐﺘـﺎﻥ‪ ،‬ﻓﺈﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺳـﺘﻜﻮﻥ ﹰ‬ ‫ﹰ‬ ‫ﻓﻤﺜـﻼ‪ ،‬ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻔﻮﻕ ﺃﻛﺴـﻴﺪ ﺍﻟﻬﻴﺪﺭﻭﺟﻴـﻦ ‪ ،HO‬ﻭﺻﻴﻐﺘـﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻫﻲ‬ ‫‪ .H2O2‬ﻻﺣﻆ ﺃﻥ ﻧﺴـﺒﺔ ﺍﻷﻛﺴﺠﻴﻦ ﺇﻟﻰ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻫﻲ‪ 1:1‬ﻓﻲ ﻛﻠﺘﺎ ﺍﻟﺼﻴﻐﺘﻴﻦ‪.‬‬ ‫ﻭﻳﻤﻜﻦ ﺍﺳـﺘﻌﻤﺎﻝ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌـﻮﻱ ﺃﻭ ﻛﺘﻞ ﺍﻟﻌﻨﺎﺻﺮ ﻓـﻲ ﻛﺘﻠﺔ ﻣﺤﺪﺩﺓ ﻣﻦ‬ ‫ﹰ‬ ‫ﻓﻤﺜـﻼ ﺇﺫﺍ ﺃﻋﻄﻴـﺖ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ‬ ‫ﺍﻟﻤﺮﻛـﺐ ﻟﺤﺴـﺎﺏ ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﻟﻠﻤﺮﻛـﺐ‪ ،‬ﻭﻋﻠﻰ ﻓﺮﺽ ﺃﻥ ﻛﺘﻠﺔ ﺍﻟﻤﺮﻛﺐ ﺍﻟﻜﻠﻴﺔ ﺗﺴـﺎﻭﻱ ‪ ،100.00g‬ﻭﺃﻥ ﺍﻟﻨﺴـﺒﺔ‬ ‫ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻜﻞ ﻋﻨﺼﺮﻣﺴﺎﻭﻳﺔ ﻟﻜﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪.‬ﻛﻤﺎ ﻓﻲ ﺍﻟﺸﻜﻞ‪،5-12‬‬ ‫ﺣﻴﺚ ﺃﻥ ﻛﻞ ‪ 100g‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ ﺗﺘﻜﻮﻥ ﻣﻦ ‪ 40.05%‬ﻣﻦ ‪ S‬ﻭ‪ 59.95%‬ﻣﻦ ‪ ،O‬ﺃﻱ‬ ‫ﺗﺤﺘﻮﻱ ‪40.05g‬ﻣﻦ ‪ S‬ﻭ ‪59.95g‬ﻣﻦ ‪ .O‬ﺛﻢ ﺗﺤﻮﻝ ﻛﺘﻠﺔ ﻛﻞ ﻋﻨﺼﺮ ﺇﻟﻰ ﻣﻮﻻﺕ‪.‬‬

‫‪S‬‬

‫= ‪ 1mol‬ﻣﻦ‬

‫‪S‬‬

‫‪ 3.747 mol‬ﻣﻦ‬ ‫‪1.249‬‬

‫‪O‬‬

‫= ‪ 3mol‬ﻣﻦ‬

‫‪O‬‬

‫ﺃﻱ ﺃﻥ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻤﻮﻻﺕ ‪ S‬ﺇﻟﻰ ‪ O‬ﻫﻲ ‪ .3:1‬ﻭﻋﻠﻴﻪ ﻓﺈﻥ ﺍﻟﺼﻴﻐﺔ‬ ‫ﹾ‬ ‫ﺍﻷﻭﻟﻴﺔ ﻫﻲ ‪.SO3‬ﻭﻓﻲ ﺑﻌﺾ ﺍﻷﺣﻴﺎﻥ‪ ،‬ﻗﺪ ﻻ ﺗﺆﺩﻱ ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻣﻮﻟﻴﺔ‬ ‫ﺇﻟـﻰ ﺃﻋـﺪﺍﺩ ﺻﺤﻴﺤﺔ‪ .‬ﻭﻓـﻲ ﻣﺜﻞ ﻫﺬﻩ ﺍﻟﺤـﺎﻻﺕ‪ ،‬ﻳﺠﺐ ﺿﺮﺏ ﻛﻞ ﻗﻴﻤـﺔ ﻣﻮﻟﻴﺔ ﻓﻲ‬ ‫ﺍﻟﻤﺜﺎﻝ‪.5-11‬‬ ‫ﺃﺻﻐﺮ ﻋﺎﻣﻞ ﻳﺠﻌﻠﻬﺎ ﻋﺪ ﹰﺩﺍ ﺻﺤﻴﺤ ﹰﺎ ﻛﻤﺎ ﻓﻲ ﺍﻟﻤﺜﺎﻝ‬ ‫‪ ?äCGôb GPÉ``e‬ﻋﺪﹼ ﺩ ﺍﻟﺨﻄﻮﺍﺕ ﺍﻟﻤﻄﻠﻮﺑﺔ ﻟﺤﺴـﺎﺏ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻣﻦ ﺍﻟﺘﺮﻛﻴﺐ‬ ‫ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ‪.‬‬

‫‪5-11 ∫Éãe‬‬

‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻣﻦ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﺣﺪﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻤﺮﻛﺐ ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 48.64%‬ﻛﺮﺑﻮﻥ‪ ،‬ﻭ‪ 8.16%‬ﻫﻴﺪﺭﻭﺟﻴﻦ‪،‬‬ ‫ﻭ‪ 43.20%‬ﺃﻛﺴﺠﻴﻦ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘـﺪ ﺃﻋﻄﻴـﺖ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌـﻮﻱ ﻟﻤﺮﻛﺐ‪ ،‬ﻭﺍﻟﻤﻄﻠﻮﺏ ﺗﺤﺪﻳـﺪ ﺻﻴﻐﺘﻪ ﺍﻷﻭﻟﻴـﺔ‪ ،‬ﻭﺑﻤﺎ ﺃﻧﻪ ﻳﻤﻜﻦ ﺍﻻﻓﺘﺮﺍﺽ ﺃﻥ ﺍﻟﻨﺴـﺐ‬ ‫ﺍﻟﻤﺌﻮﻳـﺔ ﺗﻤﺜﻞ ﻛﺘﻞ ﺍﻟﻌﻨﺎﺻﺮﻓﻲ ﻋﻴﻨﺔ ﻣﻘﺪﺍﺭﻫﺎ ‪ ،100g‬ﻟﺬﺍ ﻳﻤﻜﻦ ﺍﺳـﺘﺒﺪﺍﻝ ﺭﻣﺰﺍﻟﻨﺴـﺒﺔ ﺑﺎﻟﻮﺣﺪﺓ )‪ ،(g‬ﺛـﻢ ﹼﺣﻮﻝ ﺍﻟﺠﺮﺍﻣﺎﺕ ﺇﻟﻰ‬ ‫ﻣﻮﻻﺕ‪ ،‬ﻭﺟﺪ ﺃﺻﻐﺮ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ‪.‬‬

‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬

‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪48.64%= C‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪8.16% = H‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪43.20% = O‬‬ ‫‪2‬‬

‫ﺍﻟﻤﻄﻠﻮﺏ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ= ؟‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﹼﺣﻮﻝ ﻛﻞ ﻛﺘﻠﺔ ﺇﻟﻰ ﻣﻮﻻﺕ ﺑﺎﺳﺘﻌﻤﺎﻝ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ) ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ( ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﻟﻤﻮﻻﺕ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ ‪:‬‬ ‫‪ 1mol‬ﻣﻦ ‪C‬‬ ‫= ‪4.050mol‬ﻣﻦ ‪C‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ‪ ،C‬ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪ ،C‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ‪ 48.64g C‬ﻣﻦ ‪× C‬‬ ‫‪ 12.01g‬ﻣﻦ ‪C‬‬ ‫‪ 1mol‬ﻣﻦ ‪H‬‬ ‫= ‪8.10mol‬ﻣﻦ ‪H‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ‪ ،H‬ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪ ،H‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ‪ 8.16g H‬ﻣﻦ ‪× H‬‬ ‫‪ 1.008g‬ﻣﻦ ‪H‬‬ ‫‪ 1mol‬ﻣﻦ ‪O‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ‪ ،O‬ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺﻟـ ‪ ،O‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ‪ 43.20g O‬ﻣﻦ ‪× O‬‬ ‫‪ 16.00g‬ﻣﻦ ‪O‬‬

‫= ‪2.70mol‬ﻣﻦ ‪O‬‬

‫ﺇﺫﻥ‪ ،‬ﻓﺎﻟﻨﺴﺐ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻫﻲ‪ 4.05mol) :‬ﻣﻦ ‪ 8.10mol) (C‬ﻣﻦ ‪ 2.700mol) (H‬ﻣﻦ ‪ . (O‬ﺛﻢ ﺍﺣﺴﺐ ﺃﺑﺴﻂ‬ ‫ﻧﺴﺒﺔ ﻣﻮﻟﻴﺔ ﻟﻠﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ﺑﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻣﻮﻟﻴﺔ )‪.(2.700‬‬ ‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ C‬ﻋﻠﻰ ‪2.700‬‬ ‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ H‬ﻋﻠﻰ ‪2.700‬‬ ‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ O‬ﻋﻠﻰ ‪2.700‬‬

‫‪ 4.050 mol‬ﻣﻦ‬

‫‪C‬‬

‫= ‪ 1.5mol‬ﻣﻦ‬

‫‪H‬‬

‫= ‪ 3mol‬ﻣﻦ‬

‫‪H‬‬

‫‪O‬‬

‫= ‪ 1mol‬ﻣﻦ‬

‫‪O‬‬

‫‪2.700‬‬

‫‪ 8.10 mol‬ﻣﻦ‬ ‫‪2.700‬‬

‫‪ 2.700 mol‬ﻣﻦ‬ ‫‪2.700‬‬

‫‪C‬‬

‫ﻭﺃﺧﻴـﺮﺍ ﺍﺿﺮﺏ ﻛﻞ ﻋﺪﺩ ﺗﺸـﺘﻤﻞ ﻋﻠﻴﻪ‬ ‫ﺃﺑﺴـﻂ ﻧﺴـﺒﺔ ﻣـﻮﻻﺕ ﻫﻲ )‪ 1.5mol‬ﻣـﻦ ‪ 3mol) :(C‬ﻣﻦ ‪ 1mol) :(H‬ﻣﻦ ‪. (O‬‬ ‫ﹰ‬ ‫ﺍﻟﻨﺴﺒﺔ ﻓﻲ ﺃﺻﻐﺮ ﺭﻗﻢ‪ ،‬ﻭﻫﻮ ﻓﻲ ﻫﺬﻩ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺮﻗﻢ‪ ،2‬ﺍﻟﺬﻱ ﻳﺆﺩﻱ ﺇﻟﻰ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ‪.‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ C‬ﻓﻲ ‪ 2‬ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﻋﺪﺩ ﺻﺤﻴﺢ‪.‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ H‬ﻓﻲ ‪ 2‬ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﻋﺪﺩ ﺻﺤﻴﺢ‪.‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ O‬ﻓﻲ ‪ 2‬ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﻋﺪﺩ ﺻﺤﻴﺢ‪.‬‬

‫‪ 1.5 mol × 2‬ﻣﻦ ‪ 3mol = C‬ﻣﻦ‬ ‫‪ 3mol × 2‬ﻣﻦ ‪ 6mol = H‬ﻣﻦ‬ ‫‪ 1mol × 2‬ﻣﻦ ‪ 2mol = O‬ﻣﻦ‬

‫‪C‬‬

‫‪H‬‬ ‫‪O‬‬

‫‪67‬‬

‫ﺃﺑﺴـﻂ ﻧﺴـﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤـﺔ ﻟﻠﻤﻮﻻﺕ ﻫـﻲ )‪ 3atoms‬ﻣـﻦ ‪ 6atoms) :(C‬ﻣﻦ ‪ 2atoms) :(H‬ﻣﻦ ‪ .(O‬ﻭﻫﻜﺬﺍ‪ ،‬ﻓﺈﻥ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻫﻲ ‪.C3H6O2‬‬ ‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﻟﻠﺘﺤﻘﻖ ﻣﻦ ﺻﺤﺔ ﺍﻟﺠﻮﺍﺏ‪ ،‬ﺍﺣﺴﺐ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﺍﻟﻤﻤﺜﻞ ﺑﺎﻟﺼﻴﻐﺔ‪ ،‬ﻟﻠﻮﻗﻮﻑ ﻋﻠﻰ ﻣﺪ￯ ﺍﺗﻔﺎﻗﻪ ﻣﻊ ﻣﻌﻄﻴﺎﺕ ﺍﻟﻤﺴﺄﻟﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .58‬ﻳﻤﺜﻞ ﺍﻟﺮﺳـﻢ ﺍﻟﺒﻴﺎﻧﻲ ﺍﻟﺪﺍﺋﺮﻱ ﺍﻟﻤﺠﺎﻭﺭﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻤﺎﺩﺓ ﺻﻠﺒﺔ ﺯﺭﻗﺎﺀ‪ .‬ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ‬ ‫ﺍﻷﻭﻟﻴﺔ ﻟﻬﺬﻩ ﺍﻟﻤﺎﺩﺓ؟‬ ‫‪O‬‬ ‫‪63.16%‬‬ ‫‪ .59‬ﻣﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻤﺮﻛﺐ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ‪ 35.98%‬ﺃﻟﻮﻣﻨﻴﻮﻡ ﻭ‪ 64.02%‬ﻛﺒﺮﻳﺖ‪.‬‬ ‫‪ .60‬ﺍﻟﺒﺮﻭﺑـﺎﻥ ﻫﻮ ﺃﺣـﺪ ﺍﻟﻬﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ‪ ،‬ﻭﻫﻲ ﻣﺮﻛﺒﺎﺕ ﺗﺤﺘﻮﻱ ﻓﻘﻂ ﻋﻠﻰ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪.‬‬ ‫ﻓﺈﺫﺍ ﻛﺎﻥ ﺍﻟﺒﺮﻭﺑﺎﻥ ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 81.82%‬ﻛﺮﺑﻮﻥ ﻭ‪ 18.18%‬ﻫﻴﺪﺭﻭﺟﻴﻦ‪ ،‬ﻓﻤﺎ ﺻﻴﻐﺘﻪ ﺍﻷﻭﻟﻴﺔ؟‬ ‫ﺗﺤـﺪ‪ :‬ﺍﻹﺳـﺒﺮﻳﻦ ﻳﻌﺪ ﻣﻦ ﺃﻛﺜﺮ ﺍﻷﺩﻭﻳﺔ ﺍﺳـﺘﻌﻤﺎ ﹰ‬ ‫ﱢ‬ ‫ﻻ ﻓﻲ ﺍﻟﻌﺎﻟـﻢ‪ .‬ﻭﻳﺘﻜﻮﻥ ﻣﻦ ‪ 60.00%‬ﻛﺮﺑـﻮﻥ‪ ،‬ﻭ‪ 4.44%‬ﻫﻴﺪﺭﻭﺟﻴﻦ‪،‬‬ ‫‪.61‬‬ ‫ﻭ‪ 35.56%‬ﺃﻛﺴﺠﻴﻦ ‪.‬ﻓﻤﺎ ﺻﻴﻐﺘﻪ ﺍﻷﻭﻟﻴﺔ؟‬ ‫‪N‬‬ ‫‪36.84%‬‬

‫الصﻴﻐﺔ الﺠﺰﻳﺌﻴﺔ ‪Molecular Formula‬‬

‫ﻗـﺪ ﺗﺴـﺘﻐﺮﺏ ﺇﺫﺍ ﻋﻠﻤـﺖ ﺃﻥ ﻣﻮﺍﺩ ﻟﻬﺎ ﺧﻮﺍﺹ ﻣﺨﺘﻠﻔﺔ ﺗﻤﺎ ﹰﻣـﺎ ﻳﻤﻜﻦ ﺃﻥ ﻳﻜﻮﻥ ﻟﻬﺎ ﻧﻔﺲ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻭﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ؟‬ ‫ﺩﺍﺋﻤﺎ‬ ‫ﻭﻛﻴﻒ ﻳﻜﻮﻥ ﺫﻟﻚ ﻣﻤﻜﻨ ﹰﺎ؟ ﺗﺬﻛﺮ ﺃﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﺗﻌﻄﻲ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻟﺬﺭﺍﺕ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪ ،‬ﻭﻟﻜﻦ ﻫﺬﻩ ﺍﻟﻨﺴﺒﺔ ﻻ ﺗﻤﺜﻞ ﹰ‬ ‫ﺍﻟﻌـﺪﺩ ﺍﻟﻔﻌﻠﻲ ﻟﺬﺭﺍﺗﻪ‪ .‬ﻭﻟﺘﻌﺮﻳﻒ ﻣﺮﻛﺐ ﺟﺪﻳﺪ‪ ،‬ﻳﺤﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﻣﺎ ﻳﺴـﻤﻰ ﺑﺎﻟﺼﻴﻐـﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﻌﻄﻲ ﺍﻟﻌﺪﺩ ﺍﻟﻔﻌﻠﻲ ﻟﻠﺬﺭﺍﺕ‬ ‫ﻣـﻦ ﻛﻞ ﻋﻨﺼـﺮ ﻓﻲ ﺟـﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻤﺎﺩﺓ‪ .‬ﻓﻤﺜ ﹰ‬ ‫ﻼ‪ ،‬ﻏﺎﺯ ﺍﻷﺳـﺘﻴﻠﻴﻦ ﻭﺳـﺎﺋﻞ ﺍﻟﺒﻨﺰﻳﻦ ﻟﻬﻤﺎ ﻧﻔـﺲ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻭﺍﻟﺼﻴﻐﺔ‬ ‫ﺍﻷﻭﻟﻴﺔ )‪ ،(CH‬ﻭﻟﻜﻨﻬﻤﺎ ﻳﺨﺘﻠﻔﺎﻥ ﺗﻤﺎ ﹰﻣﺎ ﻓﻲ ﺍﻟﺨﻮﺍﺹ‪.‬‬ ‫ﻭﻟﺘﺤﺪﻳـﺪ ﺍﻟﺼﻴﻐـﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻤﺮﻛﺐ‪ ،‬ﻳﺠﺐ ﺗﺤﺪﻳﺪ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﻣﻦ ﺧﻼﻝ ﺍﻟﺘﺠـﺎﺭﺏ ﺍﻟﻌﻤﻠﻴﺔ‪ ،‬ﻭﻣﻘﺎﺭﻧﺘﻬﺎ ﺑﺎﻟﻜﺘﻠﺔ ﺍﻟﻤﻤﺜﻠﺔ‬ ‫ﺑﺎﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴﺔ‪ .‬ﻓﻤﺜ ﹰ‬ ‫ﻼ‪ ،‬ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﺳـﺘﻴﻠﻴﻦ ﻫﻲ ‪ ،26.04g/ mol‬ﻭﻛﺘﻠﺔ ﺻﻴﻐﺘـﻪ ﺍﻷﻭﻟﻴﺔ )‪ (CH‬ﻫﻲ ‪ .13.02g/ mol‬ﺇﻥ‬ ‫ﻗﺴﻤﺔ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺍﻟﻔﻌﻠﻴﺔ ﻋﻠﻰ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﺗﺒﻴﻦ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﺳﺘﻴﻠﻴﻦ ﺿﻌﻔﺎ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻼﺳﺘﻴﻠﻴﻦ‬ ‫ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ )‪(CH‬‬

‫=‬

‫‪26.04 g/mol‬‬ ‫‪13.02 g/mol‬‬

‫= ‪2.00‬‬

‫ﻭﺑﻤﺎ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﺳﺘﻴﻠﻴﻦ ﺿﻌﻔﺎ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪ ،‬ﻓﺈﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻪ ﻳﺠﺐ ﺃﻥ ﺗﺤﺘﻮﻱ ﺿﻌﻔﻲ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ‬ ‫ﻭﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﻭﻛﺬﻟـﻚ ﻋﻨـﺪ ﻣﻘﺎﺭﻧﺔ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺍﻟﻤﺤﺪﺩﺓ ﺗﺠﺮﻳﺒ ﹰﻴـﺎ ﻟﻠﺒﻨﺰﻳﻦ )‪ (78.12g/ mol‬ﺑﻜﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪ ،‬ﺳـﺘﺠﺪ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‬ ‫ﺗﺴﺎﻭﻱ ﺳﺘﺔ ﺃﺿﻌﺎﻑ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺒﻨﺰﻳﻦ‬ ‫ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ )‪(CH‬‬

‫=‬

‫‪78.12 g/mol‬‬ ‫‪13.02 g/mol‬‬

‫= ‪6.00‬‬

‫ﻭﻋﻠﻴﻪ‪ ،‬ﻓﺈﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻠﺒﻨﺰﻳﻦ ﻳﺠﺐ ﺃﻥ ﺗﻤﺜﻞ ﺳـﺘﺔ ﺃﺿﻌﺎﻑ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪ .‬ﻭﻳﻤﻜﻨﻚ‬ ‫ﺃﻥ ﺗﺴﺘﻨﺘﺞ ﺍﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻸﺳﺘﻠﻴﻦ ﻫﻲ ‪ CH × 2‬ﺃﻭ ‪.C2H2‬‬

‫‪68‬‬

‫ﻭﺃﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻠﺒﻨﺰﺑﻦ ﻫﻲ ‪ CH ×6‬ﺃﻭ ‪.C6H6‬‬ ‫ﻭﻳﻤﻜﻦ ﺗﻤﺜﻴﻞ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺑﻮﺻﻔﻬﺎ ﺃﻭﻟﻴﺔ ﻣﻀﺮﻭﺑﺔ ﻓﻲ ﻋﺪﺩ ﺻﺤﻴﺢ)ﻥ(‪.‬‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ = ﻥ )ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ(‬ ‫ﺣﻴـﺚ )ﻥ( ﺗﻤﺜـﻞ ﺍﻟﻌﺎﻣـﻞ )‪ 6‬ﻓﻲ ﻣﺜـﺎﻝ ﺍﻟﺒﻨﺰﻳﻦ( ﺍﻟﺬﻱ ﺗﻀـﺮﺏ ﺑﻪ ﺍﻷﺭﻗﺎﻡ ﻓـﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴـﺔ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬ ‫ﺍﻟﺸـﻜﻞ‪ 5-13‬ﻳﺒﻴﻦ ﺧﻄﻮﺍﺕ ﺗﺤﺪﻳﺪ ﺍﻟﺼﻴﻎ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺑﺪ ﹰﺀﺍ ﺑﺎﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﺃﻭ ﺑﻴﺎﻧﺎﺕ‬ ‫ﺍﻟﻜﺘﻠﺔ‪.‬‬ ‫ﺍﻟﺸﻜﻞ‪ 5-13‬ﺍﺳﺘﻌﻦ ﺑﻬﺬﺍ ﺍﻟﻤﺨﻄﻂ ﺍﻟﺬﻱ ﻳﺴﺎﻋﺪﻙ ﻋﻠﻰ ﺗﺤﺪﻳﺪ ﺍﻟﺼﻴﻎ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ‪.‬‬ ‫ﺻﻒ ﻛﻴﻒ ﻳﺮﺗﺒﻂ ﺍﻟﻌﺪﺩ ﺍﻟﺼﺤﻴﺢ )ﻥ( ﺑﺎﻟﺼﻴﻎ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬

‫ﻋﺒـﺮ ﻋـﻦ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳـﺔ‬ ‫ﹼ‬ ‫ﺑﺎﻟﻜﺘﻠﺔ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪.‬‬ ‫ﺟـﺪ ﻋـﺪﺩ ﺍﻟﻤـﻮﻻﺕ ﻟـﻜﻞ‬ ‫ﻋﻨﺼﺮ‪.‬‬

‫ﻛﺘﻠﺔ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻤﻜﻮﻧﺔ‬

‫ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ‬

‫ﻛﺘﻠﺔ ﻛﻞ ﻋﻨﺼﺮ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‬ ‫ﻧﺴﺒﺔ ﻣﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ‬

‫ﺍﻓﺤﺺ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬

‫ﺍﻛﺘﺐ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫ﺣــﺪﺩ ﺍﻟـﻌــﺪﺩ ﺍﻟﺼﺤﻴـﺢ‬ ‫ﺍﻟـﺬﻱ ﻳﺮﺑﻂ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫ﺑﺎﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬ ‫ﺍﺿﺮﺏ ﺍﻷﺭﻗﺎﻡ ﻓﻲ ﻗﻴﻤﺔ ﻥ‪.‬‬ ‫ﺍﻛﺘﺐ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬

‫ﺇﺫﺍ ﻟﻢ ﺗﻜﻦ ﺟﻤﻴﻊ ﺍﻷﻋﺪﺍﺩ‬ ‫ﺻﺤﻴﺤﺔ ﻓﺎﺿﺮﺏ ﻓﻲ ﺃﺻﻐﺮ ﻣﻌﺎﻣﻞ‬ ‫ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻋﺪﺍﺩ ﺻﺤﻴﺤﺔ‪.‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺟﻤﻴﻊ‬ ‫ﺍﻷﻋﺪﺍﺩ ﺻﺤﻴﺤﺔ‬

‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺍﻟﺘﺠﺮﻳﺒﻴﺔ‬ ‫ﻥ=‬ ‫ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫ﻥ) ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ(‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‬

‫‪69‬‬

‫‪5-12 ∫Éãe‬‬ ‫ﺗﺤﺪﻳـﺪ ﺍﻟﺼﻴﻐـﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻳﺸـﻴﺮ ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺤﻤﺾ ﺍﻟﺴﻜﺴـﻨﻴﻚ ﺇﻟﻰ ﺃﻧﻪ ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 40.68%‬ﻛﺮﺑـﻮﻥ‪ 5.08% ،‬ﻫﻴﺪﺭﻭﺟﻴﻦ‪،‬‬ ‫‪ 54.24%‬ﺃﻛﺴﺠﻴﻦ ﻭﻟﻪ ﻛﺘﻠﺔ ﻣﻮﻟﻴﺔ ‪ 118.1g/mol‬ﺣﺪﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻬﺬﺍ ﺍﻟﺤﻤﺾ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘﺪ ﺃﻋﻄﻴﺖ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﺤﻤﺾ ﺍﻟﺴﻜﺴـﻨﻴﻚ‪ ،‬ﺍﻓﺘﺮﺽ ﺃﻥ ﻛﻞ ﻧﺴﺒﺔ ﻣﺌﻮﻳﺔ ﻛﺘﻠﻴﺔ ﺗﻤﺜﻞ ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ ﺑـ ﹺ‪ 100g‬ﻣﻦ ﺍﻟﻌﻴﻨﺔ‪،‬‬ ‫ﻟﺬﺍ ﻳﻤﻜﻨﻚ ﻣﻘﺎﺭﻧﺔ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺍﻟﻤﻌﻄﺎﺓ )‪ (118.1g/mol‬ﺑﺎﻟﻜﺘﻠﺔ ﺍﻟﺘﻲ ﺗﻤﺜﻞ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻹﻳﺠﺎﺩ ﺍﻟﻌﺪﺩ ﺍﻟﺼﺤﻴﺢ ﻥ ‪.‬‬

‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬

‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪40.68% = C‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪5.08% = H‬‬ ‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﹺﻟـ ‪54.24% = O‬‬

‫ﺍﻟﻤﻄﻠﻮﺏ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ= ؟‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ= ؟‬

‫ﺍﻟﻜﺘﻠﻴﺔ ﺍﻟﻤﻮﻟﻴﺔ = ‪ 118.1g/mol‬ﺣﻤﺾ ﺍﻟﺴﻜﺴﻨﻴﻚ‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫‪2‬‬ ‫‪ 1mol‬ﻣﻦ ‪C‬‬ ‫= ‪3.3870mol‬ﻣﻦ ‪C‬‬ ‫ﻋـﻮﺽ ﻛﺘﻠﺔ ﺍﻟﻜﺮﺑﻮﻥ‪ ،‬ﻭﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺟﺪ ﻋﺪﺩ ‪ 40.68g‬ﻣﻦ ‪× C‬‬ ‫‪ 12.01g‬ﻣﻦ ‪C‬‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫‪ 1mol‬ﻣﻦ ‪H‬‬ ‫= ‪5.04mol‬ﻣﻦ ‪H‬‬ ‫ﻋـﻮﺽ ﻛﺘﻠـﺔ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪ ،‬ﻭﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺟﺪ ‪ 5.08g‬ﻣﻦ ‪× H‬‬ ‫‪ 1.008g‬ﻣﻦ ‪H‬‬ ‫ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‪.‬‬

‫‪ 1mol‬ﻣﻦ ‪O‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ﺍﻷﻛﺴﺠﻴﻦ‪ ،‬ﻭﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺟﺪ ﻋﺪﺩ ‪ 54.24g‬ﻣﻦ ‪× O‬‬ ‫‪ 16.00g‬ﻣﻦ ‪O‬‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬‬

‫= ‪3.39mol‬ﻣﻦ ‪O‬‬

‫ﻧﺴﺒﺔ ﺍﻟﻤﻮﻻﺕ ﻓﻲ ﺣﻤﺾ ﺍﻟﺴﻜﺴﻨﻴﻚ ﻫﻲ )‪ 3.387mol‬ﻣﻦ ‪ 5.04mol) (C‬ﻣﻦ ‪ 3.39mol) (H‬ﻣﻦ ‪ . (O‬ﻓﺎﺣﺴﺐ ﺃﺑﺴﻂ‬ ‫ﻧﺴﺒﺔ ﻟﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ ﺑﻘﺴﻤﻪ ﻣﻮﻻﺕ ﻛﻞ ﻋﻨﺼﺮ ﻋﻠﻰ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻓﻲ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﺍﻟﻤﺤﺴﻮﺑﺔ‪.‬‬ ‫‪ 3.387 mol‬ﻣﻦ ‪C‬‬ ‫= ‪ 1mol‬ﻣﻦ ‪C‬‬ ‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ C‬ﻋﻠﻰ ‪3.387‬‬ ‫‪3.387‬‬

‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ H‬ﻋﻠﻰ ‪3.387‬‬ ‫ﺍﻗﺴﻢ ﻣﻮﻻﺕ ‪ O‬ﻋﻠﻰ ‪3.387‬‬

‫‪ 5.04 mol‬ﻣﻦ‬

‫‪H‬‬

‫= ‪ 1.5mol‬ﻣﻦ‬

‫‪O‬‬

‫= ‪ 1mol‬ﻣﻦ‬

‫‪3.387‬‬

‫‪ 3.39 mol‬ﻣﻦ‬ ‫‪3.387‬‬

‫‪H‬‬

‫‪O‬‬

‫ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻣﻮﻟﻴﺔ ﻫﻲ ‪ 1 : 1.5 : 1‬ﻓﺎﺿﺮﺏ ﺟﻤﻴﻊ ﺍﻟﻘﻴﻢ ﺍﻟﻤﻮﻟﻴﺔ ﻓﻲ ‪ 2‬ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻋﺪﺍﺩ ﺻﺤﻴﺤﻴﺔ ‪.‬‬ ‫= ‪ 2mol‬ﻣﻦ ‪C‬‬ ‫‪ 1mol × 2‬ﻣﻦ ‪C‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ C‬ﻓﻲ ‪.2‬‬ ‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ H‬ﻓﻲ ‪.2‬‬ ‫‪ 1.5mol × 2‬ﻣﻦ ‪ 3mol = H‬ﻣﻦ ‪H‬‬

‫ﺍﺿﺮﺏ ﻣﻮﻻﺕ ‪ O‬ﻓﻲ ‪. 2‬‬

‫‪ 1mol × 2‬ﻣﻦ ‪ 2mol = O‬ﻣﻦ‬

‫ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻠﻤﻮﻻﺕ ﻫﻲ ‪ ،2 : 3 : 2‬ﺇﺫﻥ ﻓﺎﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻫﻲ ‪.C2H3O2‬‬ ‫ﺍﺣﺴﺐ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻜﻞ ﻋﻨﺼﺮ‪.‬‬

‫ﺍﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻜﺮﺑﻮﻥ ﻓﻲ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺗﻪ ‪.‬‬

‫‪70‬‬

‫‪ 12.01g‬ﻣﻦ ‪C‬‬ ‫‪ 2mol‬ﻣﻦ ‪× C‬‬ ‫‪ 1mol‬ﻣﻦ ‪C‬‬

‫‪O‬‬

‫= ‪24.02g‬ﻣﻦ ‪C‬‬

‫‪ 1.008g‬ﻣﻦ ‪H‬‬ ‫ﺍﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺗﻪ‪ 3mol .‬ﻣﻦ ‪× H‬‬ ‫‪ 1mol‬ﻣﻦ ‪H‬‬

‫‪ 16.00g‬ﻣﻦ ‪O‬‬ ‫ﺍﺿﺮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻓﻲ ﻋﺪﺩ ﻣﻮﻻﺕ ﺫﺭﺍﺗﻪ‪ 2mol .‬ﻣﻦ ‪× O‬‬ ‫‪ 1mol‬ﻣﻦ ‪O‬‬

‫= ‪3.024g‬ﻣﻦ ‪H‬‬ ‫= ‪32.00g‬ﻣﻦ ‪O‬‬

‫ﺍﺟﻤﻊ ﻛﺘﻞ ﺍﻟﻌﻨﺎﺻﺮ‪ .‬ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﹺ‬ ‫ﻟـ ‪59.04g/mol = 32.0g+ 3.024 g + 24.02g = C2H3O2‬‬ ‫ﻟﺘﺤﺪﻳﺪ ﻗﻴﻤﺔ ﻥ ﺍﻗﺴﻢ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺤﻤﺾ ﺍﻟﺴﻜﺴﻴﻨﻴﻚ ﻋﻠﻰ ﻛﺘﻠﺔ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤـﻮﻟﻴﺔ ﻟﺤﻤـﺾ ﺍﻟﺴﻜﺴﻴﻨﻴـﻚ‬ ‫ﻥ=‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ‪C2H3O2‬‬

‫=‬

‫‪118.1 g/mol‬‬ ‫‪59.04 g/mol‬‬

‫= ‪2.000‬‬

‫ﺍﺿﺮﺏ ﺍﻷﺭﻗﺎﻡ ﻓﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻓﻲ ‪ 2‬ﻟﺘﺤﺼﻞ ﻋﻠﻰ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ = ‪C4H6O4 = (C2H3O2) × 2‬‬ ‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺍﻟﺘﻲ ﺗﻢ ﺍﻟﺘﻮﺻﻞ ﺇﻟﻴﻬﺎ ﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻧﻔﺴﻬﺎ ﺍﻟﻤﺤﺪﺩﺓ ﺗﺠﺮﻳﺒ ﹰﻴﺎ ﻟﻠﻤﺮﻛﺐ‪.‬‬

‫‪5-13 ∫Éãe‬‬

‫ﺣﺴـﺎﺏ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻣﻦ ﺧﻼﻝ ﺍﻟﻜﺘﻠﺔ ﻳﺴـﺘﻌﻤﻞ ﻣﻌﺪﻥ ﺍﻹﻟﻤﻨﻴﺖ ﻻﺳـﺘﺨﺮﺍﺝ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ‪ .‬ﻭﻋﻨﺪ ﺗﺤﻠﻴﻞ ﻋﻴﻨﺔ ﻣﻨﻪ ﻭﺟﺪ ﺃﻧﻬﺎ ﺗﺤﻮﻱ‬ ‫‪ 5.41g‬ﻣﻦ ﺍﻟﺤﺪﻳﺪ‪ 4.64g ،‬ﻣﻦ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ‪ 4.65g ،‬ﻣﻦ ﺍﻷﻛﺴﺠﻴﻦ‪ ،‬ﺣﺪﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻬﺬﺍ ﺍﻟﻤﻌﺪﻥ‪.‬‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﺪﻳﻚ ﻛﺘﻞ ﺍﻟﻌﻨﺎﺻﺮﺍﻟﺘﺎﻟﻴﺔ ﻓﻲ ﻛﺘﻠﺔ ﻣﻌﻴﻨﺔ ﻣﻦ ﺍﻟﻤﻌﺪﻥ‪ ،‬ﻭﺍﻟﻤﻄﻠﻮﺏ ﺣﺴـﺎﺏ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻪ‪.‬ﻟﺬﺍ ﹼﺣﻮﻝ ﺍﻟﻌﻨﺎﺻﺮ ﻛﻠﻬﺎ ﺇﻟﻰ ﻣﻮﻻﺕ‪،‬‬ ‫ﺛﻢ ﺟﺪ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﺻﺤﻴﺤﺔ ﻟﻤﻮﻻﺕ ﻫﺬﻩ ﺍﻟﻌﻨﺎﺻﺮ‪.‬‬

‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬

‫ﻛﺘﻠﺔ ﺍﻟﺤﺪﻳﺪ = ‪ 5.41g‬ﻣﻦ ‪Fe‬‬ ‫ﻛﺘﻠﺔ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ= ‪ 4.64g‬ﻣﻦ ‪Ti‬‬ ‫ﻛﺘﻠﺔ ﺍﻷﻛﺴﺠﻴﻦ = ‪ 4.65g‬ﻣﻦ ‪O‬‬ ‫‪2‬‬

‫ﺍﻟﻤﻄﻠﻮﺏ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ= ؟؟‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺣﻮﻝ ﺍﻟﻜﺘﻞ ﺍﻟﻤﻌﺮﻭﻓﺔ ﺇﻟﻰ ﻣﻮﻻﺕ ﺑﺎﻟﻀﺮﺏ ﻓﻲ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﻟﻤﻮﻻﺕ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ‪ -‬ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬

‫‪ 1mol‬ﻣﻦ ‪Fe‬‬ ‫ﻋـﻮﺽ ﻛﺘﻠﺔ ﺍﻟﺤﺪﻳـﺪ‪ ،‬ﻭﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴـﺔ‪ ،‬ﻭﺟﺪ ﻋﺪﺩ ‪ 5.41g‬ﻣﻦ ‪× Fe‬‬ ‫‪ 55.85g‬ﻣﻦ ‪Fe‬‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫‪ 1mol‬ﻣﻦ ‪Ti‬‬ ‫= ‪0.0969mol‬ﻣﻦ ‪TI‬‬ ‫ﻋـﻮﺽ ﻛﺘﻠﺔ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ‪ ،‬ﻭﻣﻘﻠـﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺟﺪ ﻋﺪﺩ ‪ 4.64g‬ﻣﻦ ‪× Ti‬‬ ‫‪ 47.88g‬ﻣﻦ ‪Ti‬‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫‪ 1mol‬ﻣﻦ ‪O‬‬ ‫= ‪0.291mol‬ﻣﻦ ‪O‬‬ ‫ﻋﻮﺽ ﻛﺘﻠﺔ ﺍﻷﻛﺴﺠﻴﻦ‪ ،‬ﻭﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺟﺪ ﻋﺪﺩ ‪ 4.65g‬ﻣﻦ ‪× O‬‬ ‫‪ 16.00g‬ﻣﻦ ‪O‬‬ ‫ﺍﻟﻤﻮﻻﺕ‪.‬‬ ‫= ‪0.0969mol‬ﻣﻦ ‪Fe‬‬

‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻤﻌﺪﻥ ﺍﻹﻟﻤﻨﻴﺖ ﻫﻲ‪ 0.0969mol) :‬ﻣﻦ ‪ 0.0969mol) :(Fe‬ﻣﻦ ‪ 0.291mol) : (Ti‬ﻣﻦ ‪(O‬‬

‫ﻓﺎﻗﺴﻢ ﻛﻞ ﻗﻴﻤﺔ ﻣﻮﻟﻴﺔ ﻋﻠﻰ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻓﻲ ﺍﻟﻨﺴﺒﺔ )‪ (0.0969‬ﻟﺘﺤﺼﻞ ﻋﻠﻰ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻣﻮﻟﻴﺔ‪.‬‬

‫‪71‬‬

‫ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻣﻮﻟﻴﺔ ﻫﻲ )‪ 1mol‬ﻣﻦ ‪ 1mol) :(Fe‬ﻣﻦ ‪ 3mol): (Ti‬ﻣﻦ ‪ (O‬ﻭﺑﻤﺎ ﺃﻥ ﺟﻤﻴﻊ ﺍﻟﻘﻴﻢ ﺍﻟﻤﻮﻟﻴﺔ ﺃﻋﺪﺍﺩ ﺻﺤﻴﺤﺔ‪ ،‬ﺇﺫﻥ‬ ‫ﻓﺎﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻺﻟﻤﻨﻴﺖ ﻫﻲ ‪.FeTiO3‬‬ ‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﹰ‬ ‫ﹰ‬ ‫ﻗﻠﻴـﻼ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﻠﺘﻴﺘﺎﻧﻴﻮﻡ ﹰ‬ ‫ﺃﻳﻀـﺎ ‪ .‬ﻭﻟﻬﺬﺍ ﻓﻤﻦ‬ ‫ﻗﻠﻴـﻼ ﻣﻦ ﻛﺘﻠﺔ ﺍﻟﺘﻴﺘﺎﻧﻴـﻮﻡ‪ ،‬ﻭﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﻠﺤﺪﻳﺪ ﺃﻛﺒﺮ‬ ‫ﻛﺘﻠـﺔ ﺍﻟﺤﺪﻳـﺪ ﺃﻛﺒﺮ‬ ‫ﺍﻟﻤﻨﻄﻘـﻲ ﺃﻥ ﻳﻜـﻮﻥ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﺤﺪﻳﺪ ﻣﺴـﺎﻭ ﹰﻳﺎ ﻟﻌﺪﺩ ﻣـﻮﻻﺕ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ‪ .‬ﻛﻤﺎ ﺃﻥ ﻛﺘﻠﺔ ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ ﻣﺴـﺎﻭﻳﺔ ﺗﻘﺮﻳ ﹰﺒﺎ ﻟﻜﺘﻠﺔ ﺍﻷﻛﺴـﺠﻴﻦ‪،‬‬ ‫ﻭﻟﻜﻦ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻫﻲ ﻧﺤﻮ ﺛﻠﺚ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺘﻴﺘﺎﻧﻴﻮﻡ‪ .‬ﻭﻋﻠﻴﻪ‪ ،‬ﻓﺈﻥ ﺍﻟﻨﺴﺒﺔ ‪ 1 :3‬ﺃﻛﺴﺠﻴﻦ ﺇﻟﻰ ﺗﻴﺘﺎﻧﻴﻮﻡ ﻣﻌﻘﻮﻟﺔ‪.‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬ ‫‪ .62‬ﻭﺟﺪﺃﻥﻣﺮﻛﺒ ﹰﺎ ﻳﺤﺘﻮﻱﻋﻠﻰ‪ 49.98g‬ﻣﻦﺍﻟﻜﺮﺑﻮﻥﻭ‪ 10.47g‬ﻣﻦﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪.‬ﻓﺈﺫﺍﻛﺎﻧﺖﺍﻟﻜﺘﻠﺔﺍﻟﻤﻮﻟﻴﺔﻟﻠﻤﺮﻛﺐ‪،58.12g/mol‬‬ ‫ﻓﻤﺎ ﺻﻴﻐﺘﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ‪.‬‬ ‫‪ .63‬ﺳﺎﺋﻞ ﻋﺪﻳﻢ ﺍﻟﻠﻮﻥ ﻳﺘﻜﻮﻥ ﻣﻦ ‪ 46.68%‬ﻧﻴﺘﺮﻭﺟﻴﻦ ﻭ‪ 53.32%‬ﺃﻛﺴﺠﻴﻦ‪ ،‬ﻭﻛﺘﻠﺘﻪ ﺍﻟﻤﻮﻟﻴﺔ ‪ .60.01g/ mol‬ﻓﻤﺎ ﺻﻴﻐﺘﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ؟‬ ‫‪ .64‬ﻋﻨﺪ ﺗﺤﻠﻴﻞ ﺃﻛﺴﻴﺪ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ‪ ،‬ﻧﺘﺞ ‪ 19.55g‬ﻣﻦ ‪ ،K‬ﻭ‪ 4.00 g‬ﻣﻦ ‪ .O‬ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻸﻛﺴﻴﺪ؟‬ ‫‪H‬‬ ‫‪ .65‬ﺗﺤﺪﱢ ﻋﻨﺪ ﺗﺤﻠﻴﻞ ﻣﺎﺩﺓ ﻛﻴﻤﻴﺎﺋﻴﺔ ﺗﺴﺘﻌﻤﻞ ﻓﻲ ﺳﺎﺋﻞ ﺗﻈﻬﻴﺮ ﺍﻷﻓﻼﻡ ﺍﻟﻔﻮﺗﻮﻏﺮﺍﻓﻴﺔ ﺗﻢ ﺍﻟﺘﻮﺻﻞ ﺇﻟﻰ ﺑﻴﺎﻧﺎﺕ ﺍﻟﺘﺮﻛﻴﺐ‬ ‫‪5.45%‬‬ ‫ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﺍﻟﻤﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸـﻜﻞ ﺍﻟﻤﺠﺎﻭﺭ‪ .‬ﻓﺈﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪،110.0g/ mol‬‬ ‫‪C‬‬ ‫‪65.45%‬‬ ‫ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻪ؟‬ ‫‪O‬‬ ‫‪29.09%‬‬ ‫‪ .66‬ﻋﻨـﺪ ﺗﺤﻠﻴﻞ ﻣﺴـﻜﻦ ﺍﻵﻻﻡ ﺍﻟﻤﻌـﺮﻭﻑ ﺍﻟﻤﻮﺭﻓﻴﻦ ﺗﻢ ﺍﻟﺘﻮﺻﻞ ﺇﻟﻰ ﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﻟﻤﺒﻴﻨﺔ ﻓـﻲ ﺍﻟﺠﺪﻭﻝ ﺃﺩﻧﺎﻩ ‪ .‬ﻓﻤﺎ‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻠﻤﻮﺭﻓﻴﻦ؟‬ ‫ﻧﻴﺘﺮﻭﺟﻴﻦ‬ ‫ﺃﻛﺴﺠﻴﻦ‬ ‫ﻫﻴﺪﺭﻭﺟﻴﻦ‬ ‫ﻛﺮﺑﻮﻥ‬ ‫ﺍﻟﻌﻨﺼﺮ‬ ‫ﺍﻟﻜﺘﻠﺔ)‪(g‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪5 - 4‬‬

‫‪17.900‬‬

‫‪1.680‬‬

‫‪4.225‬‬

‫‪1.228‬‬

‫ﻗﻮﻡ‪ :‬ﺇﺫﺍ ﺃﺧﺒﺮﻙ ﺃﺣﺪ ﺯﻣﻼﺋﻚ ﺑﺄﻥ ﺍﻟﻨﺘﺎﺋﺞ‬ ‫‪ .67‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﹼ‬ ‫اﻟﺨﻼﺻﺔ‬ ‫ﺍﻟﺘﺠﺮﻳﺒﻴـﺔ ﺗﺒﻴـﻦ ﺃﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻤﺮﻛﺐ ﺗﺴـﺎﻭﻱ ﺿﻌﻒ‬ ‫ﺗﻤﺜﻞ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ ﻟﻌﻨﺼﺮ ﻓﻲ ﻣﺮﻛﺐ ﻣﺎ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ‬ ‫ﺻﻴﻐﺘﻪ ﺍﻷﻭﻟﻴﺔ ﹺﺑـ ‪ 2.5‬ﻣﺮﺓ ﻓﻬﻞ ﺇﺟﺎﺑﺘﻪ ﺻﺤﻴﺤﺔ؟ ﻓﺴـﺮﺫﻟﻚ‪.‬‬ ‫ﻣﻦ ﻛﺘﻠﺔ ﺍﻟﻤﺮﻛﺐ ﺍﻟﻜﻠﻴﺔ ﻣﻦ ﺍﻟﻌﻨﺼﺮ‪.‬‬ ‫‪ .68‬ﺍﺣﺴـﺐ‪ :‬ﻧﺘـﺞ ﻋـﻦ ﺗﺤﻠﻴـﻞ ﻣﺮﻛـﺐ ﻳﺘﻜـﻮﻥ ﻣـﻦ ﺍﻟﺤﺪﻳـﺪ‬ ‫ﺗﻤﺜـﻞ ﺍﻷﺭﻗﺎﻡ ﻓﻲ ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴﺔ ﺃﺻﻐﺮﻧﺴـﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ‬ ‫ﻭﺍﻷﻛﺴـﺠﻴﻦ‪ 174.86g ،‬ﻣـﻦ ﺍﻟﺤﺪﻳـﺪ ‪ ،Fe‬ﻭ‪ 75.14g‬ﻣﻦ‬ ‫ﻟﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫ﺍﻷﻛﺴﺠﻴﻦ ‪ .O‬ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻬﺬﺍ ﺍﻟﻤﺮﻛﺐ؟‬ ‫ﺗﻤﺜﻞ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺍﻟﻌﺪﺩ ﺍﻟﻔﻌﻠﻲ ﻟﻠﺬﺭﺍﺕ ﻣﻦ ﻛﻞ ﻋﻨﺼﺮ ﻓﻲ‬ ‫‪ .69‬ﺍﺣﺴـﺐ‪ :‬ﻳﺤﺘﻮﻱ ﺃﻛﺴـﻴﺪ ﺍﻷﻟﻤﻨﻴﻮﻡ ﻋﻠﻰ ‪ 0.545g‬ﻣﻦ ‪،Al‬‬ ‫ﺟﺰﻱﺀ ﻣﻦ ﺍﻟﻤﺎﺩﺓ‪.‬‬ ‫ﻭ‪ 0.485g‬ﻣﻦ ‪ .O‬ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻸﻛﺴﻴﺪ؟‬ ‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻫﻲ ﻣﻀﺎﻋﻒ ﺻﺤﻴﺢ ﻟﻠﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫‪ .70‬ﹼﻭﺿﺢ ﻛﻴﻒ ﺗﺮﺗﺒﻂ ﺑﻴﺎﻧﺎﺕ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻤﺮﻛﺐ‬ ‫ﺑﻜﺘﻞ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺫﻟﻚ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫‪ .71‬ﹼ‬ ‫ﻭﺿﺢ ﻛﻴﻒ ﺗﺠﺪ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻓﻲ ﻣﺮﻛﺐ ﻛﻴﻤﻴﺎﺋﻲ‪.‬‬ ‫ﻃﺒـﻖ‪ :‬ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﻤﺮﻛﺐ ﻫﻲ ﺿﻌﻒ ﺻﻴﻐﺘـﻪ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫‪ .72‬ﹼ‬ ‫ﻓﻜﻴﻒ ﺗﺮﺗﺒﻂ ﺻﻴﻐﺘﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺑﺼﻴﻐﺘﻪ ﺍﻷﻭﻟﻴﺔ؟‬ ‫‪ .73‬ﹼﺣﻠﻞ‪ :‬ﺍﻟﻬﻴﻤﺎﺗﻴﺖ )‪ (Fe2O3‬ﻭﺍﻟﻤﺎﺟﻨﺘﻴﺖ )‪ (Fe3O4‬ﺧﺎﻣﺎﻥ‬ ‫ﻳﺴـﺘﻌﻤﻼﻥ ﻣﺼﺪﺭﻳـﻦ ﻟﻠﺤﺪﻳﺪ‪ .‬ﻓﺄﻳﻬﻤﺎ ﻳﻌﻄﻲ ﻧﺴـﺒﺔ ﺃﻋﻠﻰ ﻣﻦ‬ ‫ﺍﻟﺤﺪﻳﺪ ﻟﻜﻞ ﻛﻴﻠﻮﺟﺮﺍﻡ؟‬

‫‪72‬‬

‫‪5-5‬‬ ‫ا ﻫﺪاف‬

‫ﺗﻮﺿﺢ ﺍﳌﻘﺼﻮﺩ ﺑﺎﳌﻠﺢ ﺍﳌﺎﺋﻲ‬ ‫ﻭﺗﺮﺑﻂ ﺍﺳﻤﻪ ﺑﱰﻛﻴﺒﻪ‪.‬‬ ‫ﲢﺪﺩ ﺻﻴﻐﺔ ﻣﻠﺢ ﻣﺎﺋﻲ ﻣﻦ‬ ‫ﺍﻟﺒﻴﺎﻧﺎﺕ ﺍﳌﺨﱪﻳﺔ‪.‬‬ ‫ﻣﺮاﺟﻌﺔ اﻟﻤﻔﺮدات‬

‫ﺍﻟﺸﺒﻜﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ‪ :‬ﺍﻟﱰﺗﻴﺐ‬ ‫ﺍﳍﻨﺪﳼ ﺍﻟﺜﻼﺛﻲ ﺍﻷﺑﻌﺎﺩ‬ ‫ﻟﻠﺠﺴﻴﲈﺕ‪.‬‬ ‫اﻟﻤﻔﺮدات اﻟﺠﺪﻳﺪة‬

‫ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ‬

‫ﺍﻟﺸﻜﻞ ‪ 5-14‬ﺇﻥ ﻭﺟﻮﺩ ﺍﳌﺎﺀ ﻭﺷﻮﺍﺋﺐ‬ ‫ﺍﳌﻌـﺎﺩﻥ ﺍﳌﺨﺘﻠﻔﺔ ﻳﻔﴪﺍﻥ ﺍﻟﺘﻨﻮﻉ ﺍﻟﻜﺒﲑ‬ ‫ﻷﺣﺠـﺎﺭ ﺍﻷﻭﺑـﺎﻝ ﺍﻟﻜﺮﻳﻤـﺔ‪ .‬ﻭﲢﺪﺙ‬ ‫ﺗﻐﲑﺍﺕ ﺃﺧﺮ￯ ﰲ ﺍﻟﻠﻮﻥ ﻋﻨﺪﻣﺎ ﳚﻒ‪.‬‬

‫ﺻﻴﻎ ا ﻣﻼح اﻟﻤﺎﺋﻴﺔ‬ ‫‪Formulas of Hydrates‬‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻷﻣﻼﺡ ﺍﳌﺎﺋﻴﺔ ﻫﻲ ﻣﺮﻛﺒﺎﺕ ﺃﻳﻮﻧﻴﺔ ﺻﻠﺒﺔ ﻓﻴﻬﺎ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ﳏﺘﺠﺰﺓ ‪.‬‬ ‫الرﺑﻂ ﺑﻮاﻗﻊ الحﻴاﺓ ﹸﺗﻌﺒﺄ ﺑﻌﺾ ﺍﳌﻨﺘﺠﺎﺕ‪ ،‬ﻛﺎﳌﻌﺪﺍﺕ ﺍﻹﻟﻜﱰﻭﻧﻴﺔ‪ ،‬ﰲ ﺻﻨﺎﺩﻳﻖ‬ ‫ﻣـﻊ ﺃﻛﻴـﺎﺱ ﺻﻐﲑﺓ ﻣﻜﺘـﻮﺏ ﻋﻠﻴﻬـﺎ "ﳎﻔـﻒ"‪ .‬ﻭﺗﻀﺒﻂ ﻫـﺬﻩ ﺍﻷﻛﻴـﺎﺱ ﺍﻟﺮﻃﻮﺑﺔ‬ ‫ﺑﺎﻣﺘﺼﺎﺹ ﺍﳌﺎﺀ‪ .‬ﻭﳛﺘﻮﻱ ﺑﻌﻀﻬﺎ ﻣﺮﻛﺒﺎﺕ ﺃﻳﻮﻧﻴﺔ ﺗﺴﻤﻰ ﺍﻷﻣﻼﺡ ﺍﳌﺎﺋﻴﺔ‪.‬‬

‫ﺗسمﻴﺔ اﻷمﻼﺡ الماﺋﻴﺔ ‪Naming Hydrates‬‬

‫ﻫﻞ ﺳـﺒﻖ ﺃﻥ ﺭﺍﻗﺒﺖ ﺑﻠﻮﺭﺍﺕ ﺗﺘﻜﻮﻥ ﺑﺒﻂﺀ ﻣﻦ ﳏﻠﻮﻝ ﻣﺎﺋﻲ؟ ﺗﻠﺘﺼﻖ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ ﺃﺣﻴﺎ ﹰﻧﺎ‬ ‫ﺑﺎﻷﻳﻮﻧـﺎﺕ ﺧـﻼﻝ ﺗﻜـﻮﻥ ﺍﳌـﺎﺩﺓ ﺍﻟﺼﻠﺒﺔ‪ .‬ﻭﺗﺴـﻤﻰ ﺟﺰﻳﺌﺎﺕ ﺍﳌـﺎﺀ ﺍﻟﺘﻲ ﺗﺼﺒﺢ ﺟـﺰ ﹰﺀﺍ ﻣﻦ‬ ‫ﺃﻣﻼﺣﺎ‬ ‫ﺍﻟﺒﻠﻮﺭﺓ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‪ .‬ﻭ ﹸﺗﺴـﻤﻰ ﺍﳌﻮﺍﺩ ﺍﻷﻳﻮﻧﻴﺔ ﺍﻟﺼﻠﺒﺔ ﺍﻟﺘﻲ ﲢﺘﺠﺰ ﻓﻴﻬﺎ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ‬ ‫ﹰ‬ ‫ﻣﺎﺋﻴﺔ‪ .‬ﻓﺎﳌﻠﺢ ﺍﳌﺎﺋﻲ ﻣﺮﻛﺐ ﳛﺘﻮﻱ ﻋﲆ ﻋﺪﺩ ﻣﻌﲔ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ ﺍﳌﺮﺗﺒﻄﺔ ﺑﺬﺭﺍﺗﻪ‪ .‬ﻭﻳﺒﲔ‬ ‫ﺍﻟﺸـﻜﻞ ‪ 5-14‬ﺍﳊﺠـﺮ ﺍﻟﻜﺮﻳﻢ ﺍﳉﻤﻴﻞ ﺍﳌﻌﺮﻭﻑ ﺑﺎﻷﻭﺑﺎﻝ‪ ،‬ﻭﻫﻮ ﺛﺎﲏ ﺃﻛﺴـﻴﺪ ﺍﻟﺴـﻴﻠﻴﻜﻮﻥ‬ ‫ﺍﳌﺎﺋﻲ )‪ (SiO2‬ﺍﻟﺬﻱ ﳛﺘﻮﻱ ﻋﲆ ﻣﺎﺀ‪ .‬ﻭﺍﻷﻟﻮﺍﻥ ﺍﻟﻔﺮﻳﺪﺓ ﻫﻲ ﻧﺘﻴﺠﺔ ﻭﺟﻮﺩ ﺍﳌﺎﺀ ﰲ ﺍﳌﻌﺪﻥ‪.‬‬ ‫ﻳﻜﺘـﺐ ﰲ ﺻﻴﻐـﺔ ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ‪ ،‬ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ ﺍﳌﺮﺗﺒﻄـﺔ ﺑﻮﺣﺪﺓ ﺍﻟﺼﻴﻐﺔ ﻟﻠﻤﺮﻛﺐ ﺗﺎﻟﻴ ﹰﺎ‬ ‫ﻟﻨﻘﻄـﺔ‪ ،‬ﻣﺜـﻞ ‪ .COCl2 . 6H2O‬ﻭ ﹸﻳﺴـﻤﻰ ﻫـﺬﺍ ﺍﳌﺮﻛـﺐ ﻛﻠﻮﺭﻳـﺪ ﺍﻟﻜﻮﺑﻠـﺖ )‪(II‬‬ ‫)ﺃﻱ ﳛﺘﻮﻱ ﻋﲆ ‪ 6‬ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ(‪ .‬ﻭﺗﺪﺧﻞ ﻛﺘﻠﺔ ﺍﳌﺎﺀ ﺍﳌﺮﺗﺒﻄﺔ ﺑﻮﺣﺪﺓ ﺍﻟﺼﻴﻐﺔ‬ ‫ﺳﺪﺍﳼ ﺍﳌﺎﺀ ﹾ‬ ‫ﰲ ﺣﺴﺎﺏ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪ .‬ﻭﳜﺘﻠﻒ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﰲ ﺍﻷﻣﻼﺡ ﻋﲆ ﻧﺤﻮ ﻭﺍﺳﻊ‪،‬‬ ‫ﻭﻳﺒﲔ ﺍﳉﺪﻭﻝ ‪ 5-1‬ﺑﻌﺾ ﺍﻷﻣﻼﺡ ﺍﳌﺎﺋﻴﺔ ﺍﻟﺸﺎﺋﻌﺔ‪.‬‬

‫‪á«FÉŸG ìÓeC’G ≠«°U 5-1 ∫hó÷G‬‬ ‫ﺍﳌﻘﻄﻊ‬

‫ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ‬ ‫ﺍﳌﺎﺀ‬

‫ﺛﻨﺎﺋﻲ‬

‫‪٢‬‬

‫ﺃﺣﺎﺩﻱ‬ ‫ﺛﻼﺛﻲ‬

‫‪١‬‬

‫‪٣‬‬

‫ﺍﻟﺼﻴﻐﺔ‬

‫‪(NH4)2C2O4 . H2O‬‬

‫ﺃﻛﺴﻼﺕ ﺍﻷﻣﻮﻧﻴﻮﻡ ﺃﺣﺎﺩﻳﺔ ﺍﳌﺎﺀ‪.‬‬

‫‪NaC2H3O2 . 3H2O‬‬

‫ﺃﺳﻴﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺛﻼﺛﻴﺔ ﺍﳌﺎﺀ‬

‫‪CuSO4. 5H2O‬‬

‫ﻛﱪﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ )‪ (II‬ﲬﺎﺳﻴﺔ ﺍﳌﺎﺀ‬

‫‪MgSO4 . 7H2O‬‬

‫ﻛﱪﻳﺘﺎﺕ ﺍﳌﺎﻏﻨﻴﺴﻴﻮﻡ ﺳﺒﺎﻋﻴﺔ ﺍﳌﺎﺀ‪.‬‬

‫‪CaCl2 . 2H2O‬‬

‫‪FePO4. 4H2O‬‬

‫ﺭﺑﺎﻋﻲ‬

‫‪٤‬‬

‫ﺳﺪﺍﳼ‬

‫‪٦‬‬

‫‪CoCl2.6H2O‬‬

‫ﺛﲈﲏ‬

‫‪٨‬‬

‫‪Ba (OH)2. 8H2O‬‬

‫ﲬﺎﳼ‬

‫ﺳﺒﺎﻋﻲ‬ ‫ﻋﴩﻱ‬

‫‪٥‬‬

‫‪٧‬‬ ‫‪١٠‬‬

‫ﺍﻻﺳﻢ‬

‫‪Na2CO2.10H2O‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﺛﻨﺎﺋﻲ ﺍﳌﺎﺀ‪.‬‬

‫ﻓﺴﻔﺎﺕ ﺍﳊﺪﻳﺪ )‪ (III‬ﺭﺑﺎﻋﻴﺔ ﺍﳌﺎﺀ‪.‬‬ ‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ ﺳﺪﺍﳼ ﺍﳌﺎﺀ‬

‫ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺒﺎﺭﻳﻮﻡ ﺛﲈﲏ ﺍﳌﺎﺀ‪.‬‬

‫ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﻋﴩﻳﺔ ﺍﳌﺎﺀ‬

‫‪73‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ )‪ (II‬ﺍﻟﻤﺎﺋﻲ ﺯﻫﺮﻱ‪.‬‬

‫ﻳﻤﻜﻦ ﺗﺴﺨﻴﻦ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻟﻄﺮﺩ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‪.‬‬

‫ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ )‪ (II‬ﺍﻟﻼﻣﺎﺋﻲ ﺃﺯﺭﻕ‪.‬‬

‫ﺍﻟﺸﻜﻞ ‪ 5-15‬ﻳﻤﻜﻦ ﺇﺯﺍﻟﺔ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﺑﺘﺴﺨﻴﻦ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪ ،‬ﻟﺘﻜﻮﻳﻦ ﻣﻠﺢ ﻻ ﻣﺎﺋﻲ ﻭﺍﻟﺬﻱ ﻗﺪ ﻳﺒﺪﻭ ﻣﺨﺘﻠ ﹰﻔﺎ ﺟﺪﹰ ﺍ ﻋﻦ ﺍﻟﻤﻠﺢ‬ ‫ﺍﻟﻤﺎﺋﻲ‪.‬‬

‫ﺗحلﻴﻞ اﻷمﻼﺡ الماﺋﻴﺔ ‪Analyzing a Hydrates‬‬

‫ﻋﻨـﺪ ﺗﺴـﺨﻴﻦ ﻣﻠـﺢ ﻣﺎﺋﻲ‪ ،‬ﹸﺗﻄـﺮﺩ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤـﺎﺀ ﺗﺎﺭﻛﺔ ﻭﺭﺍﺀﻫﺎ ﺍﻟﻤﻠـﺢ ﺍﻟﻼﻣﺎﺋﻲ‪ .‬ﺍﻧﻈﺮ ﺇﻟﻰ ﺍﻟﺸـﻜﻞ ‪ ،5-15‬ﺣﻴﺚ ﺗﻮﺿﺢ ﺳﻠﺴـﻠﺔ‬ ‫ﺍﻟﺼﻮﺭﺃﻧﻪ ﻋﻨﺪ ﺗﺴﺨﻴﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ )‪ (II‬ﺳﺪﺍﺳﻲ ﺍﻟﻤﺎﺀ ﺍﻟﺰﻫﺮﻱ‪ ،‬ﻳﻨﺘﺞ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ )‪ (II‬ﺍﻟﻼﻣﺎﺋﻲ ﺍﻷﺯﺭﻕ‪.‬‬

‫ﻓﻜﻴﻒ ﻳﻤﻜﻨﻚ ﺗﺤﺪﻳﺪ ﺻﻴﻐﺔ ﻣﻠﺢ ﻣﺎﺋﻲ؟ ﻳﺠﺐ ﺃﻥ ﺗﺤﺴﺐ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺮﺗﺒﻄﺔ ﺑـ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪ .‬ﺍﻓﺘﺮﺽ‬ ‫ﺃﻥ ﻟﺪﻳـﻚ ﻋﻴﻨـﺔ ﻣﻜﻮﻧﺔ ﻣـﻦ ‪ 5.00g‬ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﺒﺎﺭﻳﻮﻡ ﺍﻟﻤﺎﺋـﻲ‪ .‬ﻭﺑﻤﺎ ﺃﻧﻚ ﺗﻌﺮﻑ ﺃﻥ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﻫـﻲ ‪ ،BaCl2. xH2O‬ﻟﺬﺍ‬ ‫ﻳﺠﺐ ﺃﻥ ﺗﺤﺪﺩ ﻗﻴﻤﺔ ‪ ،x‬ﻭﻫﻲ ﻣﻌﺎﻣﻞ ‪ H2O‬ﻓﻲ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺸﻴﺮ ﺇﻟﻰ ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺮﺗﺒﻄﺔ ﺑﻤﻮﻝ ﻭﺍﺣﺪ‬ ‫ﻣﻦ ‪ .BaCl2‬ﻭﺣﺘﻰ ﺗﺠﺪ ﻗﻴﻤﺔ ‪ ،x‬ﻳﺠﺐ ﺃﻥ ﺗﺴـﺨﻦ ﺍﻟﻌﻴﻨﺔ ﻟﻠﺘﺨﻠﺺ ﻣﻦ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‪ .‬ﻭﺍﻓﺮﺽ ﺃﻧﻚ ﺑﻌﺪ ﺗﺴـﺨﻴﻨﻬﺎ ﻭﺟﺪﺕ ﺃﻥ ﻛﺘﻠﺔ‬ ‫ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ ﹺﻟـ ‪BaCl2‬ﻫﻲ )‪.(4.26g‬‬ ‫ﺇﺫﻥ‪ ،‬ﻓﻜﺘﻠﺔ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﺗﺴﺎﻭﻱ ﺍﻟﻔﺮﻕ ﺑﻴﻦ ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ )‪ (5.00g‬ﻭﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ )‪.(4.26g‬‬ ‫‪ 0.74g = 4.26g - 5.00g‬ﻣﺎﺀ‪.‬‬

‫ﻭﺑﻌـﺪ ﺃﻥ ﻋﺮﻓـﺖ ﻛﺘﻠﺔ ﻛﻞ ﻣـﻦ ‪ BaCl2‬ﻭ ‪ H2O‬ﻓﻲ ﺍﻟﻌﻴﻨﺔ‪ ،‬ﻳﻤﻜﻨﻚ ﺗﺤﻮﻳﻞ ﻫﺬﻩ ﺍﻟﻜﺘﻞ ﺇﻟﻰ ﻣﻮﻻﺕ ﺑﺎﺳـﺘﻌﻤﺎﻝ ﺍﻟﻜﺘﻞ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﹺـ ‪ BaCl2‬ﻫﻲ ‪ 208.23g/ mol‬ﻭﻟﹺـ ‪.H2O 18.02g/ mol‬‬ ‫‪ 4.26 g‬ﻣﻦ ‪× BaCl2‬‬ ‫‪ 0.74 g‬ﻣﻦ ‪× H 2O‬‬

‫‪ 1 mol‬ﻣﻦ‬

‫‪BaCl2‬‬

‫‪ 208.23g‬ﻣﻦ‬

‫‪BaCl2‬‬

‫‪ 1 mol‬ﻣﻦ ‪H 2O‬‬

‫‪ 18.02g‬ﻣﻦ ‪H 2O‬‬

‫= ‪ 0.0205 mol‬ﻣﻦ‬

‫‪BaCl2‬‬

‫= ‪ 0.041 mol‬ﻣﻦ ‪H 2O‬‬

‫ﺇﺫﻥ‪ ،‬ﻧﺴﺒﺔ ﻣﻮﻻﺕ ‪ H2O‬ﺇﻟﻰ ﻣﻮﻻﺕ ‪BaCl2‬ﻫﻲ ‪ ،1 : 2‬ﻭﻟﺬﺍ ﻓﺈﻥ ‪ 2‬ﻣﻮﻝ ‪ H2O‬ﺗﺮﺗﺒﻂ ﺑﹺـ ‪ 1‬ﻣﻮﻝ ‪.BaCl2‬‬

‫ﺃﻱ ﹼ‬ ‫ﺃﻥ ﻗﻴﻤﺔ ﺍﻟﻤﻌﺎﻣﻞ ‪ x‬ﻫﻲ ‪ ،2‬ﻭﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻫﻲ ‪.BaCl2. 2H2O‬‬ ‫ﹾ‬ ‫‪ moles‬ﻣﻦ ‪H2O‬‬

‫‪=x‬‬

‫‪ moles‬ﻣﻦ ‪BaCl2‬‬

‫=‬

‫‪ 0.041 mol‬ﻣﻦ ‪H2O‬‬

‫‪ 0.0205 mol‬ﻣﻦ‬

‫‪BaCl2‬‬

‫=‬

‫‪ 2.0 mol‬ﻣﻦ‬ ‫‪= H2O‬‬

‫‪ 1.00 mol‬ﻣﻦ‬

‫ﻓﺴﺮ ﻟﻤﺎﺫﺍ ﺗﺴﺘﻌﻤﻞ ﺍﻟﻨﻘﻄﺔ ﻓﻲ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ؟‬ ‫‪ ?äCGôb GPÉe‬ﹼ‬

‫‪74‬‬

‫‪BaCl2‬‬

‫‪2‬‬ ‫‪1‬‬

‫‪5-14 ∫Éãe‬‬ ‫ﺗﺤﺪﻳـﺪ ﺻﻴﻐـﺔ ﺍﻟﻤﻠـﺢ ﺍﻟﻤﺎﺋﻲ ﻭﺿﻌﺖ ﻋﻴﻨﺔ ﻣﻦ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ ﺍﻟﻤﺎﺋﻴﺔ ﺍﻟﺰﺭﻗـﺎﺀ ‪ CuSO4.xH2O‬ﻛﺘﻠﺘﻬﺎ ‪ 2.50g‬ﻓﻲ‬ ‫ﻭﺳ ﹼ‬ ‫ـﺨﻨﺖ‪ .‬ﻭﺑﻘـﻲ ﺑﻌـﺪ ﺍﻟﺘﺴـﺨﻴﻦ ‪ 1.59g‬ﻣﻦ ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﻨﺤـﺎﺱ ﺍﻟﻼﻣﺎﺋﻴﺔ ﺍﻟﺒﻴﻀـﺎﺀ ‪ . CuSO4‬ﻓﻤـﺎ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ‬ ‫ﹶﺟﻔﻨـﺔ ﹸ‬ ‫ﺍﻟﻤﺎﺋﻲ؟ ﻭﻣﺎ ﺍﺳﻤﻪ؟‬ ‫‪1‬‬

‫–∏«‪ádCÉ°ùŸG π‬‬

‫ﻟﻘﺪ ﺃﻋﻄﻴﺖ ﻛﺘﻠﺔ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ ﺍﻟﻤﺎﺋﻴﺔ‪ ،‬ﻭﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ ﺍﻟﻼﻣﺎﺋﻴﺔ‪ .‬ﻛﻤﺎ ﺃﻧﻚ ﺗﻌﺮﻑ ﺻﻴﻐﺔ ﺍﻟﻤﺮﻛﺐ ﺑﺎﺳﺘﺜﻨﺎﺀ ﻗﻴﻤﺔ ‪،x‬‬ ‫ﻭﻫﻲ ﻣﻌﺎﻣﻞ ‪ H2O‬ﻓﻲ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺸﻴﺮ ﺇﻟﻰ ﻋﺪﺩ ﻣﻮﻻﺕ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‪.‬‬

‫ﺍﻟﻤﻄﻠﻮﺏ‬ ‫ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ = ؟‬ ‫ﺍﺳﻢ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ = ؟‬

‫ﺍﻟﻤﻌﻄﻴﺎﺕ‬

‫ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ = ‪ 2.50g‬ﻣﻦ ‪CuSO4. xH2O‬‬ ‫ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ = ‪ 1.59g‬ﻣﻦ ‪CuSO4‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﹺـ ‪18.02g/ mol = H2O‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﹺـ ‪159.6g/ mol = CuSO‬‬ ‫‪4‬‬

‫‪2‬‬

‫‪܃∏£ŸG ÜÉ°ùM‬‬

‫ﺣﺪﺩ ﻛﺘﻠﺔ ﺍﻟﻤﺎﺀ ﺍﻟﻤﻔﻘﻮﺩ‬ ‫ﻛﺘﻠﺔ ﺍﻟﻤﺎﺀ ﺍﻟﻤﻔﻘﻮﺩ = ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ – ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ‬

‫ﺍﻃﺮﺡ ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ ‪ CuSO4‬ﻣﻦ ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ‬ ‫ﺍﻟﻤﺎﺋﻲ ‪CuSO4.XH2O‬‬

‫‪2.50g - 1.59g=0.91g‬‬

‫ﺣﻮﻝ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻌﻠﻮﻣﺔ ﻟﻠﻤﺎﺀ ﻭﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﺇﻟﻰ ﻣﻮﻻﺕ ﻣﺴﺘﻌﻤ ﹰ‬ ‫ﻼ ﻣﻌﺎﻣﻞ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺬﻱ ﻳﺮﺑﻂ ﺍﻟﻤﻮﻻﺕ ﺑﺎﻟﻜﺘﻠﺔ – ﻣﻘﻠﻮﺏ‬ ‫ﹼ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬ ‫‪ 1mol‬ﻣﻦ ‪CuSO4‬‬ ‫= ‪0.00996mol‬ﻣﻦ ‪CuSO4‬‬ ‫ﻋﻮﺽ ﺑﺪﻝ ﻛﺘﻠﺔ ‪،CuSO4‬‬ ‫‪ 1.59g‬ﻣﻦ ‪× CuSO4‬‬ ‫‪ 159.6g‬ﻣﻦ ‪CuSO4‬‬

‫ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪،‬‬ ‫ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‬

‫ﻋﻮﺽ ﺑﺪﻝ ﺍﻟﻜﺘﻠﺔ ‪ ،H2O‬ﻣﻘﻠﻮﺏ ﺍﻟﻜﺘﻠﺔ‬ ‫ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‬

‫‪ 1mol‬ﻣﻦ ‪H2O‬‬ ‫= ‪0.05mol‬ﻣﻦ ‪H2O‬‬ ‫‪ 0.91g‬ﻣﻦ ‪× H2O‬‬ ‫‪ 18.02g‬ﻣﻦ ‪H2O‬‬ ‫‪=x‬‬

‫ﻋﻮﺽ ﺑﺪﻝ ﻣﻮﻻﺕ ‪ ،H2O‬ﻭﻣﻮﻻﺕ‬ ‫‪ ،CuSO4‬ﻭﺍﺣﺴﺐ ﺃﺑﺴﻂ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ‬

‫‪=x‬‬

‫‪ moles‬ﻣﻦ ‪H2O‬‬ ‫‪ moles‬ﻣﻦ ‪CuSO4‬‬

‫‪ 0.050 mol‬ﻣﻦ‬ ‫‪ 0.00996 mol‬ﻣﻦ ‪CuSO4‬‬ ‫‪H2O‬‬

‫≈‬

‫‪ 5 mol‬ﻣﻦ‬ ‫‪ 1 mol‬ﻣﻦ ‪CuSO4‬‬ ‫‪H 2O‬‬

‫= ‪5‬‬

‫ﺇﺫﻥ‪ ،‬ﻓﺼﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻫﻲ ‪ ،CuSO4 .5H2O‬ﻭﺍﺳﻤﻪ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ )‪ (II‬ﺧﻤﺎﺳﻴﺔ ﺍﻟﻤﺎﺀ‪.‬‬

‫‪75‬‬

‫‪3‬‬

‫‪áHÉLE’G ˃≤J‬‬

‫ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻨﺤﺎﺱ )‪ (II‬ﺧﻤﺎﺳﻴﺔ ﺍﻟﻤﺎﺀ ﻭﻫﻲ ﻣﻠﺢ ﺷﺎﺋﻊ ﻭﻣﺪﻭﻥ ﻓﻲ ﺍﻟﺠﺪﻭﻝ ‪.5 -1‬‬

‫‪á«ÑjQóJ πFÉ°ùe‬‬

‫‪ .74‬ﻳﻈﻬﺮ ﻓﻲ ﺍﻟﺸﻜﻞ ﺍﻟﻤﺠﺎﻭﺭ ﺗﺮﻛﻴﺐ ﺃﺣﺪ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ‪.‬‬ ‫ﻓﻤﺎ ﺻﻴﻐﺔ ﻫﺬﺍ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ؟ ﻭﻣﺎ ﺍﺳﻤﻪ؟‬

‫‪ .75‬ﺗﺤﺪ‪ :‬ﺳﺨﻨﺖ ﻋﻴﻨﺔ ﻛﺘﻠﺘﻬﺎ ‪ 11.75g‬ﻣﻦ ﻣﻠﺢ ﻣﺎﺋﻲ ﺷﺎﺋﻊ ﻟﻜﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ )‪.(II‬‬

‫ﻭﺑﻘﻲ ﺑﻌﺪ ﺍﻟﺘﺴـﺨﻴﻦ‪ 0.0712mol ،‬ﻣﻦ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﻮﺑﻠﺖ ﺍﻟﻼﻣﺎﺋﻲ‪ .‬ﻓﻤﺎ ﺻﻴﻐﺔ ﻫﺬﺍ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ؟‬ ‫ﻭﻣﺎ ﺍﺳﻤﻪ؟‬

‫‪H 2O‬‬ ‫‪51.2%‬‬

‫‪MgSO4‬‬ ‫‪48.8%‬‬

‫اﺳﺘﻌماﻻت اﻷمﻼﺡ الماﺋﻴﺔ ‪Uses of Hydrates‬‬

‫ﻳﻜﻮﻥ ﺛﻼﺛﺔ ﺃﻣﻼﺡ‬ ‫ﻟﻸﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﺍﺳـﺘﻌﻤﺎﻻﺕ ﻣﻬﻤﺔ ﻓﻲ ﻣﺨﺘﺒﺮ ﺍﻟﻜﻴﻤﻴﺎﺀ‪ .‬ﻓﻜﻠﻮﺭﻳﺪ ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ﹼ‬ ‫ﻣﺎﺋﻴﺔ– ﺃﺣﺎﺩﻱ ﺍﻟﻤﺎﺀ‪ ،‬ﻭﺛﻨﺎﺋﻲ ﺍﻟﻤﺎﺀ‪ ،‬ﻭﺳﺪﺍﺳـﻲ ﺍﻟﻤﺎﺀ‪ .‬ﻭﻳﻮﺿﻊ ﻛﻠﻮﺭﻳﺪ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﺍﻟﻼﻣﺎﺋﻲ ﻓﻲ‬ ‫ﺍﻟﻤﺠﻔﻔﺎﺕ‪،‬ﻛﻤﺎ ﻓﻲ ﺍﻟﺸﻜﻞ‪ ،5-15‬ﺣﻴﺚ ﻳﻘﻮﻡ ﺑﺎﻣﺘﺼﺎﺹ‬ ‫ﻗﻌﺮ ﺃﻭﻋﻴﺔ ﻣﺤﻜﻤﺔ ﺍﻹﻏﻼﻕ ﺗﹸﺴـﻤﻰ ﱢ‬ ‫ﺟﻮﺍ ﺟﺎ ﹰﻓـﺎ ﻳﻤﻜﻦ ﺣﻔﻆ ﺍﻟﻤـﻮﺍﺩ ﺍﻷﺧﺮ￯ ﻓﻴﻪ‬ ‫ﺍﻟﺮﻃﻮﺑـﺔ ﻣـﻦ ﺍﻟﻬـﻮﺍﺀ ﻓﻲ ﺩﺍﺧﻞ ﱢ‬ ‫ﺍﻟﻤﺠﻔـﻒ‪ ،‬ﻭﻳﺼﻨﻊ ﹰ‬ ‫ﺟﺎﻓـﺔ‪ .‬ﻭﻛﺜﻴـﺮ ﹰﺍ ﻣﺎ ﺗﻀـﺎﻑ ﻛﺒﺮﻳﺘﺎﺕ ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﺇﻟﻰ ﺍﻟﻤﺬﻳﺒـﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﻛﺎﻹﻳﺜـﺮ ﻭﺍﻟﻜﻠﻮﺭﻓﻮﺭﻡ‬ ‫ﻟﻠﺤﻔﺎﻅ ﻋﻠﻴﻬﺎ ﺧﺎﻟﻴﺔ ﻣﻦ ﺍﻟﻤﺎﺀ‪.‬‬ ‫ﺇﻥ ﻗـﺪﺭﺓ ﺍﻟﻤﻠـﺢ ﺍﻟﻼﻣﺎﺋﻲ ﻋﻠﻰ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻤﺎﺀ ﻟﻪ ﹰ‬ ‫ﺃﻳﻀﺎ ﺑﻌـﺾ ﺍﻟﺘﻄﺒﻴﻘﺎﺕ ﺍﻟﺘﺠﺎﺭﻳﺔ‪ .‬ﻓﺎﻟﻤﻌﺪﺍﺕ‬ ‫ﺍﻹﻟﻜﺘﺮﻭﻧﻴـﺔ ﻭﺍﻟﺒﺼﺮﻳـﺔ‪ ،‬ﻭﺑﺨﺎﺻـﺔ ﺗﻠﻚ ﺍﻟﺘﻲ ﺗﹸﺸـﺤﻦ ﻋﺒـﺮ ﺍﻟﺒﺤﺎﺭ ﺑﺎﻟﺴـﻔﻦ‪ ،‬ﻭﻏﺎﻟ ﹰﺒﺎ ﻣـﺎ ﺗﹸﻌﺒﺄ ﻣﻊ‬ ‫ﺍﻟﻤﺠ ﹼﻔﻔـﺎﺕ ﺍﻟﺘﻲ ﺗﻤﻨﻊ ﺗﺄﺛﻴـﺮ ﺍﻟﺮﻃﻮﺑﺔ ﻓﻲ ﺍﻟﺪﻭﺍﺋـﺮ ﺍﻹﻟﻜﺘﺮﻭﻧﻴﺔ ﺍﻟﺪﻗﻴﻘﺔ‪ .‬ﻭﺗﺴـﺘﻌﻤﻞ‬ ‫ﺃﻛﻴـﺎﺱ ﻣـﻦ ﹸ‬ ‫ﺑﻌـﺾ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﻣﺜﻞ ﻛﺒﺮﻳﺘـﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺍﻟﻤﺎﺋﻴﺔ )‪ (Na2SO4. 10H2O‬ﻟﺨﺰﻥ ﺍﻟﻄﺎﻗﺔ‬ ‫ﺍﻟﺸﻤﺴـﻴﺔ‪ .‬ﻓﻌﻨﺪﻣﺎ ﺗ ﹼ‬ ‫ﹸﺴـﺨﻦ ﺍﻟﺸـﻤﺲ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﺇﻟﻰ ﺃﻛﺜﺮ ﻣﻦ ‪ 32‬ﺱ‪ °‬ﺗﺬﻭﺏ ‪ Na2SO4‬ﻓﻲ‬ ‫ﻣـﻮﻻﺕ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﺍﻟﻌﺸـﺮﺓ‪ ،‬ﻭﺧـﻼﻝ ﺫﻟﻚ ﻳﻤﺘﺺ ﺍﻟﻤﻠـﺢ ﺍﻟﻤﺎﺋﻲ ﺍﻟﻄﺎﻗﺔ‪ ،‬ﻭﻫـﺬﻩ ﺍﻟﻄﺎﻗﺔ ﺗﻨﻄﻠﻖ‬ ‫ﻋﻨﺪﻣﺎ ﺗﻨﺨﻔﺾ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻭﻳﺘﺒﻠﻮﺭ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﺛﺎﻧﻴﺔ‪.‬‬

‫اﻟﺘﻘﻮﻳﻢ ‪5 - 5‬‬ ‫اﻟﺨﻼﺻﺔ‬

‫ﺗﺘﻜـﻮﻥ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻣﻦ ﺻﻴﻐﺔ ﺍﻟﻤﺮﻛﺐ‬ ‫ﺍﻷﻳﻮﻧﻲ ﻭﻋﺪﺩ ﺟﺰﻳﺌـﺎﺕ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﺍﻟﻤﺮﺗﺒﻄﺔ‬ ‫ﺑﻮﺣﺪﺓ ﺍﻟﺼﻴﻐﺔ‪.‬‬ ‫ﻳﺘﻜـﻮﻥ ﺍﺳـﻢ ﺍﻟﻤﻠـﺢ ﺍﻟﻤﺎﺋـﻲ ﻣﻦ ﺍﺳـﻢ ﺍﻟﻤﺮﻛﺐ‬ ‫ﻣﺘﺒﻮﻋﺎ ﺑﻤﻘﻄﻊ ﻳﺪﻝ ﻋﻠـﻰ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ‬ ‫ﹰ‬ ‫ﺍﻟﻤﺮﺗﺒﻄﺔ ﺑﻤﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫ﺗﻜـﻮﻥ ﺍﻟﻤﻠـﺢ ﺍﻟﻼﻣﺎﺋـﻲ ﻋﻨـﺪ ﺗﺴـﺨﻴﻦ ﺍﻟﻤﻠـﺢ‬ ‫ﺍﻟﻤﺎﺋﻲ‪.‬‬

‫‪76‬‬

‫ﺍﻟﺸﻜﻞ ‪ 5-16‬ﻳﺠﻔﻒ ﻛﻠﻮﺭﻳﺪ‬ ‫ﺍﻟﻜﺎﻟﺴـﻴﻮﻡ ﺍﻟﻬﻮﺍﺀ ﻣﻦ ﺟﺰﻳﺌﺎﺕ‬ ‫ﺍﻟﻤﺎﺀ‪ .‬ﻛﻤﺎ ﻳﺴﺘﻌﻤﻞ ﻓﻲ ﺍﻟﻤﺨﺘﺒﺮ‬ ‫ﻓﻲ ﺣﻔﻆ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬ ‫ﻣﻦ ﺭﻃﻮﺑﺔ ﺍﻟﺠﻮ‪.‬‬

‫‪ .76‬اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻟﺨﺺ ﺗﺮﻛﻴﺐ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪.‬‬ ‫ﹼ‬ ‫ﺍﻟﻤﺮﻛﺐ ﺍﻟﺬﻱ ﺻﻴﻐﺘﻪ ‪SrCl2. 6H2O‬‬ ‫ﺳﻢ‬ ‫‪ .77‬ﹼ‬ ‫ﹰ‬ ‫‪ .78‬ﺻﻒ ﺍﻟﺨﻄﻮﺍﺕ ﺍﻟﻌﻤﻠﻴﺔ ﻟﺘﺤﺪﻳﺪ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻣﻌﻠﻼ ﻛﻞ ﺧﻄﻮﺓ‪.‬‬ ‫ﻃﺒﻖ‪ :‬ﻳﺤﺘﻮﻱ ﻣﻠﺢ ﻣﺎﺋﻲ ﻋﻠﻰ ‪ 0.050mol‬ﻣﻦ ﺍﻟﻤﺎﺀ ﻟﻜﻞ ‪0.00998mol‬‬ ‫‪ .79‬ﹼ‬ ‫ﻣﻦ ﺍﻟﻤﺮﻛﺐ ﺍﻷﻳﻮﻧﻲ‪ .‬ﺍﻛﺘﺐ ﺻﻴﻐﺔ ﻋﺎﻣﺔ ﻟﻠﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‪.‬‬ ‫‪ .80‬ﺍﺣﺴـﺐ ﻛﺘﻠـﺔ ﻣـﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ﺇﺫﺍ ﻓﻘﺪ ﻣﻠـﺢ ﻣﺎﺋﻲ ‪ 0.025mol‬ﻣـﻦ ﺍﻟﻤﺎﺀ ﻋﻨﺪ‬ ‫ﺗﺴﺨﻴﻨﻪ‪.‬‬ ‫‪ .81‬ﺭﺗﹼـﺐ ﺍﻷﻣـﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺗﺼﺎﻋﺪ ﹰﻳﺎ ﺣﺴـﺐ ﺗﺰﺍﻳﺪ ﺍﻟﻨﺴـﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻟﻠﻤﺤﺘﻮ￯‬ ‫ﺍﻟﻤﺎﺋﻲ‪MgSO4. 7H2O، Ba (OH)2 . 8H2O، CoCl2.6H2O :‬‬ ‫ﻓﺴـﺮ ﻛﻴﻒ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ 5-15‬ﺑﻮﺻﻔﻪ‬ ‫‪ .82‬ﹼ‬ ‫ﻃﺒﻖ‪ :‬ﹼ‬ ‫ﻃﺮﻳﻘﺔ ﺗﻘﺮﻳﺒﻴﺔ ﻟﺘﺤﺪﻳﺪ ﺍﺣﺘﻤﺎﻝ ﺳﻘﻮﻁ ﺍﻟﻤﻄﺮ‪.‬‬

‫اﻟﺘﺎرﻳﺦ ﻓﻲ ﻛﺄس ﻣﺎء‬

‫‪21‬‬

‫ﻫﻞ ﺗﺘﺬﻛﺮ ﺁﺧﺮ ﻛﺄﺱ ﻣﺎﺀ ﺗﻨﺎﻭﻟﺘﻪ؟ ﻗﺪ ﻳﺒﺪﻭ ﻏﻴﺮ ﻗﺎﺑﻞ ﻟﻠﺘﺼﺪﻳﻖ ﺃﻥ ﻭ ‪ 6× 10‬ﻛﺄﺱ ﻣـﺎﺀ ﻋﻠـﻰ ﺍﻷﺭﺽ‪ .‬ﻭﻟـﻮ ﻗﺎﺭﻧـﺖ ﺑﻴﻦ ﻫﺬﻳﻦ‬ ‫ﻧﻘـﻮﻝ ﺇﻥ ﺗﻠﻚ ﺍﻟﻜﺄﺱ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ﻗﺪ ﺍﺳـﺘﻬﻠﻜﻬﺎ ﺍﻟﺮﻗﻤﻴﻦ ﻓﺴـﺘﺮ￯ ﺃﻥ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ ﻓـﻲ ﺍﻟﻜﺄﺱ ﺍﻟﻮﺍﺣﺪﺓ‬ ‫ﺍﻟﻤﺘﻨﺒﻲ ﻣﺜ ﹰ‬ ‫ﻼ‪ ،‬ﺃﻭ ﺁﻳﻨﺸﺘﻴﻦ‪ ،‬ﺃﻭ ﺟﺎﻥ ﺩﺍﺭﻙ‪ !..‬ﻛﻴﻒ ﻳﻤﻜﻦ ﻟﻜﺄﺳﻴﻦ ﺃﻛﺜﺮ ﺑﺄﻟﻒ ﻣﺮﺓ ﻣﻦ ﻋﺪﺩ ﻛﺆﻭﺱ ﺍﻟﻤﺎﺀ ﻋﻠﻰ ﺍﻷﺭﺽ‪.‬‬ ‫ﻣـﻦ ﺍﻟﻤـﺎﺀ ﻓﻲ ﺯﻣﻨﻴﻦ ﻣﺨﺘﻠﻔﻴـﻦ ﺃﻥ ﺗﺤﻮﻳﺎ ﹰ‬ ‫ﺑﻌﻀﺎ ﻣـﻦ ﺍﻟﺠﺰﻳﺌﺎﺕ‬ ‫)‪(A‬‬ ‫)‪(B‬‬ ‫ﻧﻔﺴﻬﺎ؟ﻳﺮﻭﻱ ﻟﻨﺎ ﺍﻟﻘﺼﺔﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻭﺍﻟﺤﺴﺎﺑﺎﺕ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻤﺤﻴﻄـﺎﺕ ﻭﺍﻟﻤـﻮﻻﺕ ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻜﻠﻴـﺔ ﻟﻠﻤﺎﺀ ﻓـﻲ ﺍﻟﻤﺤﻴﻄﺎﺕ‬ ‫‪24‬‬ ‫ﻭﻏﻴﺮﻫﺎ ﺗﻘﺎﺭﺏ ‪ .1.4 × 10 g‬ﺃﻣﺎ ﺍﻟﻜﺄﺱ ﻓﻴﺤﺘﻮﻱ ﻋﻠﻰ ‪230g‬‬ ‫ﻣـﻦ ﺍﻟﻤﺎﺀ‪ .‬ﻭﺑﺎﺳـﺘﺨﺪﺍﻡ ﻫـﺬﻩ ﺍﻟﺒﻴﺎﻧـﺎﺕ ﻳﻤﻜﻨﻚ ﺣﺴـﺎﺏ ﺍﻟﻌﺪﺩ‬ ‫ﺍﻟﻜﻠﻲ ﻟﻜﺆﻭﺱ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺘﻮﺍﻓﺮﺓ ﻟﻠﺸـﺮﺏ ﻋﻠﻰ ﺍﻷﺭﺽ‪ ،‬ﻭﺍﻟﻌﺪﺩ‬ ‫ﻛﺄﺱ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﻟﺠﺰﻳﺌﺎﺕ‬ ‫ﺍﻟﻜﻠﻲ ﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ ﻓﻲ ﻫﺬﻩ ﺍﻟﻜﺆﻭﺱ‪.‬‬ ‫ﺍﻟﻤﺨﻠﻮﻃﺔ ﺃﺧﺬﺕ‬

‫ﺟﺰﻳﺌﺎﺕ‬

‫ﻣﺨﻠﻮﻃﺔ‬

‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﻗﺒﻞ‬ ‫ﺍﻟﺨﻠﻂ‬

‫ﻣـﻦ ﺍﻟﻤﻌـﺮﻭﻑ ﺃﻥ ﻛﺘﻠـﺔ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣـﻦ ﺍﻟﻤﺎﺀ ﺗﺴـﺎﻭﻱ ‪،18 g‬‬ ‫ﻭﺑﺎﺳـﺘﺨﺪﺍﻡ ﺗﺤﻠﻴﻞ ﺍﻟﻮﺣﺪﺍﺕ ﻳﻤﻜﻨـﻚ ﺗﺤﻮﻳﻞ ﺟﺮﺍﻣﺎﺕ ﺍﻟﻤﺎﺀ ‪ájhÉM »a Ö`` °üJ (AGôªëdG) (A) ¢SCɵdG øe AÉ`` ªdG äÉÄjõL 1 πµ``°ûdG‬‬ ‫‪IPƒNCɪdG (B) ¢SCɵdGh .(AÉbQõdG) ¢VQC’G ≈∏Y AɪdG äÉÄjõL πµd ™°ùàJ‬‬ ‫ﻓﻲ ﺍﻟﻜﺄﺱ ﺇﻟﻰ ﻣﻮﻻﺕ‪.‬‬ ‫‪»a âfÉc »`` àdG AɪdG äÉÄjõL ø`` e ô«¨`` °U OóY ≈∏Y …ƒàëJ AÉ`` YƒdG ø`` e‬‬ ‫‪ 230 g‬ﻣﻦ ﺍﻟﻤﺎﺀ ‪ 1 mol‬ﻣﻦ ﺍﻟﻤﺎﺀ‬ ‫×‬ ‫‪.≈dhC’G ¢SCɵdG‬‬ ‫‪ 18 g‬ﻣﻦ ﺍﻟﻤﺎﺀ‬ ‫ﻛﺄﺱ‬ ‫ﺍﻟﺤﺎﻭﻳـﺔ ﺍﻟﻌﻤﻼﻗـﺔ ﺍﻓـﺮﺽ ﺃﻥ ﺍﻟﻤـﺎﺀ ﻛﻠﻪ ﺍﻟـﺬﻱ ﻋﻠﻰ ﺍﻷﺭﺽ‬ ‫= ‪ 13 mol‬ﻣﻦ ﺍﻟﻤﺎﺀ‪ /‬ﻛﺄﺱ‬ ‫ـﺰﻥ ﻓﻲ ﺣﺎﻭﻳﺔ ﻭﺍﺣﺪﺓ ﻣﻜﻌﺒﺔ ﺍﻟﺸـﻜﻞ‪ ،‬ﻓﺈﻧﻬﺎ ﺳـﺘﻜﻮﻥ ﺣﺎﻭﻳﺔ‬ ‫ﹸﺧ ﹼ‬ ‫ﺛـﻢ ﺗﺤﻮﻳـﻞ ﻫـﺬﻩ ﺍﻟﻤـﻮﻻﺕ ﺇﻟـﻰ ﺟﺰﻳﺌـﺎﺕ ﺑﺎﺳـﺘﺨﺪﺍﻡ ﻋـﺪﺩ‬ ‫ﻋﻤﻼﻗـﺔ ﻃـﻮﻝ ﺿﻠﻌﻬﺎ ‪ 1100 Km‬ﻭﺗﺨﻴـﻞ ﺃﻧﻚ ﻣﻸﺕ ﻛﺄﺱ‬ ‫ﺃﻓﻮﺟﺎﺩﺭﻭ‪.‬‬ ‫ﻣـﺎﺀ ﻣﻦ ﻫﺬﻩ ﺍﻟﺤﺎﻭﻳﺔ‪ ،‬ﺛﻢ ﺃﻋﺪﺗﻪ ﺇﻟﻴﻬﺎ‪ ،‬ﻭﺍﻧﺘﻈﺮﺕ ﻟﻴﺨﺘﻠﻂ ﺍﻟﻤﺎﺀ‬ ‫‪23‬‬ ‫‪ 13 mol‬ﻣﻦ ﺍﻟﻤﺎﺀ ‪ 6.02×10 molecules‬ﻣﻦ ﺍﻟﻤﺎﺀ‬ ‫ﺗﻤﺎ ﹰﻣـﺎ‪ ،‬ﺛﻢ ﻣﻸﺕ ﺍﻟـﻜﺄﺱ ﻣﺮﺓ ﺃﺧﺮ￯‪ ،‬ﻓﻬﻞ ﺳـﺘﻜﻮﻥ ﺟﺰﻳﺌﺎﺕ‬ ‫×‬ ‫ﻛﺄﺱ‬ ‫‪ 1 mol‬ﻣﻦ ﺍﻟﻤﺎﺀ‬ ‫‪24‬‬ ‫ﺍﻟﻤﺎﺀ ﻓﻲ ﺍﻟﻜﺄﺱ ﺍﻷﻭﻟﻰ ﻣﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﻜﺄﺱ ﺍﻟﺜﺎﻧﻴﺔ؟‬ ‫= ‪ 8 × 10‬ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ‪/‬ﻛﺄﺱ‬ ‫ﻛﻤﺎ ﻳﻤﻜﻨﻚ ﺣﺴﺎﺏ ﻋﺪﺩ ﻛﺆﻭﺱ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺘﻮﺍﻓﺮﺓ ﻟﻠﺸﺮﺏ ﻋﻠﻰ ﻛﻤـﺎ ﻫـﻮ ﻣﻮﺿـﺢ ﻓﻲ ﺍﻟﺸـﻜﻞ ‪ ،1‬ﻣـﻦ ﺍﻟﻤﺮﺟـﺢ ﺃﻥ ﺍﻟﻜﺄﺳـﻴﻦ‬ ‫ﺳﺘﺸـﺘﺮﻛﺎﻥ ﻓـﻲ ﻋﺪﺩ ﻣـﻦ ﺟﺰﻳﺌـﺎﺕ ﺍﻟﻤـﺎﺀ‪ .‬ﻟﻤـﺎﺫﺍ؟ ﻷﻥ ﻋﺪﺩ‬ ‫ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬ ‫ﻣﺎﺀ‬ ‫‪1‬ﻛﺄﺱ‬ ‫‪21‬‬ ‫‪24‬‬ ‫ﺟﺰﻳﺌـﺎﺕ ﺍﻟﻤـﺎﺀ ﻓﻲ ﺍﻟﻜﺄﺱ ﺃﻛﺜﺮ ﺑﺄﻟﻒ ﻣـﺮﺓ ﻣﻦ ﻋﺪﺩ ﺍﻟﻜﺆﻭﺱ‬ ‫= ‪ 6 × 10‬ﻛﺄﺱ‪.‬‬ ‫‪ 1.4 × 10 g‬ﻣﺎﺀ ×‬ ‫‪ 230 g‬ﻣﺎﺀ‬ ‫ﻓﻲ ﺍﻟﺤﺎﻭﻳﺔ‪ .‬ﻭﺑﻬﺬﺍ ﺍﻟﻤﻌﺪﻝ ﻓﺈﻥ ﺍﻟﻜﺄﺱ ﺍﻟﺜﺎﻧﻴﺔ ﺳـﺘﺤﺘﻮﻱ ﻋﻠﻰ‬ ‫‪24‬‬ ‫ﺇﺫﻥ‪ ،‬ﻳﻮﺟـﺪ ‪ 8 × 10‬ﺟـﺰﻱﺀ ﻓـﻲ ﻛﺄﺱ ﻭﺍﺣـﺪﺓ ﻣـﻦ ﺍﻟﻤﺎﺀ‪ 1000 molecules ،‬ﺗﻘﺮﻳ ﹰﺒﺎ ﻛﺎﻧﺖ ﻓﻲ ﺍﻟﻜﺄﺱ ﺍﻷﻭﻟﻰ‪.‬‬ ‫‪‘ áHÉàµdG‬‬

‫اﻟﻜﻴﻤﻴﺎء‬

‫ﻗﺪﹼ ﺭ ﻳﻤﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﻃﺮﻳﻘﺔ ﺍﻟﺘﻘﺪﻳﺮ ﺍﻟﻤﺘﺒﻌﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﻘﺎﻟﺔ ﻓﻲ ﺇﺟﺮﺍﺀ‬ ‫ﺃﻧﻮﺍﻉ ﺃﺧﺮ￯ ﻣﻦ ﺍﻟﺤﺴﺎﺑﺎﺕ‪ .‬ﻟﺬﺍ‪ ،‬ﺍﺳﺘﺨﺪﻡ ﻫﺬﻩ ﺍﻟﻄﺮﻳﻘﺔ ﻟﺘﻘﺪﻳﺮ ﺍﻟﻜﺘﻠﺔ‬

‫ﺍﻟﻜﻠﻴﺔ ﻟﻠﻄﻼﺏ ﻓﻲ ﻣﺪﺭﺳـﺘﻚ‪ .‬ﻟﺘﻌـﺮﻑ ﺍﻟﻤﺰﻳﺪ ﻋﻦ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻜﺒﻴﺮﺓ‪،‬‬

‫ﺍﺭﺟﻊ ﺇﻟﻰ ‪www.obeikaneducation.com :‬‬

‫ﻗﻮﺓ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻜﺒﻴﺮﺓ‪ .‬ﹼ‬ ‫ﻓﻜﺮ ﻓﻲ ﻛﻤﻴﺔ ﺍﻟﻤﺎﺀ ﺍﻟﺘﻲ ﻣﺮﺕ ﻓﻲ ﺟﺴﻢ ﺍﻟﻤﺘﻨﺒﻲ‬ ‫ﺣﻴﺎﺗﻬﻢ– ﻭﻫـﻲ ﺃﻛﺒﺮ ﺑﻜﺜﻴﺮ ﻣﻦ‬ ‫ﺃﻭ ﺁﻳﻨﺸـﺘﺎﻳﻦ ﺃﻭ ﺟـﺎﻥ ﺩﺍﺭﻙ‪ ،‬ﺧﻼﻝ ﺣﻴﺎﺗﻬﻢ‬ ‫ﹰ‬ ‫ﻣﻔﺘﺮﺿﺎ ﺃﻥ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ ﺍﺧﺘﻠﻄﺖ ﺑﺎﻟﺘﺴـﺎﻭﻱ ﻓﻲ‬ ‫ﻛﺄﺱ ﻭﺍﺣﺪﺓ –‬ ‫ﺣﺠﻢ ﺍﻟﻤﺎﺀ ﻛﺎﻣ ﹰ‬ ‫ﻼ ﻋﻠﻰ ﺍﻷﺭﺽ‪ .‬ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﺴﺘﻮﻋﺐ ﻟﻤﺎﺫﺍ ﻳﺠﺐ‬ ‫ﺃﻥ ﺗﺤﺘﻮﻱ ﻛﺄﺱ ﺍﻟﻤﺎﺀ ﻋﻠﻰ ﺑﻌﺾ ﻫﺬﻩ ﺍﻟﺠﺰﻳﺌﺎﺕ‪.‬‬

‫‪٧٧‬‬ ‫‪77‬‬

‫‪π°üØdG‬‬

‫‪π°üØdG á©LGôe π«dO‬‬ ‫الفكرة‬

‫العامة‬

‫ﺍﳌﻮﻝ ﻳﻤﺜﻞ ﻋﺪ ﹰﺩﺍ ﻛﺒ ﹰﲑﺍ ﻣﻦ ﺍﳉﺴﻴﲈﺕ ﺍﳌﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ‪.‬‬

‫‪ 5-1‬ﻗﻴﺎس اﻟﻤﺎدة‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﺴﺘﻌﻤﻞ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻮﻥ ﺍﳌﻮﻝ ﻟﻌﺪﹼ ﺍﻟﺬﺭﺍﺕ‪،‬‬ ‫ﻭﺍﳉﺰﻳﺌﺎﺕ ﻭﺍﻷﻳﻮﻧﺎﺕ‪ ،‬ﻭﻭﺣﺪﺍﺕ‬ ‫ﺍﻟﺼﻴﻐﺔ‪.‬‬ ‫اﻟﻤﻔﺮدات‬

‫• ﺍﳌﻮﻝ‬ ‫• ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫•‬ ‫•‬ ‫•‬ ‫•‬

‫ﺍﳌﻮﻝ ﻭﺣﺪﺓ ﺗﺴﺘﺨﺪﻡ ﻟﻌﺪﱢ ﺟﺴﻴﲈﺕ ﺍﳌﺎﺩﺓ ﺑﺸﻜﻞ ﻏﲑ ﻣﺒﺎﴍ‪1 .‬ﻣﻮﻝ ﻣﻦ‬ ‫ﺍﳌﺎﺩﺓ ﺍﻟﻨﻘﻴﺔ ﳛﺘﻮﻱ ﻋﲆ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﳉﺴﻴﲈﺕ‪.‬‬ ‫ﺍﳉﺴـﻴﲈﺕ ﺗﺸـﻤﻞ ﺍﻟـﺬﺭﺍﺕ‪ ،‬ﻭﺍﻷﻳﻮﻧـﺎﺕ‪ ،‬ﻭﺍﳉﺰﻳﺌﺎﺕ‪ ،‬ﻭﻭﺣـﺪﺍﺕ ﺍﻟﺼﻴﻎ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻭﺍﻹﻟﻜﱰﻭﻧﺎﺕ‪ ،‬ﻭﺟﺴﻴﲈﺕ ﺃﺧﺮ￯ ﻣﺸﺎﲠﺔ‪.‬‬ ‫ﺍﳌﻮﻝ ﺍﻟﻮﺍﺣﺪ ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ‪ 12 -‬ﻟﻪ ﻛﺘﻠﺔ ﻣﻘﺪﺍﺭﻫﺎ ‪ 12 g‬ﲤ ﹰﺎﻣﺎ‪.‬‬ ‫ﻳﻤﻜـﻦ ﺍﺳـﺘﻌﲈﻝ ﻣﻌﺎﻣـﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﳌﻜﺘﻮﺑـﺔ ﻣﻦ ﻋﻼﻗﺔ ﻋـﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ‬ ‫ﻟﻠﺘﺤﻮﻳﻞ ﺑﲔ ﺍﳌﻮﻻﺕ ﻭﻋﺪﺩ ﺍﳉﺴﻴﲈﺕ‪.‬‬

‫‪ 5-2‬اﻟﻜﺘﻠﺔ واﻟﻤﻮل‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﳛﺘﻮﻱ ﺍﳌﻮﻝ ﹰ‬ ‫ﺩﺍﺋﲈ‬ ‫ﻋﲆ ﺍﻟﻌﺪﺩ ﻧﻔﺴﻪ ﻣﻦ ﺍﳉﺴﻴﲈﺕ‪ ،‬ﻭﻣﻊ‬ ‫ﺫﻟﻚ‪ ،‬ﻓﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﴏ ﺍﳌﺨﺘﻠﻔﺔ ﳍﺎ‬ ‫ﻛﺘﻞ ﳐﺘﻠﻔﺔ‪.‬‬ ‫اﻟﻤﻔﺮدات‬

‫• ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫•‬ ‫•‬ ‫•‬ ‫•‬

‫ﺗﺴﻤﻰ ﻛﺘﻠﺔ ﺍﳌﻮﻝ ﺍﻟﻮﺍﺣﺪ ﺑﺎﳉﺮﺍﻣﺎﺕ ﻣﻦ ﺃﻱ ﻣﺎﺩﺓ ﻧﻘﻴﺔ ﺑﺎﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻷﻱ ﻋﻨﴫ ﺗﺴﺎﻭﻱ ﻋﺪﺩ ﹰﹼﻳﺎ ﻛﺘﻠﺘﻪ ﺍﻟﺬﺭﻳﺔ‪.‬‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻷﻱ ﻣﺎﺩﺓ ﻫﻲ ﻛﺘﻠﺔ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺍﳉﺴﻴﲈﺕ ﳍﺬﻩ ﺍﳌﺎﺩﺓ‪.‬‬ ‫ﺗﺴﺘﻌﻤﻞ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ ﺍﳌﻮﻻﺕ ﺇﱃ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﻭﻳﺴﺘﻌﻤﻞ ﻣﻘﻠﻮﺏ‬ ‫ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ ﺍﻟﻜﺘﻠﺔ ﺇﱃ ﺍﳌﻮﻻﺕ‪.‬‬

‫‪ 5-3‬ﻣﻮﻻت اﻟﻤﺮﻛﺒﺎت‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﻳﻤﻜـﻦ ﺣﺴـﺎﺏ‬ ‫ﺍﻟﻜﺘﻠـﺔ ﺍﳌﻮﻟﻴـﺔ ﻟﻠﻤﺮﻛـﺐ ﻣـﻦ ﺧـﻼﻝ‬ ‫ﺻﻴﻐﺘﻪ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ ،‬ﻛﲈ ﻳﻤﻜﻦ ﺍﺳـﺘﻌﲈﻝ‬ ‫ﺍﻟﻜﺘﻠـﺔ ﺍﳌﻮﻟﻴـﺔ ﻟﻠﺘﺤﻮﻳﻞ ﻣﺎ ﺑـﲔ ﺍﻟﻜﺘﻠﺔ‬ ‫ﻭﺍﳌﻮﻻﺕ‪.‬‬

‫‪78‬‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫• ﺗﺪﻝ ﺍﻷﺭﻗﺎﻡ ﰲ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻋﲆ ﻋﺪﺩ ﻣﻮﻻﺕ ﻛﻞ ﻋﻨﴫ ﰲ ﻣﻮﻝ ﻭﺍﺣﺪ‬ ‫ﻣﻦ ﺍﳌﺮﻛﺐ‪.‬‬ ‫• ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﲢﺴﺐ ﻣﻦ ﺍﻟﻜﺘﻞ ﺍﳌﻮﻟﻴﺔ ﳉﻤﻴﻊ ﺍﻟﻌﻨﺎﴏ ﰲ ﺍﳌﺮﻛﺐ‪.‬‬ ‫• ﻣﻌﺎﻣﻼﺕ ﺍﻟﺘﺤﻮﻳﻞ ﺍﳌﺒﻴﻨﺔ ﻋﲆ ﺍﻟﻜﺘﻠﺔ ﺍﳌﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺗﺴﺘﻌﻤﻞ ﻟﻠﺘﺤﻮﻳﻞ ﺑﲔ‬ ‫ﻣﻮﻻﺕ ﺍﳌﺮﻛﺐ ﻭﻛﺘﻠﺘﻪ‪.‬‬

‫‪π°üØdG á©LGôe π«dO‬‬ ‫‪ 5-4‬اﻟﺼﻴﻐﺔ ا وﻟﻴﺔ واﻟﺼﻴﻐﺔ اﻟﺠﺰﻳﺌﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ‬ ‫ﳌﺮﻛـﺐ ﻣﺎ ﻫﻲ ﻧﺎﺗـﺞ ﴐﺏ ﺻﻴﻐﺘﻪ • ﺍﻟﻨﺴﺒﺔ ﺍﻟﻜﺘﻠﻴﺔ ﺍﳌﺌﻮﻳﺔ ﻟﻌﻨﴫ ﰲ ﻣﺮﻛﺐ ﻣﺎ ﺗﻌﻄﻲ ﺍﻟﻨﺴﺒﺔ ﺍﳌﺌﻮﻳﺔ ﻣﻦ ﻛﺘﻠﺔ ﺍﳌﺮﻛﺐ‬ ‫ﺍﻟﻜﻠﻴﺔ ﺍﻟﻨﺎﲡﺔ ﻋﻦ ﺍﻟﻌﻨﴫ‪.‬‬ ‫ﺍﻷﻭﻟﻴـﺔ ﰲ ﻋـﺪﺩ ﺻﺤﻴـﺢ‪ ،‬ﻭﺗﻀـﻢ‬ ‫• ﺍﻟﺮﻣﻮﺯ ﺍﻟﺴﻔﻠﻴﺔ ﰲ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﺗﻌﻄﻲ ﺃﺻﻐﺮ ﻧﺴﺒﺔ ﺻﺤﻴﺤﺔ ﳌﻮﻻﺕ ﺍﻟﻌﻨﺎﴏ‬ ‫ﺃﻋﺪﺍ ﹰﺩﺍ ﺻﺤﻴﺤﺔ ﻓﻘﻂ‪.‬‬ ‫ﰲ ﺍﳌﺮﻛﺐ‪.‬‬ ‫اﻟﻤﻔﺮدات‬ ‫• ﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ ﺗﻌﻄﻲ ﺍﻟﻌﺪﺩ ﺍﻟﻔﻌﲇ ﻟﻠﺬﺭﺍﺕ ﻣﻦ ﻛﻞ ﻋﻨﴫ ﰲ ﺟﺰﻱﺀ ﻣﻦ ﺍﳌﺎﺩﺓ‪.‬‬ ‫• ﺍﻟﱰﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﳌﺌﻮﻱ‬ ‫• ﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ ﻫﻲ ﻣﻀﺎﻋﻒ ﺻﺤﻴﺢ ﻟﻠﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‪.‬‬ ‫• ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬ ‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫• ﺍﻟﺼﻴﻐﺔ ﺍﳉﺰﻳﺌﻴﺔ‬

‫‪ 5-5‬ﺻﻴﻊ ا ﻣﻼح اﻟﻤﺎﺋﻴﺔ‬

‫اﻟﻔﻜﺮة اﻟﺮﺋﻴﺴﺔ ﺍﻷﻣـﻼﺡ ﺍﳌﺎﺋﻴـﺔ‬ ‫ﻫـﻲ ﻣﺮﻛﺒـﺎﺕ ﺃﻳﻮﻧﻴـﺔ ﺻﻠﺒـﺔ ﻓﻴﻬـﺎ‬ ‫ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ﳏﺘﺠﺰﺓ‪.‬‬ ‫اﻟﻤﻔﺮدات‬

‫ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ‬

‫اﻟﻤﻔﺎﻫﻴﻢ اﻟﺮﺋﻴﺴﺔ‬

‫• ﺗﺘﻜﻮﻥ ﺻﻴﻐﺔ ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ ﻣﻦ ﺻﻴﻐﺔ ﺍﳌﺮﻛﺐ ﺍﻷﻳﻮﲏ ﻭﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‬ ‫ﺍﳌﺮﺗﺒﻄﺔ ﺑﻮﺣﺪﺓ ﺍﻟﺼﻴﻐﺔ‪.‬‬ ‫ﻣﺘﺒﻮﻋﺎ ﺑﻤﻘﻄﻊ ﻳﺪﻝ ﻋﲆ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ‬ ‫• ﻳﺘﻜﻮﻥ ﺍﺳﻢ ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ ﻣﻦ ﺍﺳﻢ ﺍﳌﺮﻛﺐ‬ ‫ﹰ‬ ‫ﺍﳌﺎﺀ ﺍﳌﺮﺗﺒﻄﺔ ﺑﻤﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﳌﺮﻛﺐ‪.‬‬ ‫• ﻳﺘﻜﻮﻥ ﺍﳌﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ ﻋﻨﺪ ﺗﺴﺨﲔ ﺍﳌﻠﺢ ﺍﳌﺎﺋﻲ‪.‬‬

‫‪79‬‬

‫‪π°üØdG‬‬

‫‪á©LGôe‬‬

‫‪5-1‬‬ ‫‪º«gÉتdG ¿É≤JEG‬‬ ‫‪ .83‬ﻣﺎ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺪﺩﻳﺔ ﻟﻌﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ؟‬

‫‪ .84‬ﻛﻢ ﺫﺭﺓ ﻓﻲ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ؟‬ ‫‪ .85‬ﻣﺎ ﺃﻫﻤﻴﺔ ﻭﺣﺪﺓ ﺍﻟﻤﻮﻝ ﻟﻠﻜﻴﻤﻴﺎﺋﻲ؟‬

‫‪ .86‬ﻭﺿﺢ ﻛﻴﻒ ﻳﺴﺘﺨﺪﻡ ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻛﻤﻌﺎﻣﻞ‬ ‫ﺗﺤﻮﻳﻞ؟‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ .87‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻓﻲ ﻛﻞ ﻣﺎﺩﺓ‪:‬‬ ‫‪ 0.25mol .a‬ﻣﻦ ‪Ag‬‬ ‫‪-3‬‬

‫‪ 8.56 × 10 mol .b‬ﻣﻦ ‪NaCl‬‬ ‫‪ 35.3mol .c‬ﻣﻦ ‪CO2‬‬ ‫‪ 0.425mol .d‬ﻣﻦ ‪N2‬‬

‫‪ .88‬ﻣﺎ ﻋﺪﺩ ﺍﻟﺠﺰﻳﺌﺎﺕ ﻓﻲ ﻛﻞ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻵﺗﻴﺔ؟‬ ‫‪ 1.35mol .a‬ﻣﻦ ‪CS2‬‬

‫‪ 0.254mol .b‬ﻣﻦ ‪As2O3‬‬

‫‪ 1.25mol .c‬ﻣﻦ ‪H2O‬‬

‫‪ 150.0mol .d‬ﻣﻦ ‪HCI‬‬

‫‪ .89‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﻓﻲ ﻛﻞ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬

‫‪ 3.25 × 1020 atoms .a‬ﻣﻦ ﺍﻟﺮﺻﺎﺹ‬

‫‪ 4.96 × 1024 molecules .b‬ﻣﻦ ﺍﻟﺠﻠﻮﻛﻮﺯ‬

‫‪ .90‬ﺃﺟﺮ ﺍﻟﺘﺤﻮﻳﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪ 1.51 × 1015 atoms .a‬ﻣﻦ ‪ Si‬ﺇﻟﻰ ﻣﻮﻻﺕ‪.‬‬

‫‪ 4.25 × 10-2 mol .b‬ﻣﻦ ‪ H2SO4‬ﺇﻟﻰ ﺟﺰﻳﺌﺎﺕ‪.‬‬ ‫‪ 8.95 × 1025 molecules .c‬ﻣﻦ ‪ CCl4‬ﺇﻟﻰ ﻣﻮﻻﺕ‪.‬‬ ‫‪ 5.90mol .d‬ﻣﻦ ‪ Ca‬ﺇﻟﻰ ﺫﺭﺍﺕ ‪.‬‬

‫‪80‬‬

‫‪ .91‬ﺇﺫﺍ ﺍﺳﺘﻄﻌﺖ ﻋﺪﹼ ﺫﺭﺗﻴﻦ ﻓﻲ ﻛﻞ ﺛﺎﻧﻴﺔ‪ ،‬ﻓﻜﻢ ﺳﻨﺔ‬ ‫ﺗﺤﺘﺎﺝ ﻟﻌﺪ ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺬﺭﺍﺕ؟‬

‫‪5-2‬‬

‫‪º«gÉتdG ¿É≤JEG‬‬ ‫‪ .92‬ﻭﺿﺢ ﺍﻟﻔﺮﻕ ﺑﻴﻦ ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ ﻭﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬

‫‪ .93‬ﺃﻳﻬﻤﺎ ﻳﺤﻮﻱ ﺫﺭﺍﺕ ﺃﻛﺜﺮ‪ :‬ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻔﻀﺔ‪ ،‬ﺃﻡ‬ ‫ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺬﻫﺐ؟ ﱢ‬

‫‪ .94‬ﺃﻳﻬﻤﺎ ﺃﻛﺒﺮ ﻛﺘﻠﺔ‪ :‬ﻣﻮﻝ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺼﻮﺩﻳﻮﻡ ﺃﻡ ﻣﻮﻝ‬ ‫ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ؟ ﱢ‬ ‫‪ .95‬ﻭﺿﺢ ﻛﻴﻒ ﺗﺤﻮﻝ ﻋﺪﺩ ﺫﺭﺍﺕ ﻋﻨﺼﺮ ﺇﻟﻰ ﻛﺘﻠﺔ؟‬

‫‪ .96‬ﻧﺎﻗﺶ ﺍﻟﻌﻼﻗﺎﺕ ﺑﻴﻦ ﺍﻟﻤﻮﻝ‪ ،‬ﻭﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‪ ،‬ﻭﻋﺪﺩ‬ ‫ﺃﻓﻮﺟﺎﺩﺭﻭ‪.‬‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ .97‬ﺍﺣﺴﺐ ﻛﺘﻠﺔ ﻛﻞ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪ 5.22mol .a‬ﻣﻦ ‪He‬‬

‫‪ 2.22mol .b‬ﻣﻦ ‪Ti‬‬

‫‪ 0.0455mol .c‬ﻣﻦ ‪Ni‬‬

‫‪ .98‬ﺃﺟﺮ ﺍﻟﺘﺤﻮﻳﻼﺕ ﺍﻵﺗﻴﺔ‪:‬‬

‫‪ 3.5mol .a‬ﻣﻦ ‪ Li‬ﺇﻟﻰ ﺟﺮﺍﻣﺎﺕ‬

‫‪ 7.65g .b‬ﻣﻦ ‪ Co‬ﺇﻟﻰ ﻣﻮﻻﺕ‬

‫‪ 5.65g .c‬ﻣﻦ ‪ Kr‬ﺇﻟﻰ ﻣﻮﻻﺕ‬

‫‪ .99‬ﻣﺎ ﻛﺘﻠﺔ ﺍﻟﻌﻨﺼﺮ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ ﻓﻲ ﻛﻞ ﻣﻦ ‪:‬‬ ‫‪ 1.33 × 1022 mol .a‬ﻣﻦ ‪Sb‬‬ ‫‪ 4.75 × 1014 mol .b‬ﻣﻦ ‪Pt‬‬ ‫‪ 1.22 × 1023 mol .c‬ﻣﻦ ‪Ag‬‬ ‫‪ 9.85 × 1024 mol .b‬ﻣﻦ ‪Gr‬‬

‫‪π°üØdG á©LGôe‬‬ ‫ﻭﺍﻷﻛﺴﺠﻴﻦ ﻓﻲ ﺻﻴﻐﺔ ﻓﺴﻔﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ‪Na3PO4‬؟‬

‫‪ .100‬ﺃﻛﻤﻞ ﺍﻟﺠﺪﻭﻝ ‪:5-2‬‬ ‫ﺟﺪﻭﻝ)‪ (5-2‬ﺑﻴﺎﻧﺎﺕ ﺍﻟﻜﺘﻠﺔ‪ ،‬ﻭﺍﻟﻤﻮﻝ‪ ،‬ﻭﺍﻟﺠﺴﻴﻤﺎﺕ‬ ‫ﺍﻟﻜﺘﻠﺔ‬

‫ﺍﻟﻤﻮﻻﺕ‬

‫‪ 3.65 mol‬ﻣﻦ‬

‫ﺍﻟﺠﺴﻴﻤﺎﺕ‬ ‫‪Mg‬‬

‫‪ 29.54g‬ﻣﻦ ‪Cr‬‬ ‫‪ 3.54 × 1025 atoms‬ﻣﻦ ‪P‬‬

‫‪ 0.568 mol‬ﻣﻦ ‪As‬‬

‫‪ .101‬ﺣﻮﻝ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﻓﻴﻤﺎ ﻳﻠﻲ ﺇﻟﻰ ﺟﺮﺍﻣﺎﺕ‪:‬‬ ‫‪ 8.65 × 1025 atom .a‬ﻣﻦ ‪H‬‬

‫‪ 1.25 × 1022 atom .b‬ﻣﻦ ‪O‬‬

‫‪ .102‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﻓﻲ ﻛﻞ ﻋﻨﺼﺮ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪ 0.034g .a‬ﻣﻦ ‪Zn‬‬

‫‪ 0.124g .b‬ﻣﻦ ‪Mg‬‬

‫‪ .103‬ﺭﺗﺐ ﺗﺼﺎﻋﺪ ﹰﹼﻳﺎ ﺑﺤﺴﺐ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‪:‬‬ ‫‪ 3.00 × 1024 atoms‬ﻣﻦ ‪ Ne 4.25 mol‬ﻣﻦ ‪Ar‬‬ ‫‪ 2.69 × 1024 atoms‬ﻣﻦ‪ 65.96 g، Xe‬ﻣﻦ ‪Kr‬‬

‫‪ .104‬ﺃﻳﻬﻤﺎ ﻳﺤﻮﻱ ﺫﺭﺍﺕ ﺃﻛﺜﺮ‪ g 10.0 :‬ﻣﻦ ‪ ،C‬ﺃﻡ‬ ‫‪ 10.0g‬ﻣﻦ ‪Ca‬؟ ﻭﻛﻢ ﺫﺭﺓ ﻳﺤﻮﻱ ﻛﻞ ﻋﻨﺼﺮ ﻣﻨﻬﻤﺎ؟‬ ‫‪ .105‬ﺃﻳﻬﻤﺎ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺃﻛﺒﺮ ﻋﺪﺩ ﻣﻦ ﺍﻟﺬﺭﺍﺕ‬ ‫‪ 10.0mol‬ﻣﻦ ‪ C‬ﺃﻡ ‪ 10.0mol‬ﻣﻦ ‪Ca‬؟‬

‫‪ .106‬ﺧﻠﻴﻂ ﻣﻜﻮﻥ ﻣﻦ ‪ 0.250mol‬ﻣﻦ ‪ Fe‬ﻭ‬ ‫‪ 1.20mol‬ﻣﻦ ‪ C‬ﻣﺎ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﺍﻟﻜﻠﻲ ﻓﻲ ﻫﺬﺍ‬ ‫ﺍﻟﺨﻠﻴﻂ؟‬

‫‪5-3‬‬

‫‪ .107‬ﻣﺎ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﻟﺘﻲ ﻳﻤﻜﻨﻚ ﺍﻟﺤﺼﻮﻝ ﻋﻠﻴﻬﺎ ﻣﻦ‬ ‫ﺻﻴﻐﺔ ﻛﺮﻭﻣﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ‪K2CrO4‬؟‬

‫‪ .108‬ﻣـﺎ ﹼﻋـﺪﺩ ﻣـﻮﻻﺕ ﻛﻞ ﻣـﻦ ﺍﻟﺼﻮﺩﻳـﻮﻡ‪ ،‬ﻭﺍﻟﻔﺴـﻔﻮﺭ‪،‬‬

‫‪ .109‬ﻟﻤﺎﺫﺍ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻝ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻛﻤﻌﺎﻣﻞ‬ ‫ﺗﺤﻮﻳﻞ؟‬ ‫‪ .110‬ﺍﻛﺘﺐ ﺛﻼﺛﺔ ﻣﻌﺎﻣﻼﺕ ﺗﺤﻮﻳﻞ ﺗﺴﺘﻌﻤﻞ ﻓﻲ‬ ‫ﺍﻟﺘﺤﻮﻳﻼﺕ ﺍﻟﻤﻮﻟﻴﺔ‪.‬‬

‫‪ .111‬ﺃﻱ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺍﻟﻌﺪﺩ ﺍﻷﻛﺒﺮ‬ ‫ﻣﻦ ﻣﻮﻻﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻟﻜﻞ ﻣﻮﻝ ﻣﻦ ﺍﻟﻤﺮﻛﺐ‪:‬‬ ‫ﺣﻤﺾ ﺍﻷﺳﻜﻮﺭﺑﻴﻚ ‪ ،C6H8O6‬ﺃﻡ ﺍﻟﺠﻠﺴﺮﻳﻦ‬ ‫ﻓﺴﺮﺇﺟﺎﺑﺘﻚ‪.‬‬ ‫‪ ،C3H8O3‬ﺃﻡ ﺍﻟﻔﻨﺎﻟﻴﻦ ‪C8H8O3‬؟ ﱢ‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬

‫‪ .112‬ﻛﻢ ﻣﻮﻝ ﻣﻦ ﺍﻻﻛﺴﺠﻴﻦ ﻓﻲ ﻛﻞ ﻣﺮﻛﺐ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪ 2.5mol .a‬ﻣﻦ ‪KMnO4‬‬

‫‪ 45.9mol .b‬ﻣﻦ ‪CO2‬‬

‫‪ 1.25 ×10 mol .c‬ﻣﻦ ‪CuSO4 .5H2O‬‬ ‫‪-2‬‬

‫‪ .113‬ﻛﻢ ﺟﺰﻱﺀ ‪ ،CCl4‬ﻭﻛﻢ ﺫﺭﺓ ‪ ، C‬ﻭﻛﻢ ﺫﺭﺓ ‪،Cl‬‬ ‫ﻳﻮﺟﺪ ﻓﻲ ‪ 3mol‬ﻣﻦ ‪CCl4‬؟ ﻭﻣﺎ ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ‬ ‫ﺍﻟﻜﻠﻲ؟‬ ‫‪ .114‬ﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻜﻞ ﻣﺮﻛﺐ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪ .a‬ﺣﻤﺾ ﺍﻟﻨﻴﺘﺮﻳﻚ ‪HNO3‬‬

‫‪ .b‬ﺃﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ ‪ZnO‬‬

‫‪ .115‬ﻣﺎ ﻋﺪﺩ ﻣﻮﻻﺕ ‪ CH3OH‬ﻓﻲ ‪ 100g‬ﻣﻦ‬ ‫‪CH3OH‬؟‬

‫‪ .116‬ﻣﺎ ﻛﺘﻠﺔ ‪ 1.25 ×102 mol‬ﻣﻦ ‪Ca(OH)2‬؟‬

‫‪ .117‬ﺍﻟﺤﻔﺮ ﻋﻠﻰ ﺍﻟﺰﺟﺎﺝ ﻳﺴﺘﻌﻤﻞ ﺣﻤﺾ ﺍﻟﻬﻴﺪﺭﻭﻓﻠﻮﺭﻳﻚ‬ ‫‪ HF‬ﻟﻠﺤﻔﺮ ﻋﻠﻰ ﺍﻟﺰﺟﺎﺝ‪.‬‬ ‫ﻣﺎ ﻛﺘﻠﺔ ‪ 4.95 ×1025 Particles‬ﻣﻦ ‪HF‬؟‬

‫‪ .118‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺍﻟﺠﺰﻳﺌﺎﺕ ﻓﻲ ‪ 47.0g‬ﻣﻦ‬ ‫‪.C2H5OH‬‬

‫‪81‬‬

‫‪π°üØdG á©LGôe‬‬ ‫‪ .119‬ﻛﻢ ﻣﻮ ﹰ‬ ‫ﻻ ﻣﻦ ﺍﻟﺤﺪﻳﺪ ﻳﻤﻜﻦ ﺍﺳﺘﺨﺮﺍﺟﻪ ﻣﻦ‬ ‫‪ 100.0kg‬ﻣﻦ ‪Fe3O4‬؟‬

‫‪ .120‬ﺍﻟﻄﺒـﺦ‪ :‬ﻳﺤﺘﻮﻱ ﺍﻟﺨﻞ ﺍﻟﻤﺴـﺘﻌﻤﻞ ﻓـﻲ ﺍﻟﻄﺒﺦ ﻋﻠﻰ‬ ‫‪ 5%‬ﻣـﻦ ﺣﻤـﺾ ﺍﻟﺨﻠﻴـﻚ ‪ .CH3COOH‬ﻓﻜـﻢ‬ ‫ﺟﺰﻳﺌ ﹰﺎ ﻣﻦ ﺍﻟﺤﻤﺾ ﻳﻮﺟﺪ ﻓﻲ ‪ 25.0g‬ﻣﻦ ﺍﻟﺨﻞ؟‬ ‫‪ .121‬ﺍﺣﺴﺐ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﻴﻦ ﻓﻲ ‪ 25.0g‬ﻣﻦ‬ ‫ﺛﺎﻧﻲ ﺃﻛﺴﻴﺪ ﺍﻟﻜﺮﺑﻮﻥ‪.‬‬

‫‪5-4‬‬ ‫‪ .122‬ﻣﺎ ﺍﻟﻤﻘﺼﻮﺩ ﺑﺎﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ؟‬

‫‪ .123‬ﻣـﺎ ﺍﻟﻤﻌﻠﻮﻣـﺎﺕ ﺍﻟﺘـﻲ ﻳﺠـﺐ ﺃﻥ ﻳﺤﺼـﻞ ﻋﻠﻴﻬـﺎ‬ ‫ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺘﺤﺪﻳﺪ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻤﺮﻛﺐ ﻣﺠﻬﻮﻝ؟‬ ‫‪ .124‬ﻣﺎ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﻟﺘﻲ ﻳﺠﺐ ﺗﻮﻓﺮﻫﺎ ﻟﻠﻜﻴﻤﻴﺎﺋﻲ‬ ‫ﻟﻴﺤﺪﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﻟﻤﺮﻛﺐ؟‬

‫‪ .125‬ﻣﺎ ﺍﻟﻔﺮﻕ ﺑﻴﻦ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻭﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ؟‬ ‫ﺃﻋﻂ ﺃﻣﺜﻠﺔ ﻋﻠﻰ ﺫﻟﻚ‪.‬‬ ‫‪ .126‬ﻣﺘﻰ ﺗﻜﻮﻥ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻫﻲ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‬ ‫ﻧﻔﺴﻬﺎ؟‬

‫ﺍﻟﻨﻘﻴﺔ ﻟﻤﺮﻛﺐ ﻣﻌﻴﻦ ﻟﻬﺎ ﺍﻟﺘﺮﻛﻴﺐ‬ ‫‪ .127‬ﻫﻞ ﻛﻞ ﺍﻟﻌﻴﻨﺎﺕ ﹼ‬ ‫ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻧﻔﺴﻪ ؟ ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ . 128‬ﺍﻟﺤﺪﻳﺪ ﻳﻮﺟﺪ ﺛﻼﺛﺔ ﻣﺮﻛﺒﺎﺕ ﻃﺒﻴﻌﻴﺔ ﻟﻠﺤﺪﻳﺪ‬ ‫ﻫﻲ‪ :‬ﺍﻟﺒﺎﻳﺮﻳﺖ ‪ ،FeS2‬ﻭﺍﻟﻬﻴﻤﺎﺗﻴﺖ ‪،Fe2O3‬‬ ‫ﻭﺍﻟﺴﻴﺪﻳﺮﺍﻳﺖ ‪ .FeCO3‬ﺃﻳﻬﺎ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺃﻋﻠﻰ‬ ‫ﻧﺴﺒﺔ ﻣﻦ ﺍﻟﺤﺪﻳﺪ؟‬

‫‪ .129‬ﺍﺣﺴﺐ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻜﻞ ﻣﺮﻛﺐ ﻣﻤﺎ‬ ‫ﻳﻠﻲ‪:‬‬ ‫‪ .a‬ﺍﻟﺴﻜﺮﻭﺯ ‪C12H22O11‬‬

‫‪ .b‬ﺍﻟﻤﺎﺟﻨﺘﻴﺖ ‪Fe3O4‬‬

‫‪82‬‬

‫‪ .130‬ﺣﺪﺩ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻜﻞ ﻣﺮﻛﺐ ﻣﻤﺎ ﻳﻠﻲ‪:‬‬ ‫‪130‬‬ ‫‪ .a‬ﺍﻹﻳﺜﻠﻴﻦ ‪C2H4‬‬

‫‪ .b‬ﺣﻤﺾ ﺍﻷﺳﻜﻮﺭﺑﻴﻚ ‪C6H8O6‬‬

‫‪ .c‬ﺍﻟﻨﻔﺜﺎﻟﻴﻦ ‪C10H8‬‬

‫‪ .131‬ﻣـﺎ ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴـﺔ ﻟﻠﻤﺮﻛـﺐ ﺍﻟﺬﻱ ﻳﺤﺘـﻮﻱ ﻋﻠﻰ‬ ‫‪ 10.52g‬ﻣﻦ ‪ ،Ni‬ﻭ‪ 4.38g‬ﻣﻦ ‪ ،C‬ﻭ‪5.10g‬ﻣﻦ‪N‬؟‬

‫‪5-5‬‬ ‫‪ .132‬ﻣﺎ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ؟ ﻭﺿﺢ ﺇﺟﺎﺑﺘﻚ ﺑﻤﺜﺎﻝ؟‬ ‫‪ .133‬ﻭﺿﺢ ﻛﻴﻒ ﺗﺴﻤﻰ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ؟‬

‫‪ .134‬ﺍﻟﻤﺠ ﱢﻔﻔﺎﺕ‪ :‬ﻟﻤﺎﺫﺍ ﺗﻮﺿﻊ ﺍﻟﻤﺠﻔﻔﺎﺕ ﻣﻊ ﺍﻷﺟﻬﺰﺓ‬ ‫ﺍﻹﻟﻜﺘﺮﻭﻧﻴﺔ ﻓﻲ ﺻﻨﺎﺩﻳﻖ ﺣﻔﻈﻬﺎ؟‬ ‫‪ .135‬ﺍﻛﺘﺐ ﺻﻴﻐﺔ ﻛﻞ ﻣﻠﺢ ﻣﻦ ﺍﻷﻣﻼﺡ ﺍﻟﻤﺎﺋﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪ .a‬ﻛﻠﻮﺭﻳﺪ ﺍﻟﻨﻴﻜﻞ )‪ (II‬ﺳﺪﺍﺳﻲ ﺍﻟﻤﺎﺀ‪.‬‬

‫‪ .b‬ﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ ﺧﻤﺎﺳﻴﺔ ﺍﻟﻤﺎﺀ‪.‬‬

‫‪πFÉ°ùªdG πM ¿É≤JEG‬‬ ‫‪ .136‬ﻳﺤﺘﻮﻱ ﺍﻟﺠﺪﻭﻝ ‪ 5-3‬ﻋﻠﻰ ﺑﻴﺎﻧﺎﺕ ﺗﺠﺮﻳﺒﻴﺔ ﻟﺘﺤﺪﻳﺪ‬ ‫ﺻﻴﻐـﺔ ﻛﻠﻮﺭﻳـﺪ ﺍﻟﺒﺎﺭﻳـﻮﻡ ﺍﻟﻤﺎﺋـﻲ‪ .‬ﺃﻛﻤـﻞ ﺍﻟﺠﺪﻭﻝ‬ ‫ﻭﺣﺪﺩ ﺻﻴﻐﺘﺔ ﻭﺍﺳﻤﻪ‪.‬‬ ‫ﺟﺪﻭﻝ ‪ .5-3‬ﺑﻴﺎﻧﺎﺕ‬

‫‪BaCl2.xH2O‬‬

‫ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ ‪ +‬ﺍﻟﺠﻔﻨﺔ‬

‫‪31.35g‬‬

‫ﺍﻟﻜﺘﻠﺔ ﺑﻌﺪ ﺍﻟﺘﺴﺨﻴﻦ ﻣﺪﺓ ‪ 5‬ﻗﺎﺋﻖ‬

‫‪29.87g‬‬

‫ﻛﺘﻠﺔ ﺍﻟﺠﻔﻨﺔ ﺍﻟﻔﺎﺭﻏﺔ‬ ‫ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻤﺎﺋﻲ‬

‫ﻛﺘﻠﺔ ﺍﻟﻤﻠﺢ ﺍﻟﻼﻣﺎﺋﻲ‬

‫‪21.30g‬‬

‫ﻣﻠﺤـﺎ ﻣﺎﺋ ﹰﻴـﺎ ﻳﺤﺘـﻮﻱ ﻋﻠﻰ‬ ‫ﺗﻜـﻮﻥ ﻧﺘـﺮﺍﺕ ﺍﻟﻜـﺮﻭﻡ )‪ (III‬ﹰ‬ ‫‪ .137‬ﹼ‬ ‫‪ 40.50%‬ﻣﻦ ﻛﺘﻠﺘﻪ ﻣﺎﺀ‪ .‬ﻓﻤﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺐ؟‬ ‫‪ .138‬ﺣـﺪﹼ ﺩ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴـﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﹺـ ‪، MgCO3 .5H2O‬‬ ‫ﻭﻣﺜﹼﻞ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺑﺮﺳﻢ ﺑﻴﺎﻧﻲ ﺩﺍﺋﺮﻱ‪.‬‬

‫‪π°üØdG á©LGôe‬‬ ‫‪ .139‬ﺳﺨﻨﺖ ﻋﻴﻨﺔ ﻛﺘﻠﺘﻬﺎ ‪ 1.628g‬ﻣﻦ ﻣﻠﺢ ﻳﻮﺩﻳﺪ‬ ‫ﺍﻟﻤﺎﻏﻨﺴﻴﻮﻡ ﺍﻟﻤﺎﺋﻲ ﺣﺘﻰ ﻃﺮﺩ ﻣﻨﻬﺎ ﺍﻟﻤﺎﺀ‪،‬‬ ‫ﻓﺄﺻﺒﺤﺖ ﺍﻟﻜﺘﻠﺔ ‪ .0721g‬ﻓﻤﺎ ﺻﻴﻐﺔ ﺍﻟﻤﻠﺢ‬ ‫ﺍﻟﻤﺎﺋﻲ؟‬

‫‪áeÉY á©LGôe‬‬ ‫‪.140‬ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﺘﻠﺔ ﺫﺭﺓ ﻭﺍﺣﺪﺓ ﻣﻦ ﻋﻨﺼﺮ ﻣﺠﻬﻮﻝ‬ ‫‪-23‬‬

‫‪ ،6.66 ×10 g‬ﻓﻤﺎ ﺍﻟﻌﻨﺼﺮ؟‬

‫‪.141‬ﻳﺤﺘﻮﻱ ﻣﺮﻛﺐ ﻋﻠﻰ ‪ 6.0g‬ﻛﺮﺑﻮﻥ‪ ،‬ﻭ ‪1.0g‬‬ ‫ﻫﻴﺪﺭﻭﺟﻴﻦ ‪.‬ﻭﻛﺘﻠﺘﻪ ﺍﻟﻤﻮﻟﻴﺔ ‪ .42.0g/mol‬ﻓﻤﺎ‬ ‫ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﺴﺒﻲ ﺍﻟﻤﺌﻮﻱ ﻟﻠﻤﺮﻛﺐ؟ﻭﻣﺎ ﺻﻴﻐﺘﻪ‬ ‫ﺍﻷﻭﻟﻴﺔ؟ﻭﻣﺎ ﺻﻴﻐﺘﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ؟‬

‫‪.‬ﺃﻱ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺃﻋﻠﻰ ﻧﺴﺒﺔ ﻣﺌﻮﻳﺔ‬ ‫‪ 142‬ﹼ‬ ‫ﺑﺎﻟﻜﺘﻠﺔ ﻣﻦ ﺍﻷﻛﺴﺠﻴﻦ‪:‬‬ ‫‪TiO2 ، Al2O3 ،Fe2O3‬‬

‫‪óbÉædG ô«µØàdG‬‬ ‫ﻃﺒﻖ ﺍﻟﻤﻔﺎﻫﻴﻢ ﻟﺪ￯ ﺷﺮﻛﺔ ﺗﻌﺪﻳﻦ ﻣﺼﺪﺭﺍﻥ‬ ‫‪ .143‬ﹼ‬ ‫ﻣﺤﺘﻤﻼﻥ ﻟﻠﻨﺤﺎﺱ‪ :‬ﺟﺎﻟﻜﻮﺑﺎﻳﺮﻳﺖ )‪،(CuFeS2‬‬

‫ﻭﺟﺎﻟﻜﻮﺳـﻴﺖ )‪ .(Cu2S‬ﻓـﺈﺫﺍ ﻛﺎﻧـﺖ ﻇـﺮﻭﻑ‬ ‫ﺍﺳـﺘﺨﺮﺍﺝ ﺍﻟﻨﺤـﺎﺱ ﻣـﻦ ﺍﻟﺨﺎﻣﻴﻦ ﻣﺘﺸـﺎﺑﻬﺔ ﺗﻤﺎ ﹰﻣﺎ‪،‬‬ ‫ﻓﺄﻳﻬﻤﺎ ﺳﻴﻨﺘﺞ ﻛﻤﻴﺔ ﺃﻛﺒﺮ ﻣﻦ ﺍﻟﻨﺤﺎﺱ؟ ﻓﺴﺮ ﺇﺟﺎﺑﺘﻚ‪.‬‬

‫‪ .144‬ﺣ ﹼﻠﻞ ﻭﺍﺳﺘﻨﺘﺞ ﺟﻤﻊ ﺍﻟﻄﻼﺏ ﻓﻲ ﺭﺣﻠﺔ ﻣﻴﺪﺍﻧﻴﺔ‬ ‫ﻋﻴﻨﺎﺕ ﺻﺨﺮﻳﺔ‪ .‬ﻭﻗﺪ ﺃﻇﻬﺮ ﺍﻟﺘﺤﻠﻴﻞ ﺃﻥ ﻋﻴﻨﺘﻴﻦ‬ ‫ﻣﻨﻬﺎ ﺗﺤﺘﻮﻳﺎﻥ ﻋﻠﻰ ﺭﺻﺎﺹ ﻭﻛﺒﺮﻳﺖ‪.‬ﻭﺍﻟﺠﺪﻭﻝ‬ ‫‪ 5-4‬ﻳﺒﻴﻦ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻟﻜﺘﻠﺔ ﻛﻞ ﻣﻦ ﺍﻟﺮﺻﺎﺹ‬ ‫ﻭﺍﻟﻜﺒﺮﻳﺖ ﻓﻲ ﻛﻞ ﻋﻴﻨﺔ ‪ .‬ﻓﻤﺎ ﺍﻟﻨﺘﺎﺋﺞ ﺍﻟﺘﻲ ﻳﻤﻜﻦ ﺃﻥ‬ ‫ﻳﺴﺘﺨﻠﺼﻬﺎ ﺍﻟﻄﻼﺏ ﻋﻦ ﺍﻟﻌﻴﻨﺘﻴﻦ؟‬ ‫ﺍﻟﺠﺪﻭﻝ ‪ 5 -4‬ﻳﺤﺘﻮﻱ ﺍﻟﺮﺻﺎﺹ ﻭﺍﻟﻜﺒﺮﻳﺖ‬ ‫‪ %‬ﻛﺒﺮﻳﺖ‬ ‫‪ %‬ﺭﺻﺎﺹ‬ ‫ﻋﻴﻨﺔ ﺍﻟﺼﺨﺮ‬ ‫‪13.4%‬‬ ‫‪86.6%‬‬ ‫‪1‬‬ ‫‪23.6%‬‬ ‫‪76.4%‬‬ ‫‪2‬‬

‫ﺻﻤﻢ ﺗﺠﺮﺑﺔ ﻳﻤﻜﻦ ﺍﺳﺘﻌﻤﺎﻟﻬﺎ‬ ‫‪ .145‬ﹼ‬ ‫ﺻﻤﻢ ﺗﺠﺮﺑﺔ‪ :‬ﹼ‬ ‫ﻟﺘﺤﺪﻳﺪ ﻛﻤﻴﺔ ﺍﻟﻤﺎﺀ ﻓﻲ ﺍﻟﺸﺐ ﺍﻟﺒﻮﺗﺎﺳﻲ‬ ‫‪KAl (SO4)2 .X H2O‬‬

‫ﺗﻘﻮﻳﻢ ﺇﺿاﻓﻲ‬ ‫‪‘ áHÉàµdG‬‬

‫اﻟﻜﻴﻤﻴﺎء‬

‫‪ .146‬ﺍﻟﻐـﺎﺯ ﺍﻟﻄﺒﻴﻌـﻲ‪ :‬ﻫﻴـﺪﺭﺍﺕ ﺍﻟﻐـﺎﺯ ﺍﻟﻄﺒﻴﻌـﻲ ﻫــﻲ‬ ‫ﻣـﺮﻛـﺒـﺎﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ ﻣﺘﺒﻠﻮﺭﺓ )‪.(Clathrate hydrate‬‬ ‫ﺍﺑﺤﺚ ﻓﻲ ﻫـﺬﻩ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻭﺍﻋﺪﹼ ﻧﺸـﺮﺓ ﺗﻌﻠﻴﻤﻴﺔ ﻋﻨﻬﺎ‬ ‫ﻟﻠﻤﺴـﺘﻬﻠﻜﻴﻦ‪ .‬ﻳﺠـﺐ ﺃﻥ ﺗﻨﺎﻗـﺶ ﻫـﺬﻩ ﺍﻟﻨﺸـﺮﺓ‬ ‫ﺗﺮﻛﻴـﺐ ﻫـﺬﻩ ﺍﻟﻤﺮﻛﺒـﺎﺕ‪ ،‬ﻭﻣـﻜﺎﻥ ﻭﺟﻮﺩﻫـﺎ‪،‬‬ ‫ﻭﺃﻫﻤﻴﺘﻬـﺎ ﻟﻠﻤﺴـﺘﻬﻠﻜﻴﻦ‪ ،‬ﻭﺍﻵﺛـﺎﺭ ﺍﻟﺒﻴﺌﻴـﺔ‬ ‫ﻻﺳﺘﺨﺪﺍﻣﻬﺎ‪.‬‬

‫‪ .147‬ﻳﺸـﺘﻤﻞ ﺍﻟﺠﺪﻭﻝ )‪ (5-5‬ﻋﻠﻰ ﺑﻴﺎﻧﺎﺕ ﻋﻦ ﻭﻗﻮﺩ‬ ‫ﻣﻜﻮﻙ ﻓﻀﺎﺀ؛ ﺇﺫ ﻻﺑﺪ ﻣﻦ ﺗﻮﺍﻓﺮ ‪3٫164٫445L‬‬ ‫ﻣﻦ ﺍﻟﻮﻗـﻮﺩ ﺍﻟﺴـﺎﺋﻞ ﺍﻷﻛﺴـﺠﻴﻦ‪ ،‬ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪،‬‬ ‫ﺃﺣـﺎﺩﻱ ﻣﻴﺜﻴـﻞ ﺍﻟﻬﻴﺪﺭﺍﺯﻳـﻦ )ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ =‬ ‫‪ ،(46.07g/mol‬ﺭﺍﺑـﻊ ﺃﻛﺴـﻴﺪ ﺍﻟﻨﻴﺘﺮﻭﺟﻴـﻦ‬ ‫)ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ= ‪ ،(92.00g/mol‬ﻓـﻲ‬ ‫ﺧﺰﺍﻧـﺎﺕ ﺍﻟﻮﻗـﻮﺩ ﻟﺨﻄﺔ ﺍﻹﻗـﻼﻉ‪ .‬ﻛﺘﻠﺘﻬـﺎ ﺍﻟﻜﻠﻴﺔ‬ ‫)‪ .(727٫233 Kg‬ﺃﻛﻤـﻞ ﺍﻟﺠـﺪﻭﻝ ﺑﺤﺴـﺎﺏ‬ ‫ﻋـﺪﺩ ﺍﻟﻤـﻮﻻﺕ‪ ،‬ﺍﻟﻜﺘﻠـﺔ ﺑﺎﻟﻜﻴﻠﻮﺟـﺮﺍﻡ‪ ،‬ﻋـﺪﺩ‬ ‫ﺍﻟﺠﺰﻳﺌﺎﺕ‪.‬‬ ‫ﺟﺪﻭﻝ)‪ (5-5‬ﺑﻴﺎﻧﺎﺕ ﻭﻗﻮﺩ ﻣﻜﻮﻙ ﻓﻀﺎﺋﻲ‬

‫ﺍﻟﻤﺎﺩﺓ‬

‫ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺍﻟﻜﺘﻠﺔ )‪(g‬‬

‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬

‫‪H2‬‬

‫ﺍﻷﻛﺴﺠﻴﻦ‬

‫‪O2‬‬

‫ﺃﺣﺎﺩﻱ ﻣﻴﺜﻴﻞ‬ ‫ﺍﻟﻬﻴﺪﺭﺍﺯﻳﻦ‬

‫‪CH3 NH NH2‬‬

‫ﺭﺍﺑﻊ ﺃﻛﺴﻴﺪ‬ ‫ﺍﻟﻨﻴﺘﺮﻭﺟﻴﻦ‬

‫‪N2O4‬‬

‫‪Moles‬‬ ‫‪5.14 × 107‬‬

‫‪Molecules‬‬

‫‪1.16 × 1031‬‬

‫‪4909‬‬ ‫‪8.64 × 104‬‬

‫‪83‬‬

‫‪≤e QÉÑàNG‬‬ ‫‪Oó©àe øe QÉ«àN’G á∏Ä°SCG‬‬

‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﺮﺳﻢ ﺍﻟﺒﻴﺎﻧﻲ ﺃﺩﻧﺎﻩ ﻟﻺﺟﺎﺑﺔ ﻋﻦ ﺍﻷﺳﺌﻠﺔ ‪.4 -1‬‬ ‫ﺍﻟﻨﺴﺐ ﺍﳌﺌﻮﻳﺔ ﳌﻜﻮﻧﺎﺕ ﻣﺮﻛﺐ‬

‫‪60‬‬

‫‪54.5‬‬

‫‪54.5‬‬

‫‪53.3‬‬

‫‪52.2‬‬

‫‪40‬‬

‫‪36.4‬‬

‫‪36.4‬‬

‫‪34.8‬‬

‫‪40.0‬‬

‫ﺍﻟﻨﺴﺒﺔ ﺍﳌﺌﻮﻳﺔ ﺑﺎﻟﻜﺘﻠﺔ‬

‫‪٪C‬‬ ‫‪٪H‬‬ ‫‪٪O‬‬

‫‪50‬‬

‫‪30‬‬ ‫‪20‬‬

‫‪9.1‬‬

‫‪9.1‬‬

‫‪6.7‬‬

‫ﲪﺾ ﺍﻟﺒﻴﻮﺗﺎﻧﻮﻳﻚ ﺍﺳﺘﻴﺎﻟﺪﻫﻴﺪ‬

‫ﺍﺳﻢ ﺍﳌﺮﻛﺐ‬

‫‪13.0‬‬

‫ﻓﻮﺭﻣﺎﻟﺪﻫﻴﺪ‬

‫‪10‬‬

‫ﺍﻳﺜﺎﻧﻮﻝ‬

‫‪0‬‬

‫‪.1‬ﻳﺘﺸﺎﺑﻪ ﺍﻷﺳﻴﺘﺎﻟﺪﻫﻴﺪ ﻭﺣﻤﺾ ﺍﻟﺒﻴﻮﺗﺎﻧﻮﻳﻚ ﻓﻲ‪:‬‬ ‫‪ .a‬ﺍﻟﺼﻴﻐﺔ ﺍﻟﺠﺰﻳﺌﻴﺔ‬

‫‪C11-19C-828378‬‬

‫‪ .b‬ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ‬

‫‪ .c‬ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ‬

‫‪ .2‬ﺇﺫﺍ ﻛﺎﻧـﺖ ﺍﻟﻜﺘﻠـﺔ ﺍﻟﻤﻮﻟﻴـﺔ ﻟﺤﻤـﺾ ﺍﻟﺒﻴﻮﺗﺎﻧﻮﻳـﻚ‬ ‫‪ ،88.1g/ mol‬ﻓﻤﺎ ﺻﻴﻐﺘﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ؟‬ ‫‪C2H4O .b‬‬

‫‪C5H12O .c‬‬

‫‪C4H8O2 .d‬‬

‫‪ .3‬ﻣﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻺﻳﺜﺎﻧﻮﻝ؟‬ ‫‪C4HO3 .a‬‬

‫‪C2H6O2 .b‬‬

‫‪C2H6O .c‬‬

‫‪C4H13O2 .d‬‬

‫‪ .4‬ﺍﻟﺼﻴﻐـﺔ ﺍﻷﻭﻟﻴـﺔ ﻟﻠﻔﻮﺭﻣﺎﻟﺪﻫﻴـﺪ ﻫﻲ ﺻﻴﻐﺘـﻪ ﺍﻟﺠﺰﻳﺌﻴﺔ‬ ‫ﻧﻔﺴـﻬﺎ ‪ .‬ﻓﻜـﻢ ﺟﺮﺍﻣـ ﹰﺎ ﻳﻮﺟـﺪ ﻓـﻲ ‪ 2.00mol‬ﻣـﻦ‬ ‫ﺍﻟﻔﻮﺭﻣﺎﻟﺪﻫﻴﺪ؟‬

‫‪84‬‬

‫‪60.06g .b‬‬

‫‪200.0g .d‬‬

‫‪ .5‬ﺃﻱ ﻣﻤﺎ ﻳﻠﻲ ﻻ ﹸﻳ ﹼﻌﺪ ﻭﺻ ﹰﻔﺎ ﻟﻠﻤﻮﻝ؟‬

‫‪ .a‬ﻭﺣﺪﺓ ﺗﺴﺘﻌﻤﻞ ﻟﻠﻌﺪ ﺍﻟﻤﺒﺎﺷﺮ ﻟﻠﺠﺴﻴﻤﺎﺕ‬

‫‪ .b‬ﻋﺪﺩ ﺃﻓﻮﺟﺎﺩﺭﻭ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﻣﺮﻛﺐ‬

‫‪ .c‬ﻋﺪﺩ ﺍﻟﺬﺭﺍﺕ ﻓﻲ ‪ 12g‬ﺑﺎﻟﻀﺒﻂ ﻣﻦ ‪ C- 12‬ﺍﻟﻨﻘﻲ‬ ‫‪ .d‬ﻭﺣﺪﺓ ﺍﻟﻨﻈﺎﻡ ﺍﻟﻌﺎﻟﻤﻲ ﻟﻜﻤﻴﺔ ﺍﻟﻤﺎﺩﺓ‬

‫ﺍﺳﺘﻌﻤﻞ ﺍﻟﺮﺳﻢ ﺍﻟﺒﻴﺎﻧﻲ ﺃﺩﻧﺎﻩ ﻟﻺﺟﺎﺑﺔ ﻋﻦ ﺍﻟﺴﺆﺍﻝ ﺍﻟﺴﺎﺩﺱ‬

‫‪ .6‬ﻣﺎ ﺍﻟﺼﻴﻐﺔ ﺍﻷﻭﻟﻴﺔ ﻟﻬﺬﺍ ﺍﻟﻤﺮﻛﺐ؟‬ ‫‪C6H2N6O3 .a‬‬

‫‪C4HN5O10 .b‬‬

‫‪CH3NO2 .c‬‬

‫‪CH5NO3 .d‬‬

‫‪ .d‬ﺍﻟﺨﻮﺍﺹ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬

‫‪C3H4O3 .a‬‬

‫‪30.00g .a‬‬

‫‪182.0g .c‬‬

‫‪N‬‬ ‫‪22.95٪‬‬ ‫‪C‬‬ ‫‪19.68٪‬‬

‫‪O‬‬ ‫‪52.42٪‬‬

‫‪H‬‬ ‫‪4.96٪‬‬

‫‪ .7‬ﻣﺎ ﻧﻮﻉ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻤﻮﺿﺢ ﺃﺩﻧﺎﻩ؟‬

‫‪2HI + (NH4)2S → H2S + 2NH4I‬‬ ‫‪ .a‬ﺗﻜﻮﻳﻦ‬

‫‪ .c‬ﺇﺣﻼﻝ ﺑﺴﻴﻂ‬

‫‪C11-18C-828378-08‬‬

‫‪ .d‬ﺇﺣﻼﻝ ﻣﺰﺩﻭﺝ‬ ‫‪ .b‬ﺗﻔﻜﻚ‬ ‫‪ .8‬ﻛﻢ ﺫﺭﺓ ﻳﻮﺟﺪ ﻓﻲ ‪ 0.625mol‬ﻣﻦ ‪Ge‬؟‬ ‫)ﺍﻟﻜﺘﻠﺔ ﺍﻟﺬﺭﻳﺔ = ‪.(72.59g/ mol‬‬ ‫‪2.73 × 1025 .a‬‬

‫‪3.76 × 1023 .c‬‬

‫‪9.63 × 1023 .d 6.99 × 1025 .b‬‬ ‫‪ .9‬ﻣﺎ ﻛﺘﻠﺔ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺠﻠﻮﻛﻮﺯ ‪C6H12O6‬؟‬ ‫)ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ=‪.(180g/mol‬‬ ‫‪6.02 × 10-23 .a‬‬

‫‪2.99 × 10-22 .b‬‬

‫‪2.16 × 1025 .c‬‬

‫‪3.34 × 1021 .d‬‬

‫‪ .10‬ﻣﺎ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﻴﻦ ﻓﻲ ‪ 18.94g‬ﻣﻦ ‪Zn(NO3)2‬؟‬ ‫)ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ=‪.(189g/mol‬‬ ‫‪3.61 × 1023 .a‬‬

‫‪1.81 × 1023 .b‬‬

‫‪6.02 × 1025 .c‬‬

‫‪1.14 × 1025 .d‬‬

‫‪ .11‬ﺇﺫﺍ ﻋﻠﻤﺖ ﺃﻥ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻬﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺼﻮﺩﻳﻮﻡ‬ ‫‪ NaOH‬ﺗﺴﺎﻭﻱ ‪ .40.0g/mol‬ﻓﻤﺎ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‬ ‫ﻓﻲ ‪ 20.00g‬ﻣﻨﻪ؟‬

‫‪ .14‬ﻣﺎ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻷﺑﺎﺗﻴﺖ ﺍﻟﻔﻠﻮﺭ ‪.Ca5(PO4)3 F‬‬ ‫‪314 g/mol .a‬‬

‫‪344 g/mol .b‬‬

‫‪442 g/mol .c‬‬

‫‪504 g/mol .d‬‬ ‫‪524 g/mol .e‬‬

‫‪0.50mol .a‬‬

‫‪1.00mol .b‬‬

‫‪2.00mol .c‬‬

‫‪4.00mol .d‬‬

‫‪ .12‬ﻛﻢ ﺫﺭﺓ ﻓﻲ ‪ 116.14g‬ﻣﻦ ‪Ge‬؟‬

‫)ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ = ‪.(72.59 g/mol‬‬ ‫‪2.73 × 1025 atoms .a‬‬

‫‪6.99 × 1025 atoms .b‬‬ ‫‪3.76 × 1023 atoms .c‬‬

‫‪9.63 × 1023 atoms .d‬‬

‫‪ .13‬ﻣﺎ ﻛﺘﻠﺔ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ) ‪(BaSiF6‬‬ ‫)ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻮﻟﻴﺔ = ‪.(180 g/mol‬‬

‫‪1.68 × 1026 g .a‬‬

‫‪2.16 × 1021 g .b‬‬

‫‪4.64 × 10-22 g .c‬‬

‫‪6.02 × 10-23 g .d‬‬

‫‪85‬‬

‫المﺨاﻃر ﻭاﻻﺣﺘﻴاﻃات الﻼﺯﻡ مراﻋاﺗﻬا ﻓﻲ المﺨﺘﺒر‬ êÓ©dG

äÉWÉ«àM’G

á∏ãeC’G

ôWÉîŸG

áeÓ°ùdG RƒeQ

äɪ«∏©J ≥ah äÉjÉØædG øe ¢ü∏îJ .º∏©ŸG

‘ OGƒŸG √òg øe ¢ü∏îàJ ’ .äÓª¡ŸG á∏°S ‘ hCG á∏°ù¨ŸG

,á«FÉ«ª«µdG OGƒŸG ¢†©H .á«M äÉbƒ∏îŸGh

¢ü∏îàdG äGƒ£N ´ÉÑJG Öéj .OGƒŸG øe

á°ùeÓe çhóM ádÉM ‘ ∂ª∏©e ≠∏HCG .G kó«L ∂jój π°ùZGh ,º°ùé∏d

√ò¡d ó∏÷G á°ùeÓe ÖæŒ (áeɪc) Ék YÉæb ¢ùÑdGh ,OGƒŸG .äGRÉØbh

,ΩódG , äÉjô£ØdG ,ÉjÒàµÑdG OGƒŸG ,áXƒØëŸG ÒZ áé°ùfC’G .á«JÉÑædG

ób á«M OGƒeh äÉbƒ∏ .¿É°ùfEÓd GQk ô°V ÖÑq °ùJ

±É©°SEÓd Ék Ñ∏W ∂ª∏©e ¤EG ÖgPG .‹hC’G

.á«bGh äGRÉØb ∫ɪ©à°SG

äÉfÉî°ùdG ,πFGƒ°ùdG ¿É«∏Z ,±É÷G ó«∏÷G ,á«FÉHô¡µdG .πFÉ°ùdG ÚLhΫædG

ó∏÷G ¥ô– ób »àdG AÉ«°TC’G É¡JOhôH hCG É¡JQGôM ÖÑ°ùH .ÚJójó°ûdG

±É©°SEÓd Ék Ñ∏W ∂ª∏©e ¤EG ÖgPG .‹hC’G

™ÑJGh ,IGOC’G ™e ᪵ëH πeÉ©J Égɪ©à°SG äGOÉ°TQG

.kGQƒa ∂ª∏©e ÈNCGh ,á≤£æŸG ∑ôJG

,Ió«L ájƒ¡J OƒLh øe ócCÉJ ,Iô°TÉÑe IôîHC’G º°ûJ ’h .(áeɪc) ÉYÉæb k …óJQGh

âjȵdG ,¿ƒà°SC’G ,É«fƒeC’G å©dG äGôc ,øNÉ°ùdG .(ÚdÉãØædG)

RÉ¡÷G ≈∏Y πªà ô£N IôîHC’G øe »°ùØæàdG

IôîHC’G

∫É£YC’G ìÓ°UEG ∫hÉ– ’ á«FÉHô¡µdG äÓ«°UƒàdG øe ócCÉJ .kGQƒa ∂ª∏©e ÈNGh ,á«FÉHô¡µdG .∂ª∏©e ™e ¿hÉ©àdÉH Iõ¡LCÓd

πFGƒ°S ,í«ë°U ÒZ ¢†jQCÉJ .IGô©e q ∑Ó°SCG ,áѵ°ùæe

á≤©°üdG øe πªà ô£N ≥jô◊G hCG á«FÉHô¡µdG

AÉHô¡µdG

±É©°SEÓd Ék Ñ∏W ∂ª∏©e ¤EG ÖgPG .‹hC’G

É«k bGh (áeɪc) ÉYÉæb k óJQG ±ô°üJh ,äGRÉØbh QÉѨdG øe ™e ∂∏eÉ©J óæY ójó°T QòëH .OGƒŸG √òg

äGôc ,ìÉ≤∏dG ܃ÑM ,P’ƒØdGh ±ƒ°üdG ,å©dG äÉæéæeôH ,êÉLõdG ±É«dCG .Ωƒ«°SÉJƒÑdG

AÉ°û¨dG hCG ó∏÷G è«¡J ób OGƒe .á«°ùØæàdG IÉæ≤∏d »WÉîŸG

áé«q ¡ŸG OGƒŸG

ÈNCGh ,AÉŸÉH áHÉ°üŸG á≤£æŸG π°ùZG .∂dòH ∂ª∏©e

,äGRÉØbh ,á«bGh äGQɶf óJQG .ÈàîŸG ∞£©e ¢ùÑdGh

ó«°ùcG ¥ƒa πãe ,äÉ°†«ÑŸG ¢VɪMC’Gh ÚLhQó«¡dG óYGƒ≤dG ,∂«àjȵdG ¢†ªëc ó«°ùchQó«gh ,É«fƒeC’Éc .ΩƒjOƒ°üdG

¿CG øµÁ »àdG á«FÉ«ª«µdG OGƒŸG OGƒŸGh áé°ùfC’G ™e πYÉØàJ .É¡Ø∏àJh iôNC’G

á«FÉ«ª«µdG OGƒŸG

AÉ¡àf’G ó©H G kó«L ∂jój π°ùZG Ék Ñ∏W ∂ª∏©e ¤EG ÖgPGh ,πª©dG øe .‹hC’G ±É©°SEÓd

.∂ª∏©e äɪ«∏©J ™ÑJG

äÉÑcôŸG øe ójó©dG ,≥ÑFõdG äÉJÉÑædG ,Oƒ«dG ,ájõ∏ØdG .áeÉ°ùdG

â©∏oàHG GPEG ºª°ùàdG ÖÑ°ùJ OGƒe .â°ùŸ hCG â≤°ûæoà°SG hCG

áeÉ°ùdG OGƒŸG

ájÉØW πª©à°SGh ,GQk ƒa ∂ª∏©e ≠∏HCG π©à°ûŸG Ö¡∏dG ≥WÉæe ÖæŒ .≥jô◊G .äÉjhɪ«µdG √òg ΩGóîà°SG óæY

,¿ƒà«°S’G ,Ú°ShÒµdG ,∫ƒëµdG , Ωƒ«°SÉJƒÑdG äÉæéæeôH .ô©°ûdG ,¢ùHÓŸG

π¡°ùj äÉjhɪ«µdG ¢†©H ,Ö¡∏dG áWÉ°SƒH É¡dÉ©à°TG É¡°Vô©J óæY hCG ,Qô°ûdG hCG .IQGôë∏d

.∫ɪ©à°S’G ó©H G kó«L ∂jój π°ùZG ’h ,∞∏ÿG ¤EG ô©°ûdG §HQG ±É©°SEÓd Ék Ñ∏W ∂ª∏©e ¤EG ÖgPGh ,á°VÉØ°†ØdG ¢ùHÓŸG ¢ùÑ∏J .‹hC’G ∫É©°TEG óæY º∏©ŸG äɪ«∏©J ™ÑJGh .¬FÉØWEG hCG Ö¡∏dG

OGƒŸG ,¥QƒdG ,¢ùHÓŸG ,ô©°ûdG .∫É©à°TEÓd á∏HÉ≤dG

ÖÑ°ùj Ék MƒàØe Ö¡∏dG ∑ôJ .≥jô◊G

øjó«dG π°ùZ

ó©H ∂jój π°ùZG AÉŸÉH áHôŒ πc ´õf πÑb ¿ƒHÉ°üdGh á«bGƒdG äGQɶædG

»YÉ©°TG •É°ûf

õeôdG Gòg ô¡¶j πª©à°ùJ ÉeóæY á©°ûe OGƒe

,Úcɵ°ùdG ,äGôØ°ûdG ,äÉ°ü≤ŸG äÉ«LÉLõdGh äGhOC’G ∫ɪ©à°SG äGhOCG ,áÑHq óŸG äGhOC’G .ádƒ¡°ùH ó∏÷G ìôŒ »àdG .Qƒ°ùµŸG êÉLõdG ,íjô°ûàdG

äÉfGƒ«◊G áeÓ°S

õeôdG Gòg Ò°ûj áeÓ°S ≈∏Y ó«cCÉà∏d äÉfGƒ«◊Gh ÜÓ£dG

¢ùHÓŸG ájÉbh

≈∏Y õeôdG Gòg ô¡¶j »àdG OGƒŸG äGƒÑY ¢ùHÓŸG ™≤ÑJ ¿CG øµÁ .É¡bô– hCG

OGƒŸG øe ¢ü∏îàdG á«q M OGƒe á©ØJôe IQGôM áLQO á°†Øîæe hCG IOÉ◊G ΩÉ°ùLC’G

∫É©à°TÓd á∏HÉb OGƒe

π©à°ûŸG Ö¡∏dG

Ú©dG áeÓ°S

AGóJQG Ék ªFGO Öéj óæY á«bGh äGQɶf .ÈàîŸG ‘ πª©dG

86

‫مسرد المصطلحات‬

‫المصطلحات‬

‫)‪(CG‬‬

‫‪ Spectator Ion êôØàªdG ¿ƒjC’G‬ﺍﻷﻳﻮﻥ ﺍﻟﺬﻱ ﻻ ﻳﺸﺎﺭﻙ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫)‪(Ü‬‬

‫‪ Proton ¿ƒJhôÑdG‬ﺟﺴﻴﻢ ﻣﻦ ﻣﻜﻮﻧﺎﺕ ﻧﻮﺍﺓ ﺍﻟﺬﺭﺓ‪ ،‬ﻭﺷﺤﻨﺘﻪ ﻣﻮﺟﺒﺔ )‪.(+1‬‬

‫)‪(ä‬‬

‫‪ Single - Replacment Reaction §``«°ùÑdG ∫Ó``ME’G π``YÉØJ‬ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﻳﻨﺘﺞ ﻋﻨﺪﻣﺎ ﺗﺤـﻞ ﺫﺭﺍﺕ ﺃﺣﺪ ﺍﻟﻌﻨﺎﺻﺮ ﻣﺤﻞ‬ ‫ﺫﺭﺍﺕ ﻋﻨﺼﺮ ﺁﺧﺮ ﻓﻲ ﻣﺮﻛﺐ‪.‬‬

‫‪ Double - Replacment Reaction êhOõ``ªdG ∫Ó``ME’G π``YÉØJ‬ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﻳﻨﺘﺞ ﻋﻦ ﺗﺒﺎﺩﻝ ﺃﻳﻮﻧﺎﺕ ﻣﺎﺩﺗﻴﻦ ﻭﻳﻨﺸـﺄ ﻋﻨﻪ‬ ‫ﻏﺎﺯ‪ ،‬ﺃﻭ ﺭﺍﺳﺐ‪ ،‬ﺃﻭ ﻣﺎﺀ‪.‬‬ ‫‪ Percent Composition …ƒÄªdG »Ñ°ùædG Ö«côàdG‬ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﻜﺘﻠﻴﺔ ﻟﻜﻞ ﻋﻨﺼﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬ ‫‪ Nuclear Reaction …hƒædG πYÉØàdG‬ﺗﻔﺎﻋﻞ ﻳﺘﻀﻤﻦ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻧﻮﺍﺓ ﺍﻟﺬﺭﺓ‪.‬‬

‫‪ Combustion Reaction ¥GôàME’G πYÉØJ‬ﺗﻔﺎﻋﻞ ﻣﺎﺩﺓ ﻣﻊ ﺍﻷﻛﺴﺠﻴﻦ ﻭﻳﻨﺘﺞ ﻋﻨﻬﺎ ﻃﺎﻗﺔ ﻓﻲ ﺻﻮﺭﺓ ﺿﻮﺀ ﻭﺣﺮﺍﺭﺓ‪.‬‬

‫‪ Decomposition Reaction ∂µ``ØàdG π``YÉØJ‬ﺗﻔﺎﻋـﻞ ﻳﺤـﺪﺙ ﻧﺘﻴﺠـﺔ ﻟﺘﻔﻜﻚ ﺃﺣـﺪ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺇﻟﻰ ﻋﻨﺼﺮﻳـﻦ ﺃﻭ ﺃﻛﺜﺮ ﺃﻭ ﺇﻟﻰ‬ ‫ﻣﺮﻛﺒﺎﺕ ﺟﺪﻳﺪﺓ‪.‬‬

‫‪ Synthesis Reaction øjƒµàdG πYÉØJ‬ﺗﻔﺎﻋﻞ ﻣﺎﺩﺗﻴﻦ ﺃﻭ ﺃﻛﺜﺮ ﻹﻧﺘﺎﺝ ﻣﺎﺩﺓ ﻭﺍﺣﺪﺓ‪.‬‬

‫)‪(ê‬‬

‫‪ Periodic Table …Qhó``dG ∫hó``édG‬ﺟـﺪﻭﻝ ﻳﻨﻈﻢ ﻛﻞ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻤﻌﺮﻭﻓﺔ ﻓﻲ ﺷـﺒﻜﺔ ﻣﻦ ﺍﻟﺼﻔـﻮﻑ ﺍﻷﻓﻘﻴﺔ )ﺩﻭﺭﺍﺕ( ﻭﺍﻟﺼﻔﻮﻑ‬ ‫ﺍﻟﻌﻤﻮﺩﻳﺔ )ﻣﺠﻤﻮﻋﺎﺕ ﻣﻦ ﺍﻟﻌﺎﺋﻼﺕ( ﻣﺮﺗﺒﺔ ﺗﺼﺎﻋﺪﻳﺎ ﺣﺴﺐ ﺍﻟﻌﺪﺩ ﺍﻟﺬﺭﻱ‪.‬‬

‫)‪(Q‬‬

‫‪ Precipitate Ö°SGôdG‬ﻣﺎﺩﺓ ﺻﻠﺒﺔ ﺗﺘﻜﻮﻥ ﺧﻼﻝ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬

‫‪88‬‬

‫المصطلحات‬

‫)‪(¢U‬‬

‫‪ Empirical Formula á«dhC’G ᨫ°üdG‬ﺍﻟﺼﻴﻐﺔ ﺍﻟﺘﻲ ﺗﺒﻴﻦ ﺃﺻﻐﺮ ﻧﺴﺒﺔ ﻋﺪﺩﻳﺔ ﺻﺤﻴﺤﺔ ﻟﻤﻮﻻﺕ ﺍﻟﻌﻨﺎﺻﺮ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‪.‬‬

‫‪ Molecular Formula á``«ÄjõédG á``¨«°üdG‬ﺍﻟﺼﻴﻐـﺔ ﺍﻟﺘـﻲ ﺗﻌﻄﻲ ﺍﻟﻌﺪﺩ ﺍﻟﻔﻌﻠـﻲ ﻟﻠﺬﺭﺍﺕ ﻣﻦ ﻛﻞ ﻋﻨﺼﺮ ﻓـﻲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ‬ ‫ﺍﻟﻤﺎﺩﺓ‪.‬‬

‫)´(‬

‫‪ Avogadro's Number hQOÉLƒaCG OóY‬ﻫﻮ ‪ ،6.0221367*1023‬ﻭﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﻋﺪﺩ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻓﻲ ﻣﻮﻝ ﻭﺍﺣﺪ‪ ،‬ﻭﻳﻤﻜﻦ‬ ‫ﺗﻘﺮﻳﺐ ﻫﺬﻩ ﺍﻟﻘﻴﻤﺔ ﻟﺜﻼﺛﺔ ﻣﻨﺎﺯﻝ ‪.6.02*1023‬‬

‫)∑(‬

‫‪ Molar Mass á«dƒªdG á∏àµdG‬ﺍﻟﻜﺘﻠﺔ ﺑﺎﻟﺠﺮﺍﻣﺎﺕ ﻟﻮﺍﺣﺪ ﻣﻮﻝ ﻣﻦ ﺃﻱ ﻣﺎﺩﺓ ﻧﻘﻴﺔ‬

‫)‪(Ω‬‬

‫‪ Reactants äÓYÉØàªdG‬ﺍﻟﻤﻮﺍﺩ ﺍﻟﺘﻲ ﻳﺒﺪﺃ ﺑﻬﺎ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬

‫‪ Aqueous Solution »FɪdG ∫ƒ∏ëªdG‬ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﺬﻱ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﻣﺎﺩﺓ ﺃﻭ ﺃﻛﺜﺮ ﻣﺬﺍﺑﺔ ﻓﻲ ﺍﻟﻤﺎﺀ‪.‬‬ ‫‪ Solute ÜGòªdG‬ﻣﺎﺩﺓ ﺃﻭ ﺃﻛﺜﺮ ﻣﺬﺍﺑﺔ ﻓﻲ ﻣﺤﻠﻮﻝ‪.‬‬

‫‪ Solvent ÖjòªdG‬ﺍﻟﻤﺎﺩﺓ ﺍﻟﺘﻲ ﺗﺬﻳﺐ ﺍﻟﻤﺬﺍﺏ ﻭﺗﺤﺘﻮﻳﻪ‪.‬‬

‫‪ Complete Ionic Equation á∏eɵdG á«fƒjC’G ádOÉ©ªdG‬ﻣﻌﺎﺩﻟﺔ ﺃﻳﻮﻧﻴﺔ ﺗﻈﻬﺮ ﻛﺎﻓﺔ ﺍﻷﻳﻮﻧﺎﺕ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺑﺼﻮﺭﺗﻬﺎ ﺍﻟﻮﺍﻗﻌﻴﺔ‪.‬‬

‫‪ Net Ionic Equation á«FÉ¡ædG á«fƒjC’G ádOÉ©ªdG‬ﻣﻌﺎﺩﻟﺔ ﺃﻳﻮﻧﻴﺔ ﺗﺸﺘﻤﻞ ﻓﻘﻂ ﻋﻠﻰ ﺍﻟﺠﺴﻴﻤﺎﺕ ﺍﻟﻤﺸﺎﺭﻛﺔ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ‪.‬‬

‫‪ Chemical Equation á``«FÉ«ª«µdG á``dOÉ©ªdG‬ﺟﻤﻠـﺔ ﺗﺴـﺘﻌﻤﻞ ﻓﻴﻬـﺎ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴـﺔ ﻟﺘﺤﺪﻳﺪ ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﺸـﺎﺭﻛﺔ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ‬ ‫ﻭﻛﻤﻴﺎﺕ ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﺘﻔﺎﻋﻠﺔ ﻭﺍﻟﻨﺎﺗﺠﺔ‪.‬‬

‫‪ Hydrates »FɪdG í∏ªdG‬ﻣﺎﺩﺓ ﺃﻳﻮﻧﻴﺔ ﺻﻠﺒﺔ ﻳﺮﺗﺒﻂ ﺑﺬﺭﺍﺗﻬﺎ ﻋﺪﺩ ﻣﺤﺪﺩ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺎﺀ‪.‬‬

‫‪ Mole ∫ƒ``ªdG‬ﻭﺣـﺪﺓ ﻧﻈـﺎﻡ ﻋﺎﻟﻤﻲ ﺗﺴـﺘﻌﻤﻞ ﻓﻲ ﻗﻴﺎﺱ ﻛﻤﻴـﺔ ﺍﻟﻤﺎﺩﺓ‪ ،‬ﻭﻫﻮ ﻋﺒﺎﺭﺓ ﻋـﻦ ﻋﺪﺩ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﺍﻟﻤﻮﺟـﻮﺩﺓ ﻓﻲ ‪ 12g‬ﻣﻦ‬ ‫ﺍﻟﻜﺮﺑﻮﻥ ‪ ،‬ﻭﺍﻟﻤﻮﻝ ﺍﻟﻮﺍﺣﺪ ﻛﻤﻴﺔ ﻣﻦ ﺍﻟﻤﺎﺩﺓ ﺍﻟﻨﻘﻴﺔ ﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ‪ 6.02*1023‬ﻣﻦ ﺍﻟﺠﺴﻴﻤﺎﺕ‪.‬‬

‫)¿(‬

‫‪ Product èJÉædG‬ﺍﻟﻤﺎﺩﺓ ﺍﻟﺘﻲ ﺗﺘﻜﻮﻥ ﺧﻼﻝ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‪.‬‬

‫‪89‬‬

‫ﺟﺪاﻭﻝ مرﺟﻌﻴﺔ‬ ‫ﺟﺪاﻭﻝ مرﺟﻌﻴﺔ‬

‫اﻟﺠﺪول اﻟﺪوري ﻟﻠﻌﻨﺎﺻﺮ‬ eK� eK� t�� eK� ô

vK� dBM� q� ‚ËbM Êu� ‰b� .eK�ô Ë√ eK� t�� Ë√ «eÒÎ K� t�u�

ÆUN�UA��« s� b�Q��« bM� UN� WOzUN� ¡UL�√ —UO��« r�O�Ë ¨W��R� 118 v�≈ 112 d UMF�« “u�— ¡UL�√ ÆULN� WIKF�*« »—U���« …œU�≈ sJ1 r� t�_ ªp�– s� l�«d��« - sJ�Ë ¨ULNM�uJ� - b� 118 Ë116 s�dBMF�« Ê√ sE� ÊU�

90

‫ﺟﺪاﻭﻝ مرﺟﻌﻴﺔ‬

‫ﺟﺪاﻭﻝ مرﺟﻌﻴﺔ‬

“U�

’«u� UN�Ë ¨W�uL�� v�b� œuL� q� w� d UMF�« ÆWN�UA�� WOzUOLO� dBMF�« Í—c�« œbF�« e�d�« W??�—c??�« WK�J�« WD�u�*«

v�b� WOI�_« d UMF�« ·uH s� Í—c�« œbF�« œ«œe� Æ «—Ëœ Æ…—Ëœ q� w� 5LO�« v�≈ —U�O�« Íc??�« ÊU??J??*«v??K??� rN��« ‰b??� d UMF�« Ác� tO� l{u� Ê√ V�� v�≈ UNKI� - bI� ƉËb???'« w??� ÆÊUJLK� «Î dO�u� ‰Ëb'« qH�√

91

qzU�

VK …œU*« W�U�

lMB�Ô

‰b??� U??O??K??F??�« W??�ö??�??�« “u???�d???�« W??�—œ w??� d??B??M??F??�« W??�U??�v??K??� e�d�« ‰b??� ULMO�ÆW�dG�« …—«d??� .WFMB*« d UMF�« vK� l�«d�«

ÆdBMFK� «dL� Î ‰u�_« dOEMK� wK�J�« œbF�« u� 5�uI� ◊U;« r�d�«