سلسلة تمارين في الفيزياء حول مراقبة تطور جملة كيميائية من اعداد الاستاذ فرقاني

‫اا‬ ‫را ا ا‬ ‫ا‬ ‫ﻥ ﻡد‬‫– ا وب‬ ‫ﻥﺏ‬ ‫وزارة ا ﺏ ا‬‫ا ذ ‪ :‬ﻥ رس‬ ‫دة ا م ا‬ ‫إن‬ ‫ا‬ ‫‪:‬ا ما وا‬ ‫ا‬ ‫ا ة‪ 3:‬ت‬ ‫م‪3:‬عت‪،‬ر‪،‬تر‬‫‪Sujet : 3AS 06 - 01‬‬ ‫‪:‬ا ر آ ‪.‬‬ ‫ا يا‬‫ا ا را ‪2011/2010 :‬‬ ‫‪2011/05/31 :‬‬ ‫ر‬ ‫א ن א ول ‪(*) :‬‬ ‫ءا ‪:‬‬ ‫آ ت ا ذات ا‬ ‫‪ -1‬أ‪ -‬أآ ا ﻥ ا‬ ‫اآ‬ ‫ا‬ ‫)‪(A‬‬ ‫‪-2‬إ ﺏ ﺕ ن‪ -2-‬ول‬ ‫)‪(B‬‬ ‫)‪(C‬‬ ‫ﺏ وﺏ ن‪ -2-‬ول‬ ‫)‪(D‬‬ ‫إ ﻥل‬ ‫)‪(E‬‬ ‫)‪(F‬‬ ‫‪ -2‬ﻡ ﺏ وﺏ ﻥ‬ ‫اﻥ‬ ‫ﻡ ﻥ ات ا ﺕ‬ ‫ب‪ -‬أآ ا و ﺹ ا ا ﺕ ﻡ ا ت ا ‪:‬‬ ‫• ﺕ ا آ )‪ (C‬ﻡ ا آ )‪. (E‬‬ ‫• ﺕ ا آ )‪ (B‬ﻡ ا آ )‪. (E‬‬ ‫• ﺕ ا آ )‪ (A‬ﻡ ا آ )‪. (D‬‬‫آ ﺏ آ )‪ ، (A‬ﻡ ‪ 0.2 mol‬ﻡ آ ل )‪ ، (B‬ﻥ ا وط ا زﻡ وث‬ ‫‪ -2‬ﻥ ج ‪ 0.2 mol‬ﻡ‬‫)ا ‪ (1-‬ﺕ ات د ﻡ ت ا )‪(E‬‬ ‫ذ أ )‪ (E‬و ﻡ ء ‪ .‬ا ن ا‬ ‫ا‪،‬‬ ‫ا ﺏ اﻡ‪.‬‬ ‫ا ‪: 1-‬‬ ‫)‪nE (mol‬‬ ‫‪0.12‬‬ ‫‪t‬‬‫)‪ ، (A‬و آ ل )‪. (B‬‬ ‫وي ا ت ن ﻡ ا‬ ‫ا ل )‪. (B‬‬ ‫‪-‬ا ﺹ‬ ‫ا )‪ (A‬ﺏ ا ﻡ ‪.‬‬ ‫أ ة اﻥ ﻡ ﻡ اﺏ ا ﻡ‬ ‫‪ -3‬ﻥ ي ﺕ‬ ‫‪:‬‬ ‫)ا ‪ (2-‬ﺕ ات آ ا‬ ‫‪.‬‬ ‫ا نا‬ ‫‪،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪2:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫ا ‪: 2-‬‬ ‫)‪nA (mol‬‬ ‫‪0. 4‬‬ ‫‪0.38‬‬ ‫‪t‬‬ ‫ا ازن ‪.‬‬ ‫ا‬ ‫ا ﺏ ا ‪،‬وا آ ا‬ ‫أ‪ -‬أوﺝ ا آ ا‬ ‫ب‪ -‬أوﺝ ﺹ ا ل )‪. (B‬‬‫ا ‪ ، CH3COOH‬أوﺝ ا آ ا‬ ‫‪ -4‬ﻥ ج ‪ 3 mol‬ﻡ ا ﻥ ل ‪ CH3OH‬ﻡ ‪ 3 mol‬ﻡ‬ ‫وث ا ازن ا ‪.‬‬ ‫)‪ (A‬ﻡ ‪ 2 mol‬ﻡ آ ل )‪ (B‬ﺹ ‪، C3H7OH‬‬ ‫آﺏآ‬ ‫‪ -5‬ﻥ ج ‪ 2 mol‬ﻡ‬ ‫‪ 1.2 mol‬ﻡ أ )‪. (E‬‬ ‫آ ﻡ ة‪.‬‬ ‫ل )‪ (B‬ﻡ ذآ ا و ا‬ ‫أ‪ -‬أآ ا ﻥ ا ا‬ ‫ل )‪. (B‬‬ ‫ب‪ -‬ا ا ا‬ ‫‪ ) :‬اﻡ ن ا ا – ‪(**) ( 2009/2008‬‬ ‫א نא‬ ‫ﻥ أ ‪E‬ﺹ ‪:‬‬‫آ ل‪B‬‬ ‫ا ﻥ ‪ A‬و آ ‪ mB = 22.0 g‬ﻡ‬ ‫‪ VA = 14.3 mL‬ﻡ‬ ‫نﻡ‬ ‫اﻥ ﻡ ﻡ‬ ‫ا‬ ‫ا ا آ وا ﺏ ا ﺕ ة‪ 4‬ت‪،‬‬ ‫ﻡ إ ﺏ ر ‪ 1 mL‬ﻡ‬ ‫آ ﻡ ا ‪. mE = 21.7 g‬‬ ‫ل‪B‬وأ إ وآ ﺹ ‪.‬‬ ‫‪ -1‬أآ ا ﻥ ا‬ ‫‪ -2‬أآ ﻡ د ا ا ج ل ا ة و أ ا ا ‪.‬‬ ‫‪ -3‬أذآ ﻡ ات ه ا ا ر ‪.‬‬ ‫‪ -4‬ﻡ ا ض ﻡ ﺕ ا ‪.‬‬ ‫اﺏا ‪.‬‬ ‫‪ -5‬أوﺝ ا آ ا‬ ‫‪ -6‬أﻥ ﺝ ول ﺕ م ا ‪.‬‬ ‫‪ -7‬أوﺝ ﻡ دود ا و ا ﺹ ا ل ﻡ ا ‪.‬‬ ‫‪ -8‬أآ رة ﺏ ا ازن و ا ‪.‬‬ ‫‪ -9‬إذا ﺏ ف ا ء آ ﺕ ‪ .‬ﻡ ذا ث ؟‬ ‫‪:‬‬ ‫ا عا‬ ‫)‪ ρ (g/ml‬ا ا )‪M (g/mol‬‬ ‫اا‬ ‫‪A : CH3COOH‬‬ ‫‪60‬‬ ‫‪1.05‬‬ ‫ا ل‪B:‬‬ ‫‪88‬‬ ‫‪0.18‬‬ ‫ا ‪E:‬‬ ‫‪130‬‬ ‫‪0.87‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪3:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫א ن א ‪ ) :‬اﻡ ن ا ا – ‪(**) ( 2009/2008‬‬ ‫‪.‬‬ ‫اا‬ ‫اء ‪ . 2.07‬أوﺝ ا ا ا‬ ‫آ ﺏ آ ‪A‬آ ﺏ ر ﺏ‬ ‫‪-1‬‬‫وﺏ ن‪ -2-‬ول ‪،‬‬ ‫)‪ (A‬و ‪ 2 mol‬ﻡ آ ل ‪ B‬ه ا‬ ‫‪ -2‬ﻥ ﻡ ﻡ وي ا ت ن ﻡ ‪ 2 mol‬ﻡ‬ ‫و ﺕ ا وط ا زﻡ وث ا ‪ ،‬أ ‪ E‬و ﻡ ء ‪.‬‬ ‫أ‪ -‬أآ ﻡ د ا ا دث ‪ .‬ﻡ ﺥ ‪.‬‬ ‫ب‪ -‬أوﺝ ﻥ ا م ا ‪:‬‬ ‫‪:‬‬ ‫ﺹ ا ل أو ﻥ ي‬ ‫ا دود ‪5% 60% 67%‬‬ ‫ﻡ اﺏ ا ﻡ وي ا ت‬ ‫ﻡ ‪:‬ه ا ﺹ‬ ‫‪.‬‬ ‫ا ة ﺏ ا م ا ‪ xf‬أ‬ ‫ﺝـ‪ -‬أآ رة ﺏ ا ازن ا‬‫)‪ (E‬و ‪ 3 mol‬ﻡ‬ ‫‪ 2mol A‬ﻡ ا ل ‪ B‬و ‪ 4 mol‬ﻡ ا‬ ‫‪ -3‬ﻥ ﻡ ﺥ ن ‪ 1 mol‬ﻡ‬ ‫ا ء ﻥ ا وط ا زﻡ وث ا ‪.‬‬ ‫ر ا ) أ ة أو إﻡ ه ( ‪.‬‬ ‫أ‪ -‬ﺏ أي ﺝ‬ ‫وث ا ازن ا ‪.‬‬ ‫ب‪ -‬أوﺝ ا آ ا‬ ‫ﺝـ‪ -‬أ ﻡ دود ا ‪.‬‬ ‫א ن א א ‪ ) :‬ﺏ ر ‪ – 2009‬م ﺕ ( )**(‬ ‫ا ﻥ )‪ (CH3COOH‬و ا ﻥ ل )‪ (C2H5OH‬ﺏ د ‪:‬‬ ‫ا ﺹﺏ‬ ‫ﻥ جا لا‬ ‫‪CH3COOH + C2H5OH = CH3COOC2H5 + H2O‬‬ ‫إﻥ ء ﻡ ع داﺥ ا ﻡ ﻡ ﻡ ‪ 0.2 mole‬ﻡ‬ ‫را ﺕ ر ا ﺏ ا ﻡ ‪ ،‬ﻥ‬ ‫ﻥا‬ ‫ا ﻥ )‪ (CH3COOH‬و ‪ 0.2 mole‬ﻡ ا ل )‪ ، (C2H5OH‬ﺏ ا ج و ا‬‫ﻥ ا ‪ V0‬ﻡ ا ‪ .‬ﻥ ا ﻥ ﺏ‬ ‫يآ ﻡ‬ ‫‪ 10‬أﻥ ﺏ اﺥ ر ﻡ ﻡ ‪ 1‬إ ‪ ، 10‬ﺏ‬ ‫م ﻡ درﺝ ارﺕ ﺏ و ﻥ ا ﺕ ‪.‬‬ ‫وﺕ‬‫ﺏ ا ﻡ ل ﻡ ﻡ ه روآ ا د م‬ ‫ا ‪ t = 0‬ﻥ ج ا ﻥ ب ا ول وﻥ ا ا‬‫ﻡ ه روآ‬ ‫)‪ (Na+ + OH-‬ﺕ آ ا ‪ ، C = 1.0 mol.L-1‬م غ ﻥ ا إ‬ ‫ا ا‪.‬‬ ‫)‪ (V’be‬ا زم ة ا‬ ‫ا د م )‪(Vbe‬‬ ‫ﺏ ﻡ ة ﻥ ر ا ﻡ أﻥ ب ﺥ و ه ا ‪ ،‬ا ت ا ول ا ‪:‬‬ ‫‪ -1‬أ‪ /‬ﻡ ا ا ا ؟‬ ‫ل )‪. (C2H5OH‬‬ ‫)‪ (CH3COOH‬و ا‬ ‫ب‪ /‬أﻥ ﺝ و م ا ﺏ ا‬ ‫ا ﻥ )‪ (CH3COOH‬و ﻡ ل‬ ‫ا ج لا ﺹ ﺏ‬ ‫ﺝـ( اآ ﻡ د ا ا‬ ‫ه روآ ا د م )‪. (Na+ + OH-‬‬ ‫ا س ا زم ‪.‬‬ ‫)‪ (n‬و )‪(V’be‬‬ ‫‪-2‬أ‪ /‬أآ ا ﺏ آ ا ا‬ ‫أآ ا ول أ ‪.‬‬ ‫)‪ (x‬ﺕ م ا‬ ‫ب‪ -‬ﺏ ﻥ ﺏ ول ا م ا ﺏ أ‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪4:‬‬ ‫‪2011/2010 :‬‬ ‫)‪- (01‬‬ ‫–‬ ‫‪-‬‬ ‫ﺝـ‪ -‬أر ا ا ﻥ )‪. x = f(t‬‬ ‫د‪ -‬أ ﻥ ا م ا ‪ ، τ‬ﻡ ذا ﺕ ؟‬ ‫‪.‬‬ ‫ا ازن ﺏ ا م ا ‪ . xf‬أ‬ ‫آ ا ا ‪Qrf‬‬ ‫هـ‪-‬‬ ‫א ن א س ‪ ) :‬ﺏ ر ‪ – 2009‬ر ت ( )**(‬‫ا ﻥ ‪ CH3COOH‬و ا ﻥ ل ‪ C2H5-OH‬ﻥ ﺥ ‪ 7‬أﻥ ﺏ‬ ‫ضﻡ ﺏ ﺕ را لا ﺏ‬‫و )‪ n0(mol‬ﻡ ا ل ا ﺏ ‪.‬‬ ‫اﺥ ر و ا )‪ (t = 0‬ﻥ ج آ وا ﻡ )‪ n0(mol‬ﻡ ا‬ ‫ج ا ل ا دث ﺏ ذي ا د ‪:‬‬ ‫)‪CH3COOH(ℓ) + C2H5OH(ℓ) = CH3COOC2H5(ℓ) + H2O(ℓ‬‬ ‫ت زﻡ ﻡ ﻡ ى ا ﻥ ﺏ ا ا ﺕ ا ﺥ ﻡ أﺝ ﻡ آ‬ ‫ﻥ درﺝ ارة ﺏ و‬‫ها ﺏ ل‬ ‫ا )‪ (n‬ﺏ ا ﻡ ل ه روآ ا د م )‪. (Na+ + OH-‬‬ ‫ﻡ دة ا‬ ‫ﺝ ول ا ت ا ‪:‬‬ ‫ا )‪. (n‬‬ ‫آ ﻡ دة ا‬ ‫‪. xmax‬‬ ‫‪ -1‬أﻥ ﺝ و م ا و ا ا م ا‬ ‫)’‪ (n‬ﺏ‬ ‫‪ -2‬ا ا ا ﺕ آ ﻡ دة ا ا‬‫‪ -3‬أآ ا ول أ ‪ ،‬و ﺏ ﺥ ر ﻡ أر ا ا ي ﺕ ات آ ﻡ دة ا ا ﺏ‬ ‫ا ﻡ )‪. n’ = f(t‬‬ ‫ا ‪،t=3h‬آ ﺕ ر ا ﻡ ا ﻡ ؟ ‪.‬‬ ‫ا‬ ‫‪ -4‬أ‬ ‫م )‪ (τf‬و ﻡ ذا ﺕ ؟‬ ‫‪ -5‬أ ا ا‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

5: 2011/2010 : - (01) – - ‫أﺝ ﺏ‬Sujet : 3AS 06 - 01 . ‫ا ر آ‬: ‫ا يا‬ : ‫א ن א ول‬ : ‫ ا ﻥ ا‬-‫ أ‬-1(A) : CH3 – CH2 – CH – CH2OH(B) : |(C) :(D) : CH2(E) : |(F) : CH3 : ‫ ﺹ و ا ا ا ﺕ‬-‫ب‬ CH3 – CHOH – CH3 . ‫ إ ﻥ ات ا‬: ‫ا‬ CH3 – CH2OH ‫ إ ﻥ ات ﻡ إ‬: ‫ا‬ CH3 – CH – COOH | CH3 CH3 – COOH H – COO – CH2 – CH2 – CH2 – CH3• (C) + (E) : CH3 – COO – CH – CH3• (B) + (E) : CH3 – COO – CH – CH3 | CH3• (A) + (D) : CH3 – CH – COO – CH2 – CH – CH2 – CH3 || CH3 CH2 | CH3 ‫ إ‬-2 ، ‫ ﻡ ﺏ وﺏ ﻥ ات‬-2 : ‫ا‬ . ‫ﺏﺕ‬ : ‫ ﺹ ا ل‬-2 ‫اما‬ A +B = E +D 0 ‫ اﺏ ا‬x = 0 0.2 0.2 0 x xf ‫ اﻥ‬x 0.2 - x 0.2 - x x :– ‫ ﻥ‬xf 0.2 - xf 0.2 - xf xf : . ،، –

6: 2011/2010 : - (01) – - : ‫ﻡ دود ا‬ ‫ﻥ‬ : ‫ آ‬A ‫ إذا اﺥ‬-0.2 – x = 0 → x = 0.2 : ‫ آ‬B ‫ إذا اﺥ‬-0.2 – x = 0 → x = 0.2 . xmax = 0.2 mol : ‫إذن‬r = xf .100 . nf(E) = 0.12 mol : ‫ ﻡ ا ن‬- x max . xf = 0.12 mol : ‫ إذن‬. nf(E) = xf : ‫ ﻡ ﺝ ول ا م‬- 0.12 :r = .100 = 60 % 0.2 . ‫إذن ا ل ا ﻥ ي‬ : ‫اﺏا‬ ‫ ا آ ا‬-‫ أ‬-3 : ‫ﺏ أن ا ا ﺏ ا ﻡ وي ا ت ن‬n0(A) = n0(B) : ‫ إذن‬n0(A) = 0.4 mol : ‫ﻡ ا ن‬n0(A) = n0(B) = 0.4 mol : ‫ا ازن‬ ‫ا آ ا‬- ‫اما‬ A +B = E+ D 0 ‫ اﺏ ا‬x = 0 0.4 0.4 0 x xf ‫ اﻥ‬x 0.4 - x 0.4 - x x ‫ ﻥ‬xf 0.4 - xf 0.4 - xf xf . nf (A) = 0.38 mol : ‫ ﻡ ا ن‬- nf = 0.4 - xf : ‫ ﻡ ﺝ ول ا م‬- : ‫وﻡ‬0.4 – xf = 0.38 → xf = 0.4 – 0.38 = 0.02 mol ‫ا ازن آ‬ ‫وﻡ ن ا آ ا‬ :xf(E) = xf = 0.02 mol : (B) ‫ ﺹ ا ل‬-‫ب‬nf(H2O) = xf = 0.02 mol : ‫ﻥ ﻡ دود ا‬nf(A) = 0.4 – xf = 0.38 molnf(B) = 0.4 – xf = 0.38 molr = xf .100 x max : . ،، :– –

7: 2011/2010 : - (01) – -xmax = 0.4 mol : ‫اﺏﻥ‬ ‫ﺏا‬ 0.02 : ‫وﻡ‬r = .100 = 5% . (B) ‫إن ا ل‬ 0.4 ‫ ا آ ا‬-4 : ‫وث ا ازن ا‬ ‫ ا م ا‬CH3COOH + CH3OH = CH3COOCH3 + H2O ‫ اﺏ ا‬x = 0 3 3 0 0 ‫ اﻥ‬x 3 - x 3 - x x x ‫ ﻥ‬xf 3 - xf 3 - xf xf xf . r = 67 % : ‫ا ﺏ ا ﻡ وي ا ت ن ﻡ دود ا‬ ‫ا ﻥ ل آ ل أو و آ ن أن ا‬ :‫و‬r = xf .100 → xf = r . xmax x max 100 :‫ﺕ ن‬ ‫ﻡ ﺝ ول ا م و ﺏ ر ا‬ : ‫وﻡ‬3 - xmax = 0 → xmax = 3 mol 67 . 3 : ‫إذن‬xf = 100 = 2 molnf(CH3COOCH3) = nf(H2O) = xf = 2 molnf(CH3COOH) = nf(CH3OH) = 3 – xf = 1 mol : (B) ‫ل‬ ‫ ا ﻥ ا ا‬-‫أ‬-5 CH3 – CH2 – CH2OH CH3 – CHOH – CH3 ‫ ول‬-1-‫ﺏ وﺏ ن‬ ‫ ول‬-2-‫ﺏ وﺏ ن‬ : (B) ‫ل‬ ‫ ا ا‬-‫ب‬ : ‫ﻡ دود ا‬ ‫ﻥ‬ ‫اما‬ A +B = E+ D ‫ اﺏ ا‬x = 0 0 220 x ‫ اﻥ‬x xf ‫ ﻥ‬xf 2-x 2-x x 2 - xf 2 - xf xf . xf(E) = 1.2 mol : - . nf(E) = xf : ‫ ﻡ ﺝ ول ا م‬- . xf = 1.2 mol : ‫إذن‬: . ،، :– –

8 : 2011/2010 : - (01) – -2 – xmax = 0 → xmax = 2 mol :‫ﺕ‬ ‫ ﻡ ﺝ ول ا م و ﺏ ر ا‬-r = xf .100 : ‫( ﻥ ي ﺹ‬B) ‫إذن ا ل‬ x max 1.2r = 2 .100 = 60 % CH3 – CHOH – CH3 : ‫א نא‬ : ‫( و ا‬B) ‫ ﺹ‬-1 CH3 – CH – CH2 – CH2OH | CH3 . ‫ ول‬-1-‫ ﻡ ﺏ ﺕ ن‬-3 ( ‫)آ ل أو‬ : ‫ ﻡ د ا‬-2 CH3COOH + CH3-CH-CH2-CH2OH = CH3-COO-CH2-CH2-CH-CH3 || CH3 CH3 :‫ا‬ ‫ ﺥ‬-3 : ‫ﺏ اص ا‬ ‫هاا ه ﺕ أ ة‬ . (‫ﻡ ود ) ﺕ م‬ . ‫اري‬ .‫س‬ .‫ﺏ ء‬ . ‫ا هﺕ ا‬ ‫ ا ض ﻡ ﺕ‬-4 : ‫اﺏا‬ ‫ ا آ ا‬-5▪ n0 (A) = mA = ρ(A) . V = 1.05 .14.3 = 0.25 mol MA MA 60▪ n0 (B) = mB = 22 = 0.25 mol MA 88 : ‫ ﺝ ول ا م‬-6 ‫اما‬ A +B = E+ D 0 ‫ اﺏ ا‬x = 0 0.25 0.25 0 x xf ‫ اﻥ‬x 0.25 - x 0.25 - x x ‫ ﻥ‬xf 0.25 - xf 0.25 - xf xf : . ،، :– –

9 : 2011/2010 : - (01) – - : ‫ ﻡ دود ا وﺹ ا ل‬-7 : ‫ﻡ ﺝ ول ا م و ﺏ ر ا ﺕ م ن‬0.25 – xmax = 0 → xmax = 0.25 mol :nf (E) = mf (E) = 21.7 = 0.167 mol M(E) 130 : ‫ﻡ ﺝ ول ا م‬nf(E) = xf = 0.16 molr= xf .100 = 0.167 .100 = 67 % x max 0.25 . ‫( ﻥ ي‬B) ‫إذن ا ل‬ [E]f [H ]2O f : ‫ رة ﺏ ا ازن ا‬-8K = [A]f .[B]f nf (E) nf (H2O)K= VV = nf (E) . nf (H2O ) nf (A) nf (B) nf (A) . nf (B) VVK = ( 0.25 - xf .xf - x ) = x 2 )2 xf ) ( 0.25 (0.25 f f - xf (0.167)2K = (0.25 - 0.167)2 ≈4‫وا ل‬ ‫ و آ ﻥ ا ء ﺕ آ ا‬، (1) ‫ا ﺕ‬ ‫ إذا ﺏ ف ا ء ر ا‬-9 .‫ا‬ : ‫ و ﻡ ن‬CnH2n+1COOH : ‫ه‬ : ‫א نא‬M(A) = d . 29 = 2.07 . 29 ≈ 60 g/mol :‫ا‬ ‫ ا ا‬-1 ‫ا ﺏآ‬ ‫ا اﻡ‬ : ‫و ﻡ ﺝ أﺥ ى‬M(A) = 12 n + 2n + 1 = 12 + 16 + 16 + 1 = 14n + 46 : ‫وﻡ‬14n + 46 = 60 → n = 60 - 46 = 1 . CH3COOH : ‫( ه‬A) ‫إذن ا ا ا‬ 14 : . ،، :– –

10 : 2011/2010 : - (01) – - : ‫ ﻡ د ا‬-‫أ‬-2 : ‫ ﺹ‬2-‫ا ﺏ ﻥ ل‬ CH3-CHOH-CH3 : ‫وﻡ ا د‬ CH3COOH + CH3-CHOH-CH3 = CH3COO-CH-CH3 + H2O | CH3 :‫ا‬ ‫هاا ه ﺕ أ ة ﺏ‬ . ( ‫ﻡ ود ) ﺕ م‬ . ‫اري‬ .‫س‬ .‫ﺏ ء‬ : ‫ا ﺏ ا ﻡ وي ا ت ن‬ : ‫ ﻥ ا م ا‬-‫ب‬ ‫ آ ل ﻥ ي و آ ن أن ا‬2-‫ا وﺏ ﻥ ل‬r = 60 % → τf = 0.60 : ‫ ﺏ ا ازن ا‬-‫ﺝـ‬K = [E]f [H ]2O f : ‫ﻥ ﺝ ول ا م‬ [A]f .[B]f nf (E) nf (H2O)K= VV = nf (E) . nf (H2O ) nf (A) nf (B) nf (A) . nf (B) VV ‫اما‬ A +B = E+ D ‫ اﺏ ا‬x = 0 0 220 x ‫ اﻥ‬x xf ‫ ﻥ‬xf 2-x 2-x x 2 - xf 2 - xf xf : ‫ﻡ ﺝ ول ا م‬nf(E) = nf(H2O) = xf : ‫رة ﺏ ا ازن ا‬ ‫ﺏ‬nf(A) = nf(B) = 2 - xfK= xf 2 (2 - xf )2 : ‫ﻡ ﺝ ول ا م و ﺏ ر ا ﺕ م ن‬2 – xmax = 0 → xmax = 2 molτf = xf → xf = τf xmax x max : . ،، :– –

11 : 2011/2010 : - (01) – -xf = 0.6 . 2 = 1.2 mol : ‫وﻡ‬ (1.2)2 : ‫ ﺝ ﺕ ر ا‬-3K = (2 -1.2)2 = 2.25 . K ‫ و‬Qri ‫ﻥ رن ﺏ‬[ ]Qri ‫ و ﻡ ا‬Qri > K= [E]0 H2O 0 ‫ ا آ ا‬-‫ب‬ [A]0 .[B]0 n0(E) n0(H2O)Qri = VV = n0 (E) . n0 (H2O ) n0 (A) n0 (B) n0 (A) . n0 (B) VV 4.3 . ( ‫إﻡ ه‬ ‫ )ﺕ‬. ‫ر ا ﺕ ا آ‬Qri = 1. 2 = 6 : ‫وث ا ازن‬ ‫اما‬ E + H2O = A + B ‫ اﺏ ا‬x = 0 4312 ‫ اﻥ‬x ‫ ﻥ‬xf 4-x 3-x 1+x 2+x 4 - xf 3 - xf 1 + xf 2 + xf .‫ا ة‬ : ‫إذا ا ﻥ‬ . ‫ا ﻡه‬ ‫ ﺏ ا ازن‬: K1 ‫ ﺏ ا ازن‬: K2 11K2 = K1 = 2.25 = 0.44 :‫ن‬ [A]f .[B]f nf (A) nf (B) : ‫ﻡ ﺝ ول ا م‬ [E]f . H2O f : ‫وﻡ‬[ ]K2= = nf V nf V = nf (A) . nf (B) (E) (H2O) nf (E) nf (H2O) VVnf (A) = 1 + xfnf(B) = 2 + xfnf(E) = 4 - xfnf(H2O) = 3 - xf : . ،، :– –

12 : 2011/2010 : - (01) – -K2 = (1 + xf ) ( 2 + xf ) = 0.44 → 2 + 3xf + xf 2 = 0.44 (4 - xf ) (3 - xf ) 12 - 7xf + xf 22 + 3xf + xf2 = 5.28 – 3.08xf + 0.44xf20.56xf2 + 6.08xf – 3.28 = 0∆ = 44.31 → ∆ = 6.66xf1 = - 6.08 - 6.66 = -11.4 (‫ض‬ ‫)ﻡ‬ 2 . 0.56xf2 = - 6.08 + 6.66 ‫ل‬ ‫ﻡ‬ ) = 0.52 mol ( : ‫وث ا ازن آ‬ ‫نا آ ا‬ ‫و‬ 2 . 0.56nf (A) = 1 + 0.52 = 1.52 molnf(B) = 2 + 0.52 = 2.52 molnf(E) = 4 – 0.52 = 3.48 molnf(H2O) = 3 – 0.52 = 2.48 molr = xf .100 : ‫ ﻡ دود ا‬-‫ﺝـ‬ x max : ‫ ﻡ ﺝ ول ا م‬-4–x=0 → x=4 : ‫ آ‬E ‫▪ إذا اﺥ‬3–x=0 → x=3 : ‫ آ‬H2O ‫▪ إذا اﺥ‬ : ‫ وﻡ‬xmax = 3 mol : ‫إذن‬ 0.52r = .100 = 17.3 % :‫ﻡ‬ 3 τ(‫ = )أ ة‬1 – τ( ‫)إﻡ ه‬ r (‫ = )أ ة‬100 – r( ‫)إﻡ ه‬ : ‫א نא א‬ : ‫ إ ا ا ﺕ‬-‫ أ‬-1 . ‫إ ﻥ ات ا‬ : ‫ ﺝ ول ا م‬-‫ب‬ ‫اما‬ CH3COOH + C2H5OH = CH3COOC2H5 + H2O ‫ اﺏ ا‬x = 0 0.2 0.2 0 0 ‫ اﻥ‬x ‫ ﻥ‬xf 0.2 - x 0.2 - x x x 0.2 - xf 0.2 - xf xf xf : . ،، :– –

13 : 2011/2010 : - (01) – - : ‫ ﻡ د ا ة‬-‫ﺝـ‬ CH3COOH + (Na+ + HO-) = (CH3COO- + Na+) + H2Ona = nb : V’be ‫ ﺏ‬n ‫ رة‬-‫ أ‬-2na = CV’be :‫ا‬na = 0.2 - xf : x -‫ب‬ : ‫ﻡ ﺝ ول ا م‬CV’be = 0.2 - xfxf = 0.2 – CV’be = 0.2 – ( 1 . V’be) : ‫وﻡ‬xf = 0.2 – V’be . ‫و ﻡ ﺥ ل ه ا ﻥ ا ول‬t(h) 04 8 12 16 20 32 40 48 66x (mol) 0 0.03 0.05 0.07 0.08 0.10 0.12 0.13 0.13 0.13 : x = f(t) ‫ ا ن‬-‫ﺝـ‬ 0.14 x(mol) 0.12 0.1 0.08 0.06 0.04 0.02 t (s) 0 0 10 20 30 40 50 60 70 : . ،، :– –

14 : 2011/2010 : - (01) – - : ‫ ﻥ ا م ا‬-‫ب‬τf = xf x max :‫ﺕم ن‬ . xf = 0.13 mol : ‫ ﻡ ا ن‬- ‫ ﻡ ﺝ ول ا م و ﺏ ض أن ا‬-0.2 – xmax = 0 → xmax = 0.2 molτf = 0.13 = 0.65 0.2 . ‫ﻥ أن ا ﺕ م‬[[ ] [ ] [ ]]Qrf nf (CH3COOC2H5) nf (H2O) : ‫ و‬Qrf ‫ رة‬-‫هـ‬ CH3COOC2H5 f H2O f = CH3COOH f C2H5OH f = V V ‫و ا دا ﺝ ول ا م‬ nf (CH3COOH) nf (C2H5OH) :‫א نא س‬ VV : ‫ ﺝ ول ا م‬-1Qrf = nf (CH3COOC2H5) nf (H2O) nf (CH3COOH) nf (C2H5OH) : ‫آﺏ‬Qrf = xf . xf → Qrf = xf 2 0.2 - ( 0.2 - xf ) ( 0.2 - xf ) ( x ) 2 f (0.13)2Qrf = ( 0.2 - 0.13)2 = 3.45 ‫اما‬ CH3COOH + C2H5OH = CH3COOC2H5 + H2O ‫ اﺏ ا‬x = 0 n0 n0 0 0 ‫ اﻥ‬x ‫ ﻥ‬xf n0 - x n0 - x x x n0 - xf n0 - xf xf xfn0 – xmax = 0 → xmax = n0 = 1 mol : ‫ا ما‬- : ‫ﺏ را ﺕ‬n = n0 – xfn’ = xf → n = n0 – n’ → n’ = n0 – n → n’ = 1 - n : n ‫ ﺏ‬n’ -2 : ‫ﻡ ﺝ ول ا م‬ : . ،، :– –

15 : 2011/2010 : - (01) – - t (h) 0 12 : ‫ إآ ل ا ول‬- n’ (mol) 0 0.39 0.55 34567 0.61 0.65 0.66 0.67 0.67 : n’= f(t) ‫ ا ن‬- 0.8 n'(mol) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 t (h) 0 0 24 68 :t=3h ‫ا‬ ‫ ا‬-4 : dx : t=3h ‫ﻡ ا نو ا‬v = dt ..................................... (1) . n’= f(t) ‫ ه ﻡ ﻡ س ا‬tanαtanα = dn' ‫ و ﻡ‬n’ = x : ‫ ﻡ و ﻡ ﺥ ل ﺝ ول ا م ن‬- dt : : ‫( ن‬2) ، (1) ‫ﻡا‬tanα = dx .................................. (2) ،، :– – dt :.

16 : 2011/2010 : - (01) – -vC = tanα :t=3h ‫ﻡ ا ن ا‬tanα = 2.8 → v = 0.47 mol/h 6 ≈0.47 . ‫ﺕ ول إ ا ازن ز ا ﺕ إ أن ﺕ م‬ ‫ا ا‬-τf = xf : ‫ ا ا م‬-5 x max :‫ﻡ ا ن‬n’ = xf = 0.67 mol : ‫ﻡ ﺝ ول ا م و ﺏ ر ا ﺕ‬1 - xmax = 0 → xmax = 1 mol ‫ﻥ أن ا ﺕ م‬τf = 0.67 = 0.67 <1 1 ** ‫ ﻥ رس‬: ‫** ا ذ‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ - ‫ا وب‬ [email protected] Tel : 0771998109 . ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و‬ ‫وﺵ ا ﻡ‬ : ‫ أدﺥ ﻡ ا ذ‬. ‫ﻥ ﻡ ه ا ا ع و‬ sites.google.com/site/faresfergani : . ،، :– –

‫اا‬ ‫را ا ا‬ ‫ا‬ ‫ﻥ ﻡد‬‫– ا وب‬ ‫ﻥﺏ‬ ‫وزارة ا ﺏ ا‬‫ا ذ ‪ :‬ﻥ رس‬ ‫دة ا م ا‬ ‫إن‬ ‫ا‬ ‫‪:‬ا ما وا‬ ‫ا‬ ‫ا ة‪ 3:‬ت‬ ‫م‪3:‬عت‪،‬ر‪،‬تر‬‫‪Sujet : 3AS 06 - 02‬‬ ‫‪:‬ا ر آ ‪.‬‬ ‫ا يا‬‫ا ا را ‪2011/2010 :‬‬ ‫‪2011/05/31 :‬‬ ‫ر‬ ‫א ن א ول ‪ ) :‬ﺏ ر ‪ – 1997‬م و ة – ب ( )**(‬‫ا ﻡ ‪ 0.4 mol ، CnH2nO2‬ﻡ ﻥ أآ‬ ‫ـ ‪ 0.1 mol‬ﻡ أ )‪(E‬‬ ‫‪ -1‬ا اق ا م ﺏ آ‬ ‫ا ﺏ ن‪.‬‬ ‫أ‪ -‬أآ ﻡ د ا اق ا ‪.‬‬ ‫ب‪ -‬ا ا ا )‪. (E‬‬ ‫إﻡ ه ا )‪ (E‬ا ﻥ و آ ل )‪ (B‬ﻡ أ دي ا ‪.‬‬ ‫‪-2‬‬ ‫ل )‪. (B‬‬ ‫أ‪ -‬ا ا ا‬ ‫ب‪ -‬أآ ا ﻥ ا ‪.‬‬‫ﻡ ﻡ وي ا ت ي ‪ 0.2 mol‬ﻡ ا )‪(E‬‬ ‫)‪، (E‬‬ ‫اا‬ ‫‪-3‬‬‫وا ﻡ ا ت‬ ‫و ‪ 0.2 mol‬ﻡ ا ء ‪ ،‬و ﺏ ﺕ ر ا ا ي ي ﺵ وط ﻡ ‪،‬‬ ‫ا ا ﺕ ﺕ ات د ﻡ ت ا ا ﺏ ا ﻡ ‪.‬‬ ‫)‪n(E) (mol‬‬ ‫)‪n(E) (mol‬‬ ‫)‪(2‬‬ ‫)‪n(E) (mol‬‬ ‫)‪(3‬‬ ‫‪0.20‬‬ ‫‪0.20‬‬ ‫)‪0.20 (1‬‬ ‫‪0.12 0.14‬‬ ‫‪0.13‬‬ ‫)‪t (h) t (h‬‬ ‫)‪t (h‬‬‫أ‪ -‬ﻡ ه ﻡ ﺏ ا ت ا ‪ ،‬ا ا ي ﺕ ات د ﻡ ت ا )‪ (E‬ا ؟ ﺏ ﻥ‬‫ا ا ﺕ إﻡ ه ا ‪ 80 mL‬ﻡ ﻡ ل ا د ا ي ﺕ آ ‪. 1 mol/L‬‬ ‫ا ازن ﻡ‬ ‫)‪ (E‬و ﻡ إ ‪.‬‬ ‫ب‪ -‬ﻡ ه ا ا‬ ‫א ن א ‪(*) :‬‬‫ي )‪ (E‬آ ﺏ ر ﺏ اء‬ ‫‪ -1‬إن ا ا دي ) ﺏ ا ا د ‪(NaOH‬‬ ‫‪ d = 4‬أ ﻡ آ )‪ ، (C‬و آ ل )‪ (B‬ي ‪ % 21.62‬أآ ‪.‬‬ ‫آ ت )‪. (B) ، (C) ، (E‬‬ ‫أ‪ -‬أآ ا ا ا‬ ‫آ‪.‬‬ ‫ل )‪ (B‬ﻡ ذآ ا و ا‬ ‫ﻥا‬ ‫ب‪ -‬أآ ا ا‬ ‫ﺝـ‪ -‬أآ ﻡ د ا ‪.‬‬‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪2:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫ات ﻡ‬ ‫ا ﻥ ‪،‬ﻥ‬ ‫‪ -2‬ﻥ ﻡ ﻡ وي ا ت ن ﻡ ا آ )‪ (B‬و‬ ‫ﺕ ‪ 11.6 g‬ﻡ ا‬ ‫ﻥا‬ ‫م ﻡ درﺝ ارﺕ ﺏ ‪ .‬ﻥ‬ ‫اآ ﻥ‬ ‫ا‬ ‫آ‪.‬‬‫ا وث‬ ‫اﻥ ‪.‬‬ ‫و ﺕ ‪114 g‬ﻡ‬ ‫أ‪ -‬أآ ﻡ د ا ا ‪.‬‬ ‫ل )‪ (B‬ﻡ ذآ ا و ا‬ ‫ب‪ -‬أآ ا ا ﻥ ا ا‬ ‫ا ﺏ ا وا آ‬ ‫ﺝـ‪ -‬ﻡ ﺝ ول ا م ا ة أوﺝ ا آ ا‬ ‫ا ازن ا ‪.‬‬ ‫ا لا ‪.‬‬ ‫د‪ -‬ا ﻡ دود ا و ا‬ ‫)‪. (E‬‬ ‫ﻥا‬ ‫هـ‪ -‬أآ ا ا‬ ‫‪:‬‬ ‫‪M(O) = 16 g/mol/L ، M(H) = 1 g/mol ، M(C) = 12 g/mol‬‬ ‫‪ ) :‬اﻡ ن ا ا ‪(**) ( 2010/2009 -‬‬ ‫א نא‬ ‫اﻥ ‪،‬‬ ‫ا‪:‬‬ ‫ﻥ ﺥ ﻡ وي ا ت ي ‪ 2,0×10-2 mol‬ﻡ آ ﻡ ا‬ ‫ا ﻥ ‪ ،‬إ ﻥ ات ا د م و ﻡ ﻥ ات ا د م ﻡ أﺝ ا ل ﻡ ل ‪. V = 100mL‬‬‫ا ء ﻡ ذآ ا ت )أ س‪( /‬‬ ‫اﻥ و اﻥ‬ ‫‪ -1‬أآ ﻡ د اﻥ ل آ ﻡ‬ ‫ا اﺥ ا ‪.‬‬ ‫ا ﻥ و ﺵ ارد ا ﻥ ات ‪. CH3COOH‬‬ ‫‪ -2‬أآ ﻡ د ا ﺏ‬ ‫د هاا ‪.‬‬ ‫‪ -3‬أ ﺏ ا ازن ا ا‬ ‫‪ -4‬أ آ ا ‪ Qri‬ا ا ﺏ ا ‪.‬‬ ‫ا ﻥ أم اﺕ ﺕ ؟‬ ‫‪ -5‬ه ا ﺕ ر اﺕ ﺕ‬ ‫‪:‬‬ ‫‪( ) ( )pKa1 HCOOH / HCOO = 3,8 pKa2 CH3COOH / CH3COO = 4,7‬‬ ‫א ن א א ‪ ) :‬ﺏ ر ‪ - 2009‬ر ت ( )**(‬ ‫ج ا ل ا ا ي ﺕ د ﺏ ذي ا د ‪:‬‬ ‫)‪Al(s) + 3Ag+(aq) = Al3+(aq) + 3Ag(s‬‬‫ا د اﺵ ﺕ را آ ﺏ ﺵ ﺕ ﺏ ‪ I = 40 mA‬ﺥ ل ﻡ ة زﻡ ‪ ∆t = 300 min‬و ث ه‬ ‫ارد ‪. Ag+‬‬ ‫اآا‬ ‫ﺕ‬ ‫‪ /1‬د ا د ؟ ﺏ ر إﺝ ﺏ ‪.‬‬ ‫اﺕ ا ر ا ﺏ و اﺕ آ ا وﻥ ت ‪.‬‬ ‫‪ /2‬ﻡ ﺏ ه ا ا د ﻡ‬ ‫أآ رﻡ ا د ) ال إ (‬ ‫ا‪.‬‬ ‫‪ /3‬اآ ا د ا‬ ‫ا د ﺥ ل ‪ 300 min‬ﻡ ا ‪.‬‬ ‫‪ /4‬ا آ ا ﺏ ء ا‬ ‫‪ /5‬ﺏ ﻥ ﺏ ول ﺕ م ا و ﺏ ﻡ ة زﻡ ‪ ∆t = 300 min‬ﻡ ا ﺵ ل ‪:‬‬ ‫أ‪ /‬ا م ‪. x‬‬ ‫ب‪ /‬أ ا ن ))‪ (∆m(Al‬آ ﻡ ى ا م ‪.‬‬ ‫‪. 1F = 96500 C ، MAL = 27 g.mol-1 :‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪3:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫א ن א س ‪(**) :‬‬‫ز‬ ‫ﻡا ا‬ ‫ات ا )‪ (Ag+ + NO3-‬ﺏ ل ﻡ‬ ‫‪ -I‬ﻥ آ ﺏ ﻡ‬ ‫ا و ن را ﻡ ﻥ ا ‪.‬‬ ‫اآ‬ ‫‪ -1‬ار ﺵ ﺕ ﺏ ‪.‬‬‫ا د ا ‪ ،‬أآ ﻡ د ا ا ﺝ ا ﺝ ل ا ي‬ ‫‪ -2‬أآ ا د ا‬ ‫ا دث أ ء ا ا ﺏ ‪.‬‬ ‫‪ -3‬ﻡ ﺝ ول ﺕ م ه ا ا ‪.‬‬ ‫ا ارة ﺕ ر ﺏ ﺵ ﺕ ‪ 2.4 A‬ﺥ ل ‪ 30‬د ‪ ،‬ﺥ ل ه ا ة ا ﻡ ‪:‬‬ ‫‪-4‬‬ ‫أ‪ -‬أ آ ا ﺏ ء ‪ Q‬ا رة و ء ا ‪.‬‬ ‫ب‪ -‬أ ا م ا ي ا ‪.‬‬ ‫ﺝـ‪ -‬أ آ ا ا ‪.‬‬ ‫ا وط ا ي ن ‪. VM = 24 L/mol‬‬ ‫زا آ ا‬ ‫د‪ -‬أوﺝ‬ ‫ا‪،‬‬ ‫ل آ ت ا س )‪ (Cu2+ + SO42-‬ا ﺏ‬ ‫‪ -II‬ﻥ م ﺏ ا ﺏ‬‫‪ ،‬و ﻥ أي‬ ‫ﺏا أ‬ ‫ﺕآ وا‬ ‫ﻡ ا س ﺏ ﻡ ة ﻡ ا ﻥ أن ا‬ ‫ا‬ ‫نا ل‪.‬‬ ‫اﻥ ق زات و ﺕ‬ ‫‪ .‬ﻡ ذا ﺕ ؟‬ ‫ا‬ ‫‪ -1‬أآ ﻡ د ت ا ا‬ ‫ا ي ا دث أ ء ا ا ﺏ ‪.‬‬ ‫‪ -2‬أآ ﻡ د ا ا ﺝ ا ﺝ ل ا‬‫و ا ن ا اد ﺕ ‪،‬‬ ‫ا دن ﻡ ا ا رأ أ‬ ‫هاا عﻡ ا‬ ‫‪-3‬‬ ‫ﻡ ا ح‪.‬‬‫ﻥ آ ‪ m = 10 g‬و ا ي ي ‪ 2%‬ﻡ ا ا ا ﻡ آ ة ‪ ،‬ﺵ ة ا ر‬ ‫‪ -4‬ﻥ ﺕ‬ ‫ا ﺏ ‪. I = 2.5 A‬‬ ‫أ‪ -‬ﻡ ﺝ ول ا م ا ا ‪.‬‬ ‫هاا ‪.‬‬ ‫ب‪ -‬ﻡ ه ا ة ا ﻡ ا زﻡ‬ ‫‪:‬‬ ‫)‪(Ag+/Ag) ، (O2/H2O‬‬ ‫‪. 1F = 96500 C/mol ، M(Ag) = 108 g/mol ، M(Cu) = 63.5 g/mol‬‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪4:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫أﺝ ﺏ‬‫‪Sujet : 3AS 06 - 02‬‬ ‫رآ ‪.‬‬ ‫ا يا ‪ :‬ا‬ ‫‪3n - 2‬‬ ‫א ن א ول ‪:‬‬ ‫‪CnH2nO2 + 2 O2 = nCO2 + nH2O‬‬ ‫‪ -1‬أ‪ -‬ﻡ د ا اق ‪:‬‬ ‫‪:‬‬ ‫ب‪ -‬ا ا ا‬ ‫اما‬ ‫‪3n - 2‬‬ ‫‪nH2O‬‬ ‫‪CnH2nO + 2 O2 = nCO2 +‬‬ ‫‪ x = 0‬اﺏ ا‬ ‫‪0‬‬ ‫‪ x‬اﻥ‬ ‫ﺏ دة ‪0.1‬‬ ‫‪0‬‬ ‫‪nx‬‬ ‫‪ xf‬ﻥ‬ ‫‪n xf‬‬ ‫ﺏ دة ‪0.1 - x‬‬ ‫‪nx‬‬ ‫‪0.1 - xf‬‬ ‫ﺏ دة‬ ‫‪n xf‬‬ ‫‪-‬ا ﺕم ا ن ﻥ ا ‪:‬‬‫‪0.1 – xf = 0 → xf = 0.1 mol‬‬ ‫‪ nf(CO2) = 0.4 mol :‬و ﻡ ﺝ ول ا م ‪:‬‬‫‪nf(CO2) = n xf‬‬ ‫→‬ ‫‪n = nf (CO2 ) = 0.4‬‬ ‫‪=4‬‬ ‫‪xf 0.1‬‬ ‫)‪ (E‬ه ‪. C4H8O2 :‬‬ ‫إذن ا ا ا‬ ‫‪ -2‬ا ا ا ل ‪:‬‬‫ﻡ ﺥ ل ا ا م د ا ة أن د ذرات ا ﺏ ن ا ﻡ وي ع د ذرات ا ﺏ ن‬ ‫ا و ا ل وﻡ ‪:‬‬ ‫▪ ا ‪ :‬د ذرات ا ﺏ ن ه ‪. 4‬‬ ‫▪ ا ﻥ ‪ :‬د ذرات ا ﺏ ن ه ‪. 1‬‬ ‫إذن د ذرات آ ﺏ ن ا ل )‪ (B‬ذو ا ا ﻡ ‪ CnH2n+1OH‬ه ‪ (n = 3) 3‬و ﻡ ا ا‬ ‫ل ه ‪. C3H7OH :‬‬ ‫ا‬ ‫ب‪ -‬ا ﻥ ا ل ‪: B‬‬ ‫‪CH3 – CH2 – CH2OH‬‬ ‫‪CH3 – CHOH – CH3‬‬ ‫ﺏ وﺏ ن‪ -1-‬ول‬ ‫ﺏ وﺏ ن‪ -2-‬ول‬ ‫)آ ل ﻥ ي(‬ ‫)آ ل أو (‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪5:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫ات د ﻡ ت ا ﺏ ا ﻡ ‪:‬‬ ‫‪-3‬أ‪ -‬ا ا‬ ‫)ا ازن( و ا‬ ‫ﻥا‬ ‫دﻡ تا ا‬ ‫ا قﺏ ا ت‬‫‪nf(A) = CbVbE‬‬ ‫ﻥا ‪.‬‬ ‫دﻡ تا ا‬ ‫ااﻥ‬‫‪nf(A) = 0.08 mol‬‬ ‫‪-‬ا ‪:‬‬ ‫و ه د ﻡ ت ا ا )‪ (A‬ﻥ ا ‪.‬‬ ‫‪ -‬ﻥ ﺝ ول ا م ا ﻡ ه ‪:‬‬ ‫اما‬ ‫‪E + H2O = A + B‬‬ ‫‪ x = 0 0.2‬اﺏ ا‬ ‫‪0.2‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪ x 0.2 - x 0.2 - x‬اﻥ‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪ xf 0.2 - xf 0.2 - xf‬ﻥ‬ ‫‪xf‬‬ ‫‪xf‬‬ ‫ﻡ ﺝ ول ا م ‪:‬‬‫‪nf(A) = xf → xf = 0.08 mol‬‬ ‫ﻡ ﺝ ول ا م أ ‪:‬‬‫‪nf(E) = 0.2 – xf = 0.2 – 0.08 = 0.12 mol‬‬ ‫)‪ (1‬ا ي‬ ‫ﺕا ا‬ ‫ها‬ ‫ﺕ ات د ﻡ ت ا ﺏ ا ﻡ ‪.‬‬ ‫‪:‬‬ ‫ا‬ ‫ﺝـ‪ -‬ا‬ ‫ل و ذ ﺏ ب ﻡ دود ا ﻡ ه ‪.‬‬ ‫ا‬ ‫ﻥ‬ ‫ا‬ ‫ا‬ ‫ن‪:‬‬‫‪r = xf .100‬‬ ‫‪x max‬‬‫‪0.2 - xmax = 0 → xmax = 0.2 mol‬‬ ‫ﻡ ﺝ ول ا م و ﺏ ض أن ا ﺕ‬ ‫وﻡ ‪:‬‬ ‫‪0.08‬‬ ‫ه‪:‬‬‫‪r = .100 = 40%‬‬ ‫إذن ا ل ﻥ ي و ا ا‬ ‫‪0.2‬‬ ‫إ ‪ :‬ﻡ ﻥ ات ﻡ إ‬ ‫أو ‪ :‬ﻡ ﻥ ات إ وﺏ وﺏ‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪6:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫א نא ‪:‬‬ ‫‪ -1‬ا ا ا ـ ‪: B ، C ، E‬‬ ‫ا ‪:E‬‬‫‪▪ M(E) = d . 29 = 4 . 29 = 116 g/mol‬‬ ‫ه ‪ CnH2nO2 :‬وﻡ ن ‪:‬‬ ‫‪-‬ا ا ﻡ‬‫‪M(E) = 12 n + 2n + 32 = 14n + 32‬‬ ‫وﻡ ‪:‬‬‫‪14n + 32 = 116‬‬ ‫→‬ ‫‪116 - 32‬‬ ‫‪n= =6‬‬ ‫ا ه ‪. C6H12O2 :‬‬ ‫إذن‬ ‫‪14‬‬ ‫ا ل )‪: (B‬‬ ‫ا ل ا ﻡ ‪ CnH2nOH‬أو ‪ CnH2n+2O‬ا ن ‪:‬‬‫‪M(B) 12n 2n +1 16‬‬ ‫==‬ ‫=‬‫‪100% C% H% O%‬‬‫‪M(B) 16‬‬ ‫=‬‫‪100% O%‬‬‫)‪M(B‬‬ ‫‪16‬‬ ‫→‬ ‫‪16 .100‬‬ ‫‪≈ 74 g/mol‬‬ ‫=‬ ‫= )‪M(B‬‬‫‪100% 21.62 %‬‬ ‫‪21.62‬‬ ‫و ﻡ ﺝ أﺥ ى ‪:‬‬‫‪M(B) = M(CnH2n+1OH) = 12n + 2n + 1 + 16 + 1 = 14n + 18‬‬ ‫وﻡ ‪:‬‬‫‪14n + 18 = 74‬‬ ‫→‬ ‫‪74 -18‬‬ ‫‪n= =4‬‬ ‫‪14‬‬ ‫ل )‪ (B‬ه ‪. C4H9OH :‬‬ ‫إذن ا ا ا‬ ‫دي )ﻡ ا د( ﺕ ن آ ‪:‬‬ ‫ا‬ ‫ا آ )‪: (C‬‬ ‫‪RCOOR’ + NaOH = RCOONa + R’OH‬‬ ‫ا داﻡ‬ ‫)‪(E) (C) (B‬‬‫د ذرات ا ﺏ ن ا )‪ (E‬ﻡ و د ذرات ا ﺏ ن ا آ )‪ (C‬ﻡ ف إ د ذرات‬ ‫و‬ ‫ا ﺏ ن ا ل )‪ (B‬و ﻡ ‪:‬‬ ‫▪ د ذرات آ ﺏ ن ا )‪ (E‬ه ‪6 :‬‬ ‫▪ د ذرات آ ﺏ ن ا ل )‪ (B‬ه ‪4 :‬‬ ‫)‪ (C‬ا ﻡ ‪ RCOONa‬أي ‪ CnH2n+1Na‬ﺕ ن ا‬ ‫إذن د ذرات آ ﺏ ن )‪ (C‬ه ‪ 2‬و آ ن أن‬ ‫)‪ (C‬ه ‪. CH3COONa :‬‬ ‫عا‬ ‫ا‬ ‫ﺝـ‪ -‬ﻡ د ا ‪:‬‬‫)‪ (E‬آ ‪:‬‬ ‫آﺏ ا ا ا‬ ‫اﻡ ا ﺏ‬ ‫ا دا ﻡ د ا‬ ‫آ‪:‬‬ ‫‪ CH3COOC4H9‬وﻡ ﺕ ن ﻡ د ا‬ ‫‪CH3COOC4H9 + NaOH = CH3COONa + C4H9OH‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

7 : 2011/05/31 - (02) – - CH3-CH2-CH2-CH2OH : (B) ‫ل‬ ‫ﻥا‬ ‫ا‬ ‫ ا‬-‫ب‬ ‫ ول‬-1-‫ﺏ ﺕ ن‬ CH3-CH2-CHOH-CH3 ( ‫)آ ل أو‬ ‫ ول‬-2-‫ﺏ ﺕ ن‬ (‫)آ ل ﻥ ي‬ CH3-CH-CH2OH CH3-COH-CH3 CH3 CH3 ‫ ﻡ‬-2 ‫)آ‬ ‫ ول‬-1-‫ ﻡ ﺏ وﺏ ن‬-1 ‫ ول‬-2-‫ﺏ وﺏ ن‬ ( ‫)آ ل أو‬ (‫ل‬ :‫م‬ ‫ ﺝ ول ا‬-‫ﺝـ‬ ‫اما‬ A+B =E = H2O ‫ اﺏ ا‬x = 0 0 n0 n0 0 x ‫ اﻥ‬x n0 - x n0 - x x xf ‫ ﻥ‬xf n0 - xf n0 - xf xf ‫اﺏا و‬ : ‫ا ازن‬ ‫ا آ ا‬-▪ nf (E) = mf (E) : ‫ول ا م‬ M(E) ‫إذن ا آ ا‬M(E) = 116 g/mol 11.6nf (E) = 116 = 0.1 mol▪ nf (A) = mf (A) M(A)M(E) = 116 g/mol 114nf (E) = 60 = 1.9 mol▪ nf(E) = xf → xf = 0.1 mol ‫اﺏا و‬▪ nf(A) = n0 – xf → n0 = nf(A) + xf = 1.9 + 0.1 = 2 mol : ‫ا ازن ن ا ل ا‬ ‫ﻡ اﺏ ا‬ ‫ا ازن‬ n(A) 2 mol 1.9 mol n(B) 2 mol 1.9 mol n(C) 0 0.1 mol n(D) 0 0.1 mol : . ،، :– –

8 : 2011/05/31 - (02) – - ‫ ﻡ دود ا‬-‫د‬r = xf .100 ‫اآ‬ :‫ل‬ ‫وا ا‬ x max :‫ﺕ‬ ‫ﻡ ﺝ ول ا م و ﺏ ض أن ا‬n0 – xmax = 0 → xmax = n0 = 2 mol ‫اﻥ‬ 0.1 ‫إذن ا ل ا‬r = .100 = 5% 2 : CH3-COH-CH3 CH3 : (E) ‫ﻥا‬ ‫ا‬ ‫ ا‬-‫هـ‬ ‫ا اﺕ‬ ‫ﺏ ءا‬: ‫اﻥ آ‬ ‫ﺕ هاا لﻡ‬ ‫ا لا ﺏ ﺕ ن‬ CH3 CH3COO-C-CH3 CH3 : ‫א نא‬ : ‫ا ءوا تا ا‬ ‫ ﻡ د اﻥ ل ا ﻥ و ا ﻥ‬-1 HCOOH + H2O = HCOO- + H3O+ /‫ا ت )أ س‬ CH3COOH + H2O = CH3COO- + H3O+ : ‫( ا اﺥ ا ه‬ HCOOH/HCOO- , CH3COOH/CH3COO- : ‫ا ﻥ و ﺵ ارد ا ﻥ ات‬ ‫ ﻡ د ا ﺏ‬-2: ‫ا وﺕ ﻥ‬ ‫ا ﻥ ﺵ ارد ا ﻥ ات اﻥ ﻡ ا د ا‬ ‫ﺏ‬ ‫ﻡد ا‬ ‫ﻥ‬ CH3COO-(aq) + HCOOH(aq) = CH3COOH(aq) + HCOO-(aq) : ‫ا ازن‬ ‫ ﺏ‬-3[ ] [ ] [ ] [ ] [[ ]][ ] [ ]K= HCOO‫ ـ‬f CH3COOH f → K = HCOO‫ ـ‬f CH3COOH f H3O+ f f f H3O+ f [HCOOH]f CH3COO‫ـ‬ [HCOOH]f CH3COO‫ـ‬[ ] [ ] [ ][ ][ ]K = HCOO‫ ـ‬f H3O+ f CH3COOH f = Ka1 CH3COO‫ ـ‬H3O+ f Ka 2 [HCOOH]fpKa1 = 3.8 → Ka1 = 10-3.8 = 1.58 . 10-4pKa2 = 4.7 → Ka2 = 10-4.7 = 2.00 . 10-5 : . ،، :– –

‫‪9:‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫‪1.58 . 10-4‬‬ ‫ا اﺏا ‪:‬‬ ‫‪ -4‬آ ا‬‫‪K = 2.00 .10-5 = 7.9‬‬‫‪[[ ] ][ ]Qr,i‬‬ ‫‪2,0 ×10 2‬‬ ‫‪2,0 ×10 2‬‬‫=‬ ‫‪ i‬ـ‪HCOO‬‬ ‫‪CH3COOH i‬‬ ‫=‬ ‫‪V‬‬ ‫‪V‬‬ ‫‪=1‬‬ ‫ـ ‪CH3COO‬‬ ‫‪i‬‬ ‫‪[HCOOH]i‬‬ ‫‪2,0 ×10 2‬‬ ‫‪2,0 ×10 2‬‬ ‫‪V‬‬ ‫‪V‬‬ ‫‪ -5‬ﺝ ﺕ ر ا ‪:‬‬‫‪ : Qr,i < K‬آ ا ﻥ ﺏ ا ازن و داد إ أن ‪ K‬و ﺕ ر ا ا ﺕ‬ ‫اﻥ‪.‬‬ ‫ا ﺵ ‪ ،‬إذن‬ ‫א نא א ‪:‬‬ ‫د‪:‬‬ ‫‪ -1‬ﺕ ا‬ ‫ﻡا د‬ ‫‪ Ag+‬أرﺝ ‪ ،‬و ﻥ ﻥ أن ا‬ ‫أن ﺵ ارد ا‬ ‫أن ا م ‪ Al‬ﺕ آ‬‫إرﺝ ع ا )ا ا ( ‪ ،‬إذن‬ ‫ا ﺏ ﺕث‬ ‫)ا ا ﺝ ( ‪ ،‬و‬ ‫أآ ة ا‬ ‫ﻡ ىا م‬ ‫ا ﺝ‪.‬‬ ‫د‪،‬وﻡ ىا‬ ‫اا‬ ‫‪ -2‬ا ‪:‬‬ ‫▪ ﺕ ن ﺝ ا ر ﻡ ﻡ ى ا ﻥ ﻡ ى ا م )ﺥ رج ا د( ‪.‬‬ ‫• رﻡ ا د ‪:‬‬ ‫)‪(-) Al3+/Al // Ag+/Ag (+‬‬ ‫‪ -3‬ا د ا ‪:‬‬ ‫‪(-) Al = Al3+ + 3e-‬‬ ‫‪(+) 3Ag+ + 3e- = 3Ag‬‬ ‫ا د ﺥ ل ‪: 300 min‬‬ ‫‪ -4‬آ ا ﺏ ء ا‬‫‪Qِ = I . ∆t‬‬‫‪Q = 40 . 10-3 . 300 . 60 = 720 C‬‬ ‫‪ -5‬أ‪ -‬ا م ‪: x‬‬ ‫ﻥ )‪ x(300‬ه ﻡ ار ا م ﻡ ور ‪ 300 min‬ﻡ اﺵ ل ا د ‪ ،‬و )‪ Q(300‬ه آ ا ﺏ ء ا‬ ‫ا د ه ا ةا ﻡ ‪ .‬ن‪:‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

10 : 2011/05/31 - (02) –-Q(300) = z . x(300) . F → x (300) = Q(300) ‫ﻥ أن د ا‬ z.F‫ ا‬3 ‫ا آ و ا ﺝ ﻡ أﺝ ﺕ وا ه‬ ‫دﺏ‬ ‫وﻥ ت ا‬ : ‫ﻡا د ا‬ Al3+ + 3Ag : ‫ وﻡ‬z = 3x (300) = 3 . 720 = 2.5 .10-4 mol 00 96500 x 3x ‫ ا ن ا‬-‫ب‬ xf 3xf : ‫ﻥ ﺝ ول ا م‬ ‫اما‬ Al + 3Ag+ = ‫ اﺏ ا‬x = 0 n1 n2 ‫ اﻥ‬x ‫ ﻥ‬xf n1 - x n2 - x n1 - xf n2 - xf : ‫ ه‬n(300)(Al) ‫ و‬300 min ‫ﻡ ﺝ ول ا م د ﻡ ت ا م ا )ا ( ﻡ ور‬n(300)(Al) = x(300) = 2.5 . 10-4 moln(300) (Al) = ∆m → ∆m = n(300) (Al). M(Al) M(Al)∆m = 2.5 .10-4 . 27 = 6.75 .10-2 g = 67.5 mg : ‫א نא س‬ : ‫ ا‬-1 -I II O2 NO3- Ag : ‫ ا د ا‬-2 :‫ا‬ Ag+ :‫ا‬ I : ‫ا د اﺝ‬ 4Ag+ + 4e- = 4Ag 2H2O = 4H+ + O2 + 4e- 4Ag+ + 2H2O = 4Ag + 4H+ + O2 : . ،، :– –

‫‪11 :‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫‪ -3‬ﺝ ول ا م ‪:‬‬ ‫ام ا‬ ‫‪4Ag+ + 2H2O = 4H+ + 4Ag + O2‬‬ ‫‪ x = 0‬اﺏ ا‬ ‫‪n0‬‬ ‫ﺏ دة‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪ 4x‬ﺏ دة ‪ x n0 - x‬اﻥ‬ ‫‪4x‬‬ ‫‪x‬‬ ‫ﻥ‬ ‫‪xf‬‬ ‫‪n0 - xf‬‬ ‫ﺏ دة‬ ‫‪4xf‬‬ ‫‪4xf‬‬ ‫‪xf‬‬‫‪Q = I ∆t‬‬ ‫‪ -4‬أ‪ -‬آ ا ﺏ ء ‪ Q‬ا رة ا ء ‪:‬‬‫‪Q = 2.4 . 30 . 60 = 4320 C‬‬ ‫ب‪ -‬ا م ا ي ا ‪:‬‬‫‪Q=z.x.F → x= Q‬‬ ‫ﺝـ‪ -‬آ ا ا ‪:‬‬ ‫‪z.F‬‬ ‫ﻡ ﺝ ول ا م د ﻡ ت ا‬‫‪x = 4320 = 1.12 .10-2 mol‬‬ ‫‪4 . 96500‬‬ ‫ا ه‪:‬‬ ‫‪ 30‬د ﻡ‬ ‫(ﺏ‬ ‫ا ﺕ )ا‬‫‪n(Ag) = 4x = 4 . 1.12 . 10-2 = 4.48 . 10-2 mol‬‬‫)‪n(Ag‬‬ ‫=‬ ‫)‪m(Ag‬‬ ‫→‬ ‫)‪m(Ag) = n(Ag). M(Ag‬‬ ‫)‪M(Ag‬‬‫‪m(Ag) = 4.48 .10-2 .108 = 4.48 g‬‬ ‫ﺵ وط ن ‪:VM = 24 L/mol‬‬ ‫زا آ ا‬ ‫ب‪-‬‬ ‫ا ا ﺏ ه‪:‬‬ ‫ﻡ ﺝ ول ا م د ﻡ ت ز ا آ ا ﺕ )ا ( ﺏ ‪ 30‬د ﻡ‬‫‪n(O2) = x = 1.12 . 10-2 mol‬‬‫‪n(O2‬‬ ‫)‬ ‫=‬ ‫‪m(O 2‬‬ ‫)‬ ‫→‬ ‫‪V(O2 ) = n(O2 ) . VM‬‬ ‫‪VM‬‬‫‪V(O2 ) = 1.12 .10-2 . 24 = 0.27 L‬‬ ‫ا‪:‬‬ ‫‪ -1 -II‬ﻡ د ت ا ا‬ ‫ا‪:‬‬ ‫ا ا ي ه ﻡ ا س ﺕ آ ‪ ،‬ه ا ل أن ﻥ س ا ﺕ آ و ا د ‪:‬‬ ‫‪Cu = Cu2+ + 2e-‬‬ ‫ا‪:‬‬ ‫ل ﺕ ا س ا ﺕ إرﺝ ع ﺵ راد ا س و ا د ‪:‬‬ ‫ا‬ ‫ﺕ ا با‬ ‫‪Cu2+ + 2e- = Cu‬‬ ‫ا‪:‬‬ ‫ا وآنا سا آ ا‬ ‫)ا آ ( ﺕ ﻡ ﺝ‬ ‫ا‬ ‫ﻥ أن ا س ا ي اﺥ‬ ‫ﻥإا ‪.‬‬ ‫ﺕ‬ ‫ا ي ا دث أ ء ا ا ﺏ ‪:‬‬ ‫‪ -2‬أآ ﻡ د ا ا ﺝ ا ﺝ ل ا‬ ‫‪Cu + Cu2+ = Cu2+ + Cu‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬

‫‪12 :‬‬ ‫‪2011/05/31‬‬ ‫)‪- (02‬‬ ‫–‬ ‫‪-‬‬ ‫‪،‬ا‬ ‫ﻥ ا سا آ إ ا‬ ‫(‬ ‫‪:‬‬ ‫‪ -4‬ﻡ ن و ا ن ا اد ﺕ‬ ‫ا‬ ‫ﻥ إ ا ﻥ ‪،‬ﺏ ﻥ‬ ‫ث‬ ‫)ا ا‬ ‫آ ذآ ﻥ ﺏ ‪ ،‬ا ه ا‬ ‫ﺕﻥ‬ ‫ﻡ ث ن )ا آ ( و ا‬ ‫و ءا ا ﺏ‬ ‫ﺝ ا ن ا اد ﺕ ه ﻡ‬ ‫ﺏ ا ا اي‬ ‫ا إا ﺏ ا‬ ‫‪.‬‬ ‫ﺝا ا نا يﺕ ﺕ‬ ‫‪ -4‬أ‪ -‬ﺝ ول ا م ‪:‬‬ ‫اما‬ ‫‪Cu‬‬ ‫‪+ Cu+2‬‬ ‫=‬ ‫‪Cu2+ + Cu‬‬ ‫‪ x = 0‬اﺏ ا‬ ‫‪n1‬‬ ‫‪n2‬‬ ‫‪00‬‬ ‫‪n1 - x‬‬ ‫‪xx‬‬ ‫‪ x‬اﻥ‬ ‫‪n1 - xf‬‬ ‫‪n2 - x‬‬ ‫‪xf xf‬‬ ‫‪ xf‬ﻥ‬ ‫‪n2 - xf‬‬ ‫‪:‬‬ ‫اا‬ ‫ب‪ -‬ا ة ا ﻡ ا زﻡ‬‫آ ا سا‬ ‫ﻥ ‪ 98%‬و ﺏ‬ ‫‪ ،‬نا سا‬ ‫ﺏ أن ا ا ﺕ ‪ 2%‬ﻡ آ ا س ا‬ ‫آ ذآ ﻥ ﺏ ه ‪:‬‬ ‫ا إ ا وا ي‬ ‫ا‬ ‫‪98‬‬ ‫ا‬ ‫و دا تا ا وا ﺕ أ دﻡ تا سا‬‫‪m(Cu) = 10 . = 9.8 g‬‬ ‫‪100‬‬ ‫ه‪:‬‬ ‫‪m(Cu) 9.8‬‬‫‪n(Cu) = M(Cu) = 63.5 = 0.15 mol‬‬ ‫ا ه‪:‬‬ ‫ا دا ﺝ ول ا م د ﻡ ت ا س ا‬ ‫و ﻡ ﺝ أﺥ ى ‪:‬‬‫‪n(Cu) = x → x = 0.15 mol‬‬‫‪Q = I.∆t = z . x . F → ∆t = z . x . F‬‬ ‫‪I‬‬‫‪∆t = 2 . 0.15 . 96500 = 11580 s = 3h ,13 min‬‬ ‫‪2.5‬‬ ‫** ا ذ ‪ :‬ﻥ رس **‬ ‫ﻥ ﺝ إﺏ‬ ‫ﻥ ﻡد ﻥ ﺏ‬ ‫ا وب ‪-‬‬ ‫‪[email protected]‬‬ ‫‪Tel : 0771998109‬‬ ‫ا ا وﻥ ﺏ ي ﺥ ا روس أو ا ر و ‪.‬‬ ‫وﺵ ا ﻡ‬ ‫ﻥ ﻡ ه ا ا ع و ‪ .‬أدﺥ ﻡ ا ذ ‪:‬‬ ‫‪sites.google.com/site/faresfergani‬‬ ‫‪:‬‬ ‫‪. ،،‬‬ ‫–‪:‬‬ ‫–‬