تصحيح الامتحان التجريبي الموحد : ثانوية طارق التأهيلية خنيفرة

( ‫ ان ا دة اء ) ر‬

ni ( I 2 ) = nr ( I 2 ) + n( I 2 ) ⇒ n( I 2 ) = ni ( I 2 ) − nr ( I 2 ) :  ‫و‬ :MB ‫ ا *د ا‬4%N ‫ دة‬B‫ آ‬4‫ ه‬n( I 2 )  n ( I 2 ) = n ( AH ) = 3, 35.10 − 5 mol 4) (P ‫ ا‬Q-  ‫ و‬10mL 4) ‫*ﺝ*دة‬B ‫ ا‬B= ‫ ا‬OB‫ ﺕ‬B= ‫ه?> ا‬ n( AH ).82 nt ( AH ) = = 2,89.10−4 mol . 78mL :*‫  ه‬2‫ا &(ﺕ‬ 10 . m = nt ( AH ).M ( AH ) = 0, 05 g : 4‫ ه‬M= ‫و ا‬ :‫  اازن‬-4 − p K A1 K A1 10 K = = = 1 0 − p K A 1 + p K A 2 = 1, 4 1 -1 K A 2 1 0 − p K A2 .‫*ر‬R‫ ﺕ‬I *B-B ‫ أي أن ا‬Qr ,i = K = 1, 41  -2

:‫ا!یء‬ 14 6

C.V − xm

--------

:‫وی‬#‫ ا‬$%‫ا‬ C → 174 N + − 10 e :S ‫  د  ا‬-1

. N (t ) = N o .e− λ .t :4 TU‫ ا‬VK ‫ﻥ*ن ا‬K -2 N . N (t 1 2 ) = o  JX ‫ ه* ا‬:WP ‫( ا‬B W( ‫ ﺕ‬-3 2 W(  ‫ ا‬73  − λ .t 1 N 1 N (t 12 ) = N o .e 2 = o ⇒ e − λ.t 12 = 2 2 ln(2) λ= ⇐ λ.t 1 2 = ln(2) : -‫ ﻥ‬ln ‫ل‬Y‫د‬Z‫و ﺏ‬ t 12 ln(2) .λ = = 1, 242.10−4 ans −1 : λ ‫ب‬3 5580ans :*‫  ﻥ_ط  ه‬-4 a(t ) = λ.N (t ) = λ.No .e−λ.t = ao .e−λ.t ⇒ a(t ) = ao .e−λ.t a(t ) = ao .e− λ .t  -5  ‫و‬

:‫اء‬ -1 − + C6 H 8O6 + H 2O C6 H 7O6 + H 3O : ‫  د  ا‬-1 n( AH ) m( AH ) AH = C6 H 8O6  C = =  -2 V M ( AH ).V 0,5 .C = = 1, 42.10−2 mol.l −1 :  ‫و‬ 176.0.2 -3 − + C 6 H 8O6 + H 2 O C 6 H 7 O6 + H 3O C.V %& ‫  ا‬$ ‫ا‬ ‫ﺏ*)(ة‬ 0 0 -------C.V − x f xf x f &(-‫ﺕ‬.‫ن‬. 

 a (t )  a (t ) = e − λ .t ⇒ ln   = −λ .t ⇐ ao  ao 

 a (t )  1 t = ln   . = 3341, 49ans  ao  λ ‫ أي‬3341, 49ans − 1995ans = 1346ans ‫ ش )(*ن‬-6 .a‫& ا ر‬K ‫ ﺱ‬1346 4) ‫اﻥ ش‬ :&‫اﻥ‬ :  J‫ *ﺕ‬ 4‫ﻥ‬O ‫ا‬ ‫ﻥ*ن‬2 ‫ا‬ b&R‫ﺏ‬ -r1 uuur ur ur : -‫ ﻥ‬AX ‫*ر‬$B ‫ ا‬LM c2‫ﺱ‬Z‫ و ﺏ‬P + R = m.a ‫ و‬Rx = 0 :‫ )ن‬MBd ‫=ك‬I‫ أن ا‬B‫ و ﺏ‬Px + Rx = m.ax

:  ‫ و‬Px = m.g.sin (α )

‫(آ‬$ ‫ و  ا‬ax = g .sin α = cte ⇐ m.g.sin α = m.ax . ax = 5m.s −2 : ‫رع‬3 ‫م و ا‬/‫(ة ﺏﻥ‬h B23 1 : x = a.t 2 + Vo .t + xo :X ‫ د  ا‬B ‫ ا‬-2 2 x = 2,5.t 2 :‫ أذن‬Vo = VA = 0 ‫ و‬xo = xA = 0  ‫ و‬VB = 5.tB :(3 ‫ب ا‬3 -3

xm

xm

:‫ )ن‬45* ‫ول ا‬- ‫ ا‬73 ‫ و‬τ =

Qr ,eq

(/‫ﻥ‬.‫ن‬.  xf xm

:‫م‬2 ‫& ا‬3‫ﻥ‬

 H 3 O +  ⇐ τ= ⇐ x f =  H 3O +  .V ‫ و‬xm = C.V C 10− pH .‫ود‬$  ‫ ⇐ ا‬τ = = 0, 07 = 7% C 2  H 3O +  .  A−   H 3O +  eq eq = = ‫ رج ا‬-3 C −  H 3O +  [ AH ]eq

(10 ) − pH

2

= 8,9.10−5 : ‫أي أن‬ C − 10− pH .. pK A = 4, 05 ⇐ K A = Qr ,eq = 8,9.10−5 - II : ‫ ﻡد ا‬1-2 C 6 H 8 O 6 + O H − → C 6 H 7 O 6− + H 2 O : <)= ‫  ا‬2-2 − >?‫ ه‬n(C6 H 8O6 ) = n(OH ) = CB .VBE = 2,88.10−4 mol CB ‫ دة‬B‫ ⇐ آ‬VA = 10mL 4) ‫ ﺕ*ﺝ‬4 ‫ ا‬B= ‫ ا‬4‫ ه‬B= ‫ا‬ ‫ و‬Vs = 100mL 4) ‫?اﺏ‬B ‫دة ا‬B ‫ ا‬B‫ آ‬4‫(ص ه‬2 ‫ ا‬4) F‫ﺱ=*رﺏ‬I‫ا‬  ‫ و‬Qr ,eq =

ns = 2,88.10−3 mol ⇐

2,88.10−4 mol → VA = 10mL

:  ns → Vs = 100mL :4‫(ص ه‬2 ‫ ا‬4) ‫*اﺝ‬B ‫ ا‬C J ‫ دة ا‬B‫آ‬ ns = 2,88.10−3 mol : mg ‫ ا ب‬3-2

ns =

m ⇒ m = ns .M ( AH ) = 0,5 g = 500mg M ( AH ) . 500mg LM ‫*ي‬$ C J) ‫(ص‬K ‫أي آ‬ : ni ( I 2 ) %& ‫ ا *د ا‬4%N ‫ دة‬B‫ آ‬:  -3

‫ أن‬B‫ ﺏ‬ni ( I 2 ) = C2 .V2 = 5,3.10−3.10.10−3 = 5,3.10−5 mol n( I 2 ) MB ‫ ا *د ا‬4%N ‫ دة‬B‫ ا *د *ﺝ ﺏ*)(ة )ن آ‬4%N . 2‫ ا &(ﺕ‬4) ‫*ﺝ*دة‬B ‫ ا‬F‫ﺱ=*رﺏ‬I‫ ا‬CB n( AH ) ‫ دة‬B‫وي آ‬3‫ﺕ‬ .MB ‫ ا *د ا‬4%N ‫ دة‬B‫د آ‬$ ::2&B ‫ ا‬nr ( I 2 ) ‫ ا *د‬4%N ‫ دة‬B‫ آ‬$‫ ﺕ‬J=B ‫ﻥ‬O ‫ (ة ا‬B ‫ﺏ‬ nr ( I 2 ) =

n( S 2O32− ) 5.10−3.7,8.10−3 = = 1,95.10−5 mol 2 2

:$%‫(ب ا‬+ -4

VB = 5m.s −2 : ‫ و‬. ‫ع‬.‫ ت‬t B =

2 xB = 1s ⇐ xB = 2,5.t B 2,5

1 ur r Ee1 = C.uC2 = 0.5.0, 2.10−6.102 = 10−5 J ⇐ t = 0  P = m . a :J‫ *ﺕ‬4‫ﻥ‬O ‫ﻥ*ن ا‬2 ‫ ا‬b&R‫ ﺏ‬: JX ‫ ا‬J ‫ د‬B ‫ ا‬1 -4 2 ax = 0  Px = m.ax = 0 Ee 2 = 0,5.0, 2.10−6.52 = 25.10−7 J ⇐ uC = 5V ‫ أن‬t = 4T  ⇐  a y = g  Py = m.a y = mg ∆Ee = Ee1 − Ee 2 r - II ‫*ر‬$B ‫ ا‬LM B/ B23 ( B, i ) ‫*ر‬$B ‫ ا‬LM ‫(آ‬$ ‫و  ا‬ r -1 x = V Bx t + x o :X ‫ د  ا‬B ‫ ( و  ا‬B, i ) ‫&دة‬B ‫ ا‬KR ‫ ا‬C* ‫ ﺏ‬F ‫ﻥ ا ?ﺏ?ﺏت و ذ‬5 *‫*  ه‬B ‫دور ا‬ r ‫م‬/‫(ة ﺏﻥ‬h B23 ( B, j ) ‫*ر‬$B ‫ ا‬LM ‫(آ‬$ ‫و ا‬ . r  T* ‫و ا‬2 4) ‫ *ل ﺝ*ل‬B‫ﺏ‬  ‫ د‬B ‫ ا‬L ‫ إ‬5*‫ ﻥ‬2‫ﺏ‬3 ‫ ا‬2(R ‫ ا‬t‫ ﺏ‬:Mr ‫ د  ا‬B ‫ ا‬-2 1 x y = a y t 2 + V By t + y o 2 d u 1 r − K duC 2 . 2C + + .uC = 0 ⇐ :  ‫ ا‬Mr ‫ا‬ ur dt L dt LC VB :  =_ ‫ ا‬73 . K = r ‫ أن =*ن‬7- -3 V B x = V B co s α x = 0 - III y ‫و‬ o t = 0   V = V sin α y = 0  B y B o   . uR (t ) OB Y2 ‫ و‬Z f R ‫ن‬I u (t ) OB Y1 -1  x = VB cos α .t U 4  x = 4,3.t   ‫ و‬Z o = m .R = .20 = 40Ω  -2 :  ‫و‬   1 2 U Rm 2 2  y = 5.t + 2,5.t  y = gt + VB sin α .t  2 . Z o = R + r ⇒ r = Z o − R = 40 − 20 = 20Ω x 1 ‫ و‬t= :n3‫ ﻥ‬x X ‫ د  ا‬B ‫ ا‬J :‫ ﻡد ا(ر‬2-4 ⇐ Lω = :J‫  ا (ﻥ‬: L BK ‫ب‬3 VB cos α Cω g 1 1 y= x 2 + x.tan α : ‫ إذن‬y  ‫ د‬B ‫ ا‬4) C*  ‫ﺏ‬ = = 0,5 H N o = 50 Hz  . L = 2 2 2 2 2VB .cos (α ) C.ω C.(2π .N o ) .Q-MT J ‫* &رة‬d) U Um . Io = = Rm = 0, 07 A :‫ة ا ر‬T -4 :‫*ط‬23 ‫ ة ا‬3-4 2.Z o 2.R :  ‫ و‬xI = HI = 4,3m  ‫ و‬x = 2,15.t : WO=B ‫ ا‬4R‫ (ﺏ‬J‫ ا *ﺕ( ا  ل ﺏ‬BK -5 x 4, 3 1 1 I R2 ‫ ا‬L ‫*ل إ‬5*M K(h3B ‫ة ا‬B ‫ ا‬4‫ ه‬t = = = 1s U c = Z c .I o = .I o = .I o 4, 3 4, 3 Cω C.2π .N o : )(‫ ا‬4-4 2 U L , r = r 2 + ( Lω ) .I o = :  T* ‫ ا‬4R‫ (ﺏ‬J‫ا *ﺕ( ﺏ‬ ur Vx = x& = 4,3 t = 1s ‫ )ن‬I R2 ‫  ا‬V  2 1  Vy = y& = g.t + VB sin α  ⇐ M ‫و‬ B J‫ﺏ‬ (‫ا *ﺕ‬ 6 Z BM = r2 +  Lω −  Cω  ur Vx = 4,3m.s −1 14243 V ‫و‬ I  =0 −1 Vy = 10.1 + 5.1 = 15m.s U BM = r.I o  ‫ و‬J‫ه(ة ا (ﻥ‬u VI = Vx2 + Vy2 = 4,32 + 152 = 15, 6m.s −1 :‫ا*)ء‬ uc + uL = 0 :  ‫) ا *ﺕ(ات‬r‫ﻥ*ن إ‬K b&R‫ ﺏ‬1-

d 2uC r duC 1 di + + .uC = 0 ⇐ uC + ri + LC = 0 ⇐ 2 dt L dt LC dt duc 1 1  ‫ و‬i = C ‫ و‬Et = C.uC2 + L.i 2 :M= ‫ ا‬KR ‫ ا‬-2 dt 2 2 2 1 1  duC  2 . Et = C.uC + L  C.  2 2  dt  dEt du d 2u duC = C.uC . + LC 2 . C . 2C :M= ‫ ا‬KR ‫ق ا‬2T‫ ﺏ‬-3 dt dt dt dt 2 d u uC . + LC 2C = −r.i :  Mr ‫ د  ا‬B ‫ ا‬73 dt dE du d 2u . t = C. C (uC . + LC 2C ) = i. ( −r.i ) = −r.i 2 ⇐ dt dt dt