إختبار الثلاثي الثاني في العلوم الفيزيائية للسنة...
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Transcript of إختبار الثلاثي الثاني في العلوم الفيزيائية للسنة...
-
5/19/2018 +
1/16
1
2014-2015
: :
:3
(:(04
= 60 C1 = 11 m0 = 1,25 3
.
:
lO2-
aq3g2
-
aq3aq3 H+COO-CH+CO=HCO+COOH-CH
- (CO2 (g -1-
.
1- 1,3525
CO2 (g) .
-2 .
3- .
-4
-5 C2 = 21
:3) = 84 g/mol = 8,32 ; (
-
5/19/2018 +
2/16
2
-1: .
CO2 (g)
st 400
:
:
atmatmgd() PPPP
st 0 molng 0PaPg .0
PaPPP atmatmeldifferenti .00
-2-
CO2 (g)
.
CO2 (g) :TRnVP ...
moln
TR
VP.105,1
36,2479
26,37
2527332,8
1035,1106,27
.
. 233
--3 .
:molVCn .10611060 231
:molM
mn .105,1
84
25,1 20
-
5/19/2018 +
3/16
3
-4- .
lO2-
aq3g2
-
aq3aq3 H+COO-CH+CO=HCO+COOH-CH
mol
002105,1 2106 0.
xxx 2105,1x 2106
x.
fxfxfx 2105,1fx
2106
fx.
: 33 11 HCONamolCOOHCHmol
33 .6.6 HCONamolCOOHCHmol
mol.6
molmol .6105,1 2
: 3HCONa
2: 3HCONa
:
molxx .105,10105,1 2
maxmax
2
molxxf
.105,1 2
max
molx 22max
2105,410.5,16106
mol2105,4 (3COOH(aq
:
molx .105,1 2
max
--5
molxCOnCOOCHn .105,1 2max23
2-
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5/19/2018 +
4/16
4
(:(04
32
tN
dt
tdN. .
teNtN .0.
: tN0N-1
32- jt 2,142
1
2- st .00
N
3- gm 1 2
1.2 tt
-4 -2 32
-5 43-1
6- st .0-3
23
10023,6 AN
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5/19/2018 +
5/16
5
-1: tN: t
:
0N: st .0
:2- /t
000/ .4
3
4N
NNtN
/...
00
/
.ln4
3
ln4
3
.4
3 ///
teeeNNtN ttt
29,029,038,109,14ln3ln.
4
3ln //
tt
:
1
2
1 .048.02,14
693,02ln jt
:jt .04,6048,0
29,029,0/
-3
2
1t
:
2
0
21
NtN
42
2
2.2 0
0
2
1
2
1
NNtN
tN
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5/19/2018 +
6/16
6
4
..2..2. 0
2
1
2
1
NtNtAtN
dt
tdNtA
:M
Nm
NNm
NMNAAA
0
000
M
NmN A
2023
00 102,188
32
10023,61
171.1055,5
606024
048,0.048.0
sj
BqN
tA
111320
70
106,11012,1614
102,188
1055,54.
-
:4 -2
jt .8,20
-1 : tNtA .
7
15
8
10102
102
st .10
7
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5/19/2018 +
7/16
7
-3
tNtN .ln 0
tNtN .lnln 0
1.047,0
1070
28,51 j
jt .2,21
NN-6 2000 1047,18628.51ln
-4
te
N
tN .
0
100
NtNt
je
N
tNt .8,2037,0
.1
0
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5/19/2018 +
8/16
8
(:(04
:
= 0k. EXAO -2-
1- .
2-
.
3-
.
-4 .
- =
(1 ) .
-6 RL.
-7
1-
-2 ti
01000
6.60max r
rR
EmAII
T
00/////.6.10106 0033 rIVIrEUU LR
-2:
-
5/19/2018 +
9/16
9
:-3
Etirdt
tdiLtiREUU LR ...
TRrR ///
L
E
dt
tditi
L
rREUU LR
.
R
L
rR
L
........
0.1
.
L
Eti
dt
tdi
L
E
dt
tditi
L
rR
=
(1 )
1 .........................
tt
L
R
eL
Ee
R
E
L
R
dt
tdi
...
.
2...............
tt
eL
E
L
Etie
R
E
R
Eti
..1
.
12 -
0..
L
Ee
L
Ee
L
E
L
E tt
-6:
R
L
rR
L
-7:
:msR
L.5,1
1000
5,1
:t
AmAmeR
Ei ..8,3.78,3
1000
6.63,01
ms5,1
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5/19/2018 +
10/16
10
:(04)
1- -2 ,RT-3
-3-1 RT,,MT
3-2-
3-3-
RT,,MT
3-4-.
-4
-4-1
4-2- .
:MT = 5,98 1024, = 35800,RT = 6400Km, = 6,67 1011
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5/19/2018 +
11/16
11
-1: :
(
)
:-2 ,RT
:-
:- :-
-
: hRVaaT
N
2
:-3-1 RT,,MT
22.
...
hR
MGaam
hR
mMGF
T
T
s
T
sT
-3-2:
13
.31,3067606024
1035800640014,32..2
sm
T
hRV T
sm
hR
MGV
hR
MG
hR
Va
T
T
T
T
T
/.3,307410.640035800
1098,51067,6..3
2411
2
2
:-3-3
RT,,MT
2
22
2
2
2.4..
T
hR
hR
MGV
hR
MG
hR
Va T
T
T
T
T
T
-
5/19/2018 +
12/16
12
sMG
hRT
T
hR
hR
MG
T
TT
T
T .1014,8620.
..2.4. 1
3
2
22
ssT .1062,8.1014,8620 41
:-3-4
TTT
T
T
MGhR
T
T
hR
hR
MG
.
.4.4. 2
3
2
2
22
:
:-3-4
A
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5/19/2018 +
13/16
13
:(04)
160
.
:
25
:14
10
Ke
:PKa NH4 (aq)+ /3 ( ) = 9,2
:
4,4-3,16,8-5,27,6-610-8,2
(b) Cb = 2 1 021
= 10,75
:
1-1- .rf-1-2 Cb. .1-3- PKa NH4 (aq )
+ /3 ( )
2-
b = 30 (b
),
b
, (a) C = 2 1 021
-2-1
2-2- -3- a(a).
-2-2-1 .
b-2-2-2.
,,-2-2-3 .
-
5/19/2018 +
14/16
14
a1-2-2-4
NH4+ = 15[3 ] .
-1-1:
.
0303max .
.
NH
OH
VNH
VOH
x
xff
25,3
75,10
1414
3
3 10
10
10
10
10.
PHOH
KeOHOHOHKe
1108,2102
106,5 22
4
03
NH
OHf
--3
03
NH
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5/19/2018 +
15/16
15
-1-2: rf Cb.
f
bf
bfb
bfbff
fr
c
cc
cc
NH
NHOH
x
xQ
1
.
.
... 2
3
4
max
52
6
2
4222
1062,11019,97
1079,15
1081,21
1081,2102
1
.
f
bf
fr
cQ
-1-3:
PKa
NH4 (aq )+ /3 ( )
f
f
NH
NHLogPKaPH
4
3
124243
.1094,11062,5102 lmolNHcNH
fbf
:
ff OHNH 4
2,91062,5
1094,14
2
LogPHPKa
:-1-4
Laqaqaqaqaq OHClNHClOHNH 2433
:-1-5
7,5,4,22 EE PHmlVE
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5/19/2018 +
16/16
16
b:-1-6
lmolCVCVC baEabb /105,130
4,22102 22
//
-1-5:
8,62,57,5 EPH
a1:-1-5
NH4+ = 15[3 ]
4
3
NH
NHLogPKaPH
4
3
15
1
NH
NH
8
15
12,9 PHLogPH
mlVa 2,211