3as.ency-educationΒ Β· ii 2759 - π½ β 2760 β2608 π+β10 β’ 28 π60 β’ π61 28 π62...
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(π»3π+ + πΆπβ)π΄π
π΄π3+π‘ = 0π0 = 0.81π
π = 100ππΏπΆ = 0.3 πππ πβ
π‘(πππ)ππ»2
(ππ)
[π΄π3+](π πππ πβ )
-
-
(π΄π3+ π΄πβ )(π»3π+ π»2β )
- π₯πππ₯
[π΄π3+](π‘)[π΄π3+](π‘) =2ππ»2
(π‘)
3πΓππ
ππ»2(π‘)ππ
[π΄π3+] = π(π‘)
[π΄π3+]ππ‘ = 8πππ
- π‘π£ =1
2Γ
π[π΄π3+]
ππ‘
π‘ = 2πππ π‘ = 6πππ
π»3π+
- π‘1 2β
- π‘ = 5πππ
π(π΄π) = 27π. πππβ1ππ = 24πΏ. πππβ1
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πΆπ60
I. 1(π β π)
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-
-
-
πΆπ2760
-
60
II. 60
π0
2π΄
1-
2-
π΄(π‘) = π΄0πβππ‘
3- π΄0
π‘ = 0
4- π‘ = π
ππ
5- π0
6- 5
- π΄
- 100ππ΅π
ππ΄ = 6.02 Γ 1023πππβ1
1
2
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Ψ§ΩΨͺΩ Ψ±ΩΩ Ψ§Ψ£ΩΩΩ:
-
-
π΄π = π΄π3+ + 3πβ
2π»3π+ + 2πβ = π»2 + 2π»2π
2π΄π(π ) + 6π»3π+(ππ) = 2π΄π3+
(ππ) + 3π»2(π) + 6π»2π(π)
-
π0(π΄π) =π
π=
0.81
27= 0.03πππ
π0(π»3π+) = πΆπ = 0.3 Γ 0.1 = 0.03πππ
2π΄π(π ) + 6π»3π+(ππ) = 2π΄π3+
(ππ) + 3π»2(π) + 6π»2π(π)
00π0(π»3π+)π0(π΄π)
3π₯2π₯π0(π»3π+) β 6π₯π0(π΄π) β 2π₯
3π₯π2π₯ππ0(π»3π+) β 6π₯ππ0(π΄π) β 2π₯π
{π0(π΄π) β 2π₯πππ₯ = 0
π0(π»3π+) β 6π₯πππ₯ = 0βΉ {
π₯πππ₯ =π0(π΄π)
2= 0.015πππ
π₯πππ₯ =π0(π»3π+)
6= 0.005πππ
βΉ
π₯πππ₯ = 0.005πππ
[π΄π3+](π‘)[π΄π3+](π‘) =2ππ»2
(π‘)
3πΓππ
[π΄π3+](π‘) =2π₯
π
π(π»2) =ππ»2
(π‘)
ππ= 3π₯ βΉ π₯ =
ππ»2(π‘)
3ππ
βΉ [π΄π3+](π‘) =2π₯
π=
2ππ»2(π‘)
3π Γ ππ
[π΄π3+](π‘) =2ππ»2
(π‘)
3π Γ ππ=
2ππ»2(π‘)
3 Γ 0.1 Γ 24=
2
7.2Γ ππ»2
(π‘)
π‘(πππ)ππ»2
(ππ)
[π΄π3+](π πππ πβ )
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β’
[π΄π3+]π
[π΄π3+]π =2π₯π
π=
2 Γ 0.005
0.1= 0.1πΏ
[π΄π3+]π = 100ππππ/π
- π‘ = 8πππ
-
π£ =1
πΓ
ππ₯
ππ‘
[π΄π3+] =2π₯
πβΉ π₯ =
[π΄π3+]π
2
βΉ π£ =1
πΓ
ππ₯
ππ‘=
1
πΓ
π ([π΄π3+]π
2 )
ππ‘=
1
2Γ
π[π΄π3+]
ππ‘
π£2 =1
2Γ
π[π΄π3+]
ππ‘=
1
2Γ
(54 β 18) Γ 10β3
3 β 1= 0.009πππ/π. πππ
π£6 =1
2Γ
π[π΄π3+]
ππ‘=
1
2Γ
(75 β 63) Γ 10β3
7 β 5= 0.003πππ/π. πππ
β’
π»3π+
π£β² = βπππ»3π+
ππ‘= β
π(π0(π»3π+) β 6π₯)
ππ‘= 6
ππ₯
ππ‘= 6 Γ
π
πΓ
ππ₯
ππ‘= 6 Γ π Γ π£
π£β²2 = 6 Γ π Γ π£ = 6 Γ 0.1 Γ 0.009 = 0.0054πππ/πππ
π£β²6 = 6 Γ π Γ π£ = 6 Γ 0.1 Γ 0.003 = 0.0018πππ/πππ
- π‘1 2β
π‘ = π‘1 2β βΉ π₯ =π₯π
2 βΉ [π΄π3+] =
[π΄π3+]π
2=
100
2= 50ππππ/π
βΉ π‘1 2β β 3.5πππ
- π‘ = 5πππ
π‘ = 5πππ βΉ π₯ =[π΄π3+]π
2=
63.19 Γ 10β3 Γ 0.1
2= 3.16 Γ 10β3πππ
[π΄π3+] = 0.063πππ/π
[π»3π+] =π0(π»3π+) β 6π₯
π=
0.03 β 6 Γ 3.16 Γ 10β3
0.1= 0.11πππ/π
[πΆπβ] = 0.3πππ/π
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II.
- πΆπ2759
β’ πΆπ2760 β ππ28
60 + πβ10π½β
β’ 60ππ2861ππ28
62ππ2858
III.
7-
8- π(π‘) = π0πβππ‘
β’
π΄ = β ππ(π‘)
ππ‘β π΄ = β
π(π0πβππ‘)
ππ‘= ππ0πβππ‘
π‘ = 0 β π΄0 = ππ0
β π΄ = ππ0πβππ‘ β π΄(π‘) = π΄0πβππ‘
9- π΄0 = 29.22 Γ 1010π΅π.
10- π‘ = π
π‘ = π β π΄(π) = π΄0πβππ = π΄0πβ1 = 0.37π΄0
β π΄(π) = 0.37 Γ 29.22 Γ 1010 = 10.8 Γ 1010π΅π
β’ π = 7.6πππ .
π =1
π=
1
7.6= 0.131πππ β1
π =1
π=
1
7.6 Γ 365.25 Γ 24 Γ 3600β π = 4.16 Γ 10β9π β1
11- π0
π΄0 = ππ0 = π Γπ0 Γ ππ΄
πβ π0 =
π΄0 Γ π
π Γ ππ΄
β π0 =π΄0 Γ π
π Γ ππ΄=
29.22 Γ 1010 Γ 60
4.16 Γ 10β9 Γ 6.02 Γ 1023β π0 = 7 Γ 10β3π
12- π΄
π΄(π‘) = π΄0πβππ‘ = 29.22 Γ 1010 Γ πβ0.131Γ5 = 15.17 Γ 1010π΅π
π΄(π‘) = π΄0πβππ‘ βπ΄(π‘)
π΄0= πβππ‘ β ππ
π΄(π‘)
π΄0= βππ‘ β π‘ =
1
πΓ ππ
π΄0
π΄(π‘)= π ππ
π΄0
π΄(π‘)
π‘ = π πππ΄0
π΄(π‘)= 7.6 Γ ππ (
29.22 Γ 1010
100 Γ 103) = 113.14πππ
β’ 113.14 β 5 = 108.14 πππ
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