When mixing 20.0 mL 0.500 M KOH, 20.0 ml 0.100 M NaCl and 40.0 ml 0.220 M Co(NO,)2 solutions, Co(OH), precipitates. Under these circumtances calculate the solubility of Co(OH), using activities. Ksp = 2.5 x10-16 «со+23 600 рт COH-= 350 pm

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When mixing 20.0 mL 0.500 M KOH, 20.0 ml 0.100 M NaCl and 40.0 ml 0.220 M
Co(NO3)2 solutions, Co(OH), precipitates. Under these circumtances calculate the
solubility of Co(OH), using activities. Ksp = 2.5 x10-16
Co+2= 600 pm
COH-= 350 pm
Transcribed Image Text:When mixing 20.0 mL 0.500 M KOH, 20.0 ml 0.100 M NaCl and 40.0 ml 0.220 M Co(NO3)2 solutions, Co(OH), precipitates. Under these circumtances calculate the solubility of Co(OH), using activities. Ksp = 2.5 x10-16 Co+2= 600 pm COH-= 350 pm
Values of t for various levels of probability
Values of Q for various levels of probability
Degrees of freedom 80%
90%
95%
99%
Confidence levels
(N-1)
No. of observations 90%
95%
99%
1
3.08
6.31
12.7
63.7
3
0.941 0.970 0.994
2
1.89
2.92
4.30
9.92
4
0.765 0.829 0.926
3.
1.64
2.35
3.18
5.84
0.642 0.710 0.821
4
1.53
2.13
2.78
4.60
0.560 0.625 0.740
1.48
2.02
2.57
4.03
7
0.507 0.568 0.680
6.
1.44
1.94
2.45
3.71
8
0.468 0.526 0.634
7
1.42
1.90
2.36
3.50
9.
0.437 0.493 0.598
1.40
1.86
2.31
3.36
10
0.412 0.466 0.568
9.
1.38
1.83
2.26
3.25
19
1.33
1.73
2.10
2.88
59
1.30
1.67
2.00
2.66
1.29
1.64
1.96
2.58
Standard deviation
N
i=1
j=1
S =
N-1
Confidence Interval H=x±ts
-0.51(Z;)?V
1+(ªV#/305
logy
2
Transcribed Image Text:Values of t for various levels of probability Values of Q for various levels of probability Degrees of freedom 80% 90% 95% 99% Confidence levels (N-1) No. of observations 90% 95% 99% 1 3.08 6.31 12.7 63.7 3 0.941 0.970 0.994 2 1.89 2.92 4.30 9.92 4 0.765 0.829 0.926 3. 1.64 2.35 3.18 5.84 0.642 0.710 0.821 4 1.53 2.13 2.78 4.60 0.560 0.625 0.740 1.48 2.02 2.57 4.03 7 0.507 0.568 0.680 6. 1.44 1.94 2.45 3.71 8 0.468 0.526 0.634 7 1.42 1.90 2.36 3.50 9. 0.437 0.493 0.598 1.40 1.86 2.31 3.36 10 0.412 0.466 0.568 9. 1.38 1.83 2.26 3.25 19 1.33 1.73 2.10 2.88 59 1.30 1.67 2.00 2.66 1.29 1.64 1.96 2.58 Standard deviation N i=1 j=1 S = N-1 Confidence Interval H=x±ts -0.51(Z;)?V 1+(ªV#/305 logy 2
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