1.Calculate the rate of reaction for the solution to turn from colorless to blue
ID: 988965 • Letter: 1
Question
1.Calculate the rate of reaction for the solution to turn from colorless to blue-black after each addition.
Volume Added to Solution 1
Reaction time (s)
Total Volume (L)
Rate (mol/L·s)
25 mL (NH4)2S2O8
69
.100
.032
1 mL Na2S2O3
105
.101
9.96X10^-4
1 mL Na2S2O3
80
.102
.0013
1 mL Na2S2O3
75
.103
.0014
2.What is the average rate of the reaction for Solution 1 (include units)? .009 mol/L*s
3.What is the concentration of KI and (NH4)2S2O8 in the reaction flask?
M KI M (NH4)2S2O8
How do I figure out #3?
There is
25.0 mL of .2 M KI
25.0 mL of .2 M (NH4)2S2O8
Volume Added to Solution 1
Reaction time (s)
Total Volume (L)
Rate (mol/L·s)
25 mL (NH4)2S2O8
69
.100
.032
1 mL Na2S2O3
105
.101
9.96X10^-4
1 mL Na2S2O3
80
.102
.0013
1 mL Na2S2O3
75
.103
.0014
Explanation / Answer
3. Given the volume of KI = 25.0 mL = 0.025 L
Concentration of KI = 0.2M
Hence moles of KI in the solution = MxV = 0.2 M x 0.025L = 0.005 mol
Given the total volume of the solution, Vt = 0.100 L
Concentration of KI = moles of KI / Vt = 0.005mol / 0.100 = 0.05 M(answer)
Given the volume of (NH4)2S28 = 25.0 mL = 0.025 L
Concentration of (NH4)2S28 = 0.2M
Hence moles of (NH4)2S28 in the solution = MxV = 0.2 M x 0.025L = 0.005 mol
Given the total volume of the solution, Vt = 0.100 L
Concentration of (NH4)2S28 = moles of (NH4)2S28 / Vt = 0.005mol / 0.100 = 0.05 M(answer)
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