Since CO2 makes up 0.0355% of the atmosphere (on the average) calculate the molar fraction of [CO2•H2O] in the unpolluted rain water (please use constants in table attached) Selected Henry's Law Constants and Equilibrium Constants at 298 K.К (М)H (mole L-1 atm-1)EquilibriumH20 → H* + OH-1.0 x 10-14CO2 + H20 → CO2·H2O3.4 x 10-2CO2·H20 > H* + HCO34.3 x 10-7HCO3 > H* + CO32-4.7 x 10-11SO2 + H20 > SO2·H2OSO2·H2O H* + HSO3"1.231.32 x 102HSO3 H* + SO3²-6.42 х 108NH3 + H20 > NH3'H2O62NH3 H2O NH4* + OH1.8 x 105HNO3 + H20 €→ HNO3-H202.1 x 105HNO3·H2O → H* + NO315.4H2O2 + H20 → H2O2·H2OH2O2·H2O → H* + HO2O3 + H20 → O3·H2O7.45 x 1042.5 x 10-120.011

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Answered: Since CO2 makes up 0.0355% of the…

Since CO2 makes up 0.0355% of the atmosphere (on the average) calculate the molar fraction of [CO2•H2O] in the unpolluted rain water

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Since CO2 makes up 0.0355% of the atmosphere (on the average) calculate the molar fraction of [CO2•H2O] in the unpolluted rain water (please use constants in table attached)

Selected Henry's Law Constants and Equilibrium Constants at 298 K.
К (М)
H (mole L-1 atm-1)
Equilibrium
H20 → H* + OH-
1.0 x 10-14
CO2 + H20 → CO2·H2O
3.4 x 10-2
CO2·H20 > H* + HCO3
4.3 x 10-7
HCO3 > H* + CO32-
4.7 x 10-11
SO2 + H20 > SO2·H2O
SO2·H2O H* + HSO3"
1.23
1.32 x 102
HSO3 H* + SO3²-
6.42 х 108
NH3 + H20 > NH3'H2O
62
NH3 H2O NH4* + OH
1.8 x 105
HNO3 + H20 €→ HNO3-H20
2.1 x 105
HNO3·H2O → H* + NO3
15.4
H2O2 + H20 → H2O2·H2O
H2O2·H2O → H* + HO2
O3 + H20 → O3·H2O
7.45 x 104
2.5 x 10-12
0.011

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