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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52

Problem 33 Easy Difficulty

Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given
(a) $\operatorname{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)$
(b) $\operatorname{CdS}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q)$
(c) $\mathrm{Hg}^{2+}(a q)+4 \mathrm{Br}^{-}(a q) \rightleftharpoons\left[\mathrm{HgBr}_{4}\right]^{2-}(a q)$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$

Answer

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Chemistry 102

Chemistry

Chapter 17

Electrochemistry

Related Topics

Electrochemistry

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Problem 10
Problem 11
Problem 12
Problem 13
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Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
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Problem 21
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Problem 25
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Problem 48
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Problem 50
Problem 51
Problem 52

Video Transcript

for a ready Casey Expression I and in a Casey expression, we do not include solids, products of a reactant. So we have a TSH to cubed over h 20 cubed. We could also rate a K p expression, which will be the partial pressure of H two. Cubed over the partial pressure of H 20 cubed relationship between the two k p is equal to K. C R T on Delta End here would be zero. So then that would just simplify. The K P is equal to K C for B. We can write a K C expression. We have a single liquid called a pure liquid, which is not included in a new equilibrium expression. So we've got products each too squared over. 02 k p is equal to the partial pressure of each two squared times. A partial pressure of old too. K p is equal to K C R T. And then end here would be, ah, one minus two. So this would be negative one. We're sorry. Um, it would be three. There's three moles of gas, the product side zero on the react inside. So the relationship would be to the power three Chrissy, we've got ah solid which is not included. R K C expression is going to be each CEO to the fourth. As I see 04 in each two squared KP is going to be equal to partial pressure of HCL to the fourth partial pressure of S T I C 04 times a partial pressure of H two squared K p is equal to K C R T and the Delta End will be four minus three, which is going to be one for d. K. C is equal to one over HD two two plus over CEO minus squared. Since there are no gas is involved, there is no K c expression. Ah, our sorry KP expression. So this would be a K P is equal to K C R T to the zero, which would mean that

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

Chemistry

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