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MrRissso [65]
3 years ago
7

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate sol

ubilities. (If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.) fill in the blank 1 1. manganese(II) sulfide A. CuS fill in the blank 2 2. calcium fluoride B. FeS C. PbCl2 D. CaCrO4 Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water. manganese(II) sulfide calcium fluoride Ksp
Chemistry
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate solubilities. (If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.) fill in the blank 1 1. manganese(II) sulfide A. CuS fill in the blank 2 2. calcium fluoride B. FeS C. PbCl2 D. CaCrO4 Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water. manganese(II) sulfide calcium fluoride Ksp

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<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

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dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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