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WARRIOR [948]
3 years ago
8

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp = 9.05×10−18?

Chemistry
1 answer:
saveliy_v [14]3 years ago
5 0
<span>M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9</span>
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4 0
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Read 2 more answers
Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide:
NeTakaya

Answer:

The partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C Is 0.103 atm.

The correct option is A.

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NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°C

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Therefore = 0.103 +0.112= 0.215

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7 0
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