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antoniya [11.8K]
4 years ago
12

A solution contains 0.10 m sodium cyanide and 0.10 m potassium hydroxide. solid zinc acetate is added slowly to this mixture. wh

at is the formula of the substance that precipitates first?
Chemistry
1 answer:
Brrunno [24]4 years ago
6 0
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)

Ksp{Zn(OH)₂}=1.2*10⁻¹⁷

2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)

Ksp{Zn(CN)₂}=2.6*10⁻¹³


Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}

Zn(OH)₂ precipitates first

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Mumz [18]

Answer:

54 grams of H₂O

Explanation:

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Ratio is 4:6.

Let's convert the mass of ammonia in moles

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34 g / 17 g/m = 2 moles

If 4 moles of ammonia are needed to produce 6 mol of water

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8 0
3 years ago
Suppose that 0.25 mole of gas C was added to the mixture without changing the total pressure of the mixture. How does the additi
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Here we have to get the effect of addition of 0.25 moles of gas C on the mole fraction of gas A in a mixture of gas having constant pressure.

On addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

The partial pressure of gas A can be written as P_{A} = x_{A}×P (where x_{A} is the mole fraction of gas A present in the mixture and P is the total pressure of the mixture.

The mole fraction of gas A in a mixture of gas A and C is = \frac{moles of gas A}{moles of gas A + moles of gas C} and \frac{moles of gas C}{moles of gas A + moles of gas C} respectively.

Thus on addition of 0.25 moles of C gas, the mole fraction of gas A will be  \frac{moles of gas A}{moles of gas A + 0.25}.

Which is different from the initial state.

4 0
3 years ago
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<h2>Answer:</h2>

<em>Given data:</em>

1 kilogram = 1,000 grams

how many kilograms is 1,216 grams?

<em>Solution:</em>

1000 grams = 1 kilogram

1 gram = 1/1000 kilograms

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