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OleMash [197]
2 years ago
10

If 13 grams of copper sulfate is reacted with zinc how much of each product is produced?

Chemistry
1 answer:
Sladkaya [172]2 years ago
7 0

Answer:

No. Of Moles of zinc = m/Ar

= 13/ 65.38 = 0.198 moles

From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1

So only 0.198 moles of CuSO4 reacts, it is in excess

Mass = no of Moles X Mr

Mass = 0.198 X 159.5 = 31.59 grams

Volume = mass m denisty

Volume j 31.59 / 3.6 = 8.78 ml

Explanation:

i think this wrong

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Answer:

pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%

Explanation:

For the equilibrium buffer of NH₃:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵

kb = [NH₄⁺] [OH⁻] / [NH₃] <em>(1)</em>

When 0,245 M of NH₃ is added, the equilibrium concentrations are:

[NH₃] = 0,245 - x

[NH₄⁺] = x

[OH⁻] = x.

Replacing this values in (1)

1,8x10^{-5} = \frac{x^2}{0,245 - x}

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Thus [OH⁻] in equilibrium is 0,00211 M.

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It is possible to calculate the percent ionization thus:

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Replacing:

0,00211 / 0,245 × 100 = <em>0,86%</em>

I hope it helps!

6 0
2 years ago
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