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adelina 88 [10]
4 years ago
8

The cycle of copper reactions starts with solid copper metal and ends with solid copper metal. Suppose you start with 0.502 g of

Cu. After performing all the reactions, you have 0.446 g of Cu. What percentage of the copper did you recover?
Chemistry
1 answer:
Marianna [84]4 years ago
7 0

Answer: percentage of the copper recovered is 88.8 %

Explanation:

Given : Initial amount of copper = 0.502 g

Recovered amount of copper = 0.446 g

percentage of copper recovered = \frac{\text {recovered amount}}{\text {initial amount}}\times 100\%

Putting in the values :

percentage of copper recovered = \frac{0.446}{0.502}\times 100\%=88.8\%

Thus percentage of the copper recovered is 88.8 %

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Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate·x-hydrate (CuSO4·xH2O), where x is an integer. Part of their
iris [78.8K]

Answer:

5

Explanation:

We can obtain the value of x by doing the following:

Mass of hydrated salt (CuSO4.xH2O) = 1.50g

Mass of anhydrous salt (CuSO4) = 0.96g

Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g

Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x

Mass of water(xH2O) molecules in the hydrate salt is given by:

xH2O/CuSO4.xH2O = 0.54/1.5

18x/(159.5 + 18x) = 0.36

Cross multiply to express in linear form

18x = 0.36 (159.5 + 18x)

18x = 57.42 + 6.48x

Collect like terms

18x — 6.48x = 57.42

11.52x = 57.42

Divide both side by 11.52

x = 57.42/11.52

x = 5

Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O

8 0
4 years ago
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