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brilliants [131]
2 years ago
9

Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which

value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%
Chemistry
1 answer:
prohojiy [21]2 years ago
8 0

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

<u>Step 1:</u> Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

<u>Step 2:</u> Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

<u>Step 3:</u> Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

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The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

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C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

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