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saul85 [17]
3 years ago
12

Clch2ch2chclch2ch3 +NaI (1.0 equiv)acetone

Chemistry
1 answer:
Margaret [11]3 years ago
5 0

No thanks. I dont buy drugs

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A sample of raw mining ore contains a hydrated salt called copper sulfate tetrahydrate, CuSO4.4H2O, along with other impurities.
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Answer:

88.5 % of the raw ore is CuSO4*4H2O

Explanation:

Step 1: Data given

Mass of the ore = 10.854 grams

Mass of water = 2.994 grams

Step 2: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 2.994 grams / 18.02 g/mol

Moles H2O = 0.166 moles

Step 3: Calculate moles CuSO4*4H2O

For 1 mol CuSO4*4H2O we have 4 moles H2O

For 0.166 moles H2O we have 0.166/4 = 0.0415 moles CuSO4*4H2O

Step 4: Calculate mass CuSO4*4H2O

Mass CuSO4*4H2O = moles * molar mass

Mass CusO4*4H2O = 0.0415 moles * 231.67 g/mol

Mass CuSO4*4H2O = 9.61 grams

Step 5: Calculate mass % of CuSO4*4H2O

Mass % = (9.61 grams / 10.854 grams )*100 %

Mass % = 88.5 %

88.5 % of the raw ore is CuSO4*4H2O

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