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%

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

Explanation:

Step 1: Data given

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

Molar mass of CuSO4 = 160 g/mol

Step 2: Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

Step 3: 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

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

since electrons are negative the answer is neagtaive