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NNADVOKAT [17]
3 years ago
5

"suggest a simple solution to this problem that does not involve the use of timestamps and show how would your solution solve th

is impersonating attack"
Chemistry
1 answer:
kirill [66]3 years ago
8 0
I'm not sure i need more information.

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In an experiment, a student gently heated a hydrated copper compound to remove the water of hydration. The following data was re
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The percent of water in the hydrated salt will be 55.08 %

<h3>What is mass?</h3>

Mass is a numerical measure of quantity, which is a basic feature of all matter. The kilogram is the international system of units of mass.

Given data;

Mass of crucible, cover, and contents before heating = 23.4 g.

Mass of empty crucible and cover = 18.82 g.

Mass of crucible, cover, and contents after heating to constant mass = 20.94 g.

⇒Mass of hydrated salt used =  Mass of the crucible, cover, and contents before heating - Mass of the empty crucible and cover                    

⇒Mass of hydrated salt used =   = 23.54 g – 18.82 g

⇒Mass of hydrated salt used =  = 4.72 g

⇒Mass of dehydrated salt after heating = Mass of the crucible, cover, and contents after heating to constant mass-Mass of the empty crucible and cover

⇒Mass of dehydrated salt after heating = 20.94 g – 18.82 g

⇒Mass of dehydrated salt after heating = 2.12 g

⇒ Mass of water liberated from salt = Mass of hydrated salt used -   Mass of dehydrated salt after heating

⇒  Mass of water liberated from salt = 4.72 g – 2.12 g    

⇒  Mass of water liberated from salt = 2.60 g  

   

Water % in the hydrated salt is found as;

% water =(mass of water/  mass of hydrated salt)× 100

% water =(2.60/4.72) × 100

% water =55.08 %

Hence the percent of water in the hydrated salt will be 55.08 %

To learn more about the mass, refer to the link.\

brainly.com/question/13073862

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