Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
We know that density = mass /volume
so
mass=volume*density
volume=mass/density
so c is wrong
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Answer: it would release heat because the thermal energy it absorbed to become a gas. so it would release heat. hope this helps :)
Explanation:
