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)
The reaction equation is:
CaF₂ + H₂SO₄ → 2HF + CaSO₄
The molar ratio between fluorite and hydrogen fluoride is 1 : 1.
The moles of fluorite supplied are:
Moles = 15.6 / 78.07
Moles = 0.200
The moles of hydrogen fluoride produced will be 0.2.
Now, we may use the ideal gas equation to determine the temperature:
PV = nRT
T = PV/nR
T = (899 * 7.4) / (0.2 * 62.36)
T = 533.40 K
The temperature will be 260.25 °C
Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the sugars can’t be separated byfiltration. The category is insoluble.
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