In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=

%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
C)
<em>O</em><em>n</em><em>l</em><em>y</em><em> </em><em>v</em><em>a</em><em>lence</em><em> </em><em>electrons</em><em> </em><em>are involved in forming chemical bonds betwen two atoms</em><em>.</em>
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Answer:
dG will be the same -20 kcal/mol
Explanation:
The dG can be expressed in terms of the G(products) - G(reactants). If the amount of enzyme is doubled the Gibbs energy of the reactants and products will be the same, so the substraction dG has the same value
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