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%
Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,

Answer:
Lithosphere, mantle, outer core, inner core
Explanation: