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Suppose $10^{14} kg$ of carbon dioxide was released into the atmosphere and absorbed completely, what is the percentage change of carbon dioxide concentration? Take initial atmospheric mass mixing ratio to be $5.7 \times 10^{−4} kg/kg$.

Using density of air as around $29g/mol$, the weight of the atmosphere is about $10^{18}kg$. The percentage change is $10^{14} \times 10^{18} \times 100=0.01$? Where does the mixing ratio come in?

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    $\begingroup$ This looks like a homework question... the "mixing ratio" tells you how much carbon dioxide was already there; when you ask about the percentage change, it is relative to this initial ratio. So if you go from 0.01 to 0.02, that would be a 100% change. Is that enough to get you going? $\endgroup$ – Floris Jun 11 '15 at 1:29
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    $\begingroup$ So the i find the new ratio by using $\frac{10^{14}}{10^{18} + 10^{14}}$ which is about $10^{-4}$. So the percentage change is $18%$? $\endgroup$ – user44840 Jun 11 '15 at 1:32
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Following up on your comment:

If you initially have $5.7 \cdot 10^{-4} \mathrm{kg/kg}$ of carbon dioxide, and the total mass of the atmosphere is $10^{18} \mathrm{kg}$, you can compute the mass of $\mathrm{CO_2} = 5.7 \cdot 10^{14} \mathrm{kg}$. Adding $10^{14}\mathrm{kg}$ to that does indeed give a $\frac{1}{5.7} = 18\%$ change as you calculated in your comment.

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  • $\begingroup$ Sorry, which method are you referring to? Is the percentage change 18%? $\endgroup$ – user44840 Jun 11 '15 at 1:45
  • $\begingroup$ Yes the percentage is 18%. $\endgroup$ – Floris Jun 11 '15 at 2:02

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