Yes, this is correct. The answer is 1.3 x 10-8, or 0.000000013%.

[bergausstein]

Atmospheric concentrations are usually measured
in parts per billion, by volume (ppbv). So, a concentration of carbon dioxide equal to 13 ppbv means that, for every billion volume units (say,milliliters) of the atmosphere, there are 13 units of carbon dioxide.
Convert an atmospheric concentration of carbon
dioxide of 13 ppbv to a percent (that is, what per-cent of this atmosphere would be carbon dioxide?)

my solution

$\frac{13}{10^9}=1.3\times 10^{-8}$

is this correct. if not please help me arrive at the correct answer. thanks!

 

[MarkFL]

We know that $$\frac{1}{2}=50\%$$ or:

$$\frac{1}{2}=\left(\frac{1}{2}\cdot100 \right)\%=50\%$$

We see that when converting a fraction to a percentage, we must treat the percent sign as division by 100, because per cent literally means "for each 100." So if we attach a percent sign, we must multiply the fraction by 100 so that we are in effect multiplying by:

$$1=\frac{100}{100}=100\cdot\frac{1}{100}=100\%$$

Hence:

$$\frac{a}{b}=\frac{100a}{b}\%$$

Can you proceed?

 

[bergausstein]

I still don't get it.

it's because of parts per billion thing. I'm confused. how am I suppose to convert 13ppbv to a percent? please bear please bear with me.

 

[MarkFL]

What you need to do is:

$$\frac{13}{10^9}=\frac{13\cdot10^2}{10^9}\%$$

Now simplify. :D