Question

Question #e74c7

Answer 1

`1.1" micromoles"`

Explanation:

You should consider the equation

`n = cv`

where `n` is the number of moles (or micromoles), `c` is the concentration and `v` is the volume.

The question gives us a volume of `0.35L` or `0.35"dm"^3` (which are the same) and a concentration of `3.1xx10^-6M` or `3.1xx10^-6"moldm"^-3`.

Therefore,

`n = 0.35"dm"^3 xx 3.1 xx 10^-6"moldm"^-3 = 1.085 xx 10^-6"mol"`

So we know there are `1.085 xx 10^-6"mol Hg"_2"Cl"_2` in the solution, which is about `1.1` micromoles of mercury(I) chloride, since there are `10^6"micromol"`s in a mol.