Question #e74c7

Question

Question #e74c7

1 Answer

Impact of this question

905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-22T21:11:31

$1.1 micromoles$ $1.1" micromoles"$

Explanation:

You should consider the equation

$n = c v$ $n = cv$

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

The question gives us a volume of $0.35 L$ $0.35L$ or $0.35 {dm}^{3}$ $0.35"dm"^3$ (which are the same) and a concentration of $3.1 \times 10^{- 6} M$ $3.1xx10^-6M$ or $3.1 \times 10^{- 6} {moldm}^{- 3}$ $3.1xx10^-6"moldm"^-3$ .

Therefore,

$n = 0.35 {dm}^{3} \times 3.1 \times 10^{- 6} {moldm}^{- 3} = 1.085 \times 10^{- 6} mol$ $n = 0.35"dm"^3 xx 3.1 xx 10^-6"moldm"^-3 = 1.085 xx 10^-6"mol"$

So we know there are $1.085 \times 10^{- 6} {mol Hg}_{2} {Cl}_{2}$ $1.085 xx 10^-6"mol Hg"_2"Cl"_2$ in the solution, which is about $1.1$ $1.1$ micromoles of mercury(I) chloride, since there are $10^{6} micromol$ $10^6"micromol"$ s in a mol.

Science

Math

Humanities

... and beyond

Question #e74c7

1 Answer

Explanation:

Related questions

Impact of this question