You have 4.5 x 10^24 particles of C3H8. How many grams are present?

Chemistry

1 answer:

Citrus2011 [14]3 years ago

3 0

Answer:

33 g.

Explanation:

Hello there!

In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

$1mol=44.11g=6.022x10^{23}molecules$

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

$4.5x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} *\frac{44.11g}{1mol}\\\\33g$

Best regards!

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If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef

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Answer:

Explanation:

Recall the law of effusion:

$\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }$

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be r₁ and the rate of hydrogen be r₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for r₂:

$\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\ r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}$