Can u please help me with these 2 answers!

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

andrezito [222]

Answer:

B

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:

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

Because there are 5 moles of hydrogen gas:

$\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}$

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0

2 years ago

An open flask sitting in a lab refrigerator looks empty, but it is actually filled with a mixture of gases called air. If the fl

ira [324]

Answer:

9.39 × 10²² molecules

Explanation:

We can find the moles of gases (n) using the ideal gas equation.

P . V = n . R . T

where,

P is the pressure (standard pressure = 1 atm)

V is the volume

R is the ideal gas constant

T is the absolute temperature (standard temperature = 273.15 K)

$n=\frac{P.V}{R.T} =\frac{1atm.3.50L}{(0.08206atm.L/mol.K).273.15K} =0.156mol$

There are 6.02 × 10²³ molecules in 1 mol (Avogadro's number). Then,

$0.156mol.\frac{6.02\times10^{23}molecules}{mol} =9.39\times10^{22}molecules$

4 0

3 years ago

The psychoactive drug sold as a methamphetamine - "speed" (c10h15n) - undergoes a series of reactions in the body; the net resul

Phoenix [80]

Answer: The coefficient of nitrogen in the given equation is 2.

Explanation: The reaction for the oxidation of methamphentamine with oxygen gas in the body is given by:

$4C_{10}H_{15}N(s)+55O_2(g)\rightarrow 40CO_2(g)+30H_2O(l)+2N_2(g)$

By Stoichiometry,

4 moles of methamphentamine reacts with 55 moles of oxygen gas to produce 40 moles of carbon dioxide gas, 30 moles of water and 2 moles of nitrogen gas.

Coefficient of $C_{10}H_{15}N=4$