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Arada [10]
3 years ago
8

The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, w

hereas nitrogen gas required 48 s. The molar mass of the gas is ________
Chemistry
1 answer:
Marizza181 [45]3 years ago
4 0

Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }

6,61 = √M₂

44g/mol = M₂

<em>The molar mass of the gas is 44 g/mol</em>

<em></em>

I hope it helps!

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Acetic acid, CH3COOH, can be produced by bubbling oxygen gas into acetaldehyde, CH3CHO, in the presence of
slamgirl [31]

Explanation:

The balanced equation for the reaction is given as;

2CH3CHO + O2 → 2CH3COOH  

If 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel, (a)

how many grams of acetic acid will be produced?

First thing's first, we have to find he limiting reactant. This is done by comparing the number of moles of the reactants.

From the equation of the reaction;

2 mol of CH3CHO reacts with 1 mol of O2

From the masses given;

Number of moles = Mass / Molar mass

CH3CHO;

Number of moles = 20 / 44.0526 = 0.454 mol

O2;

Number of moles = 10 / 32 = 0.3125 mol

The limiting reactant is CH3CHO because O2 would be in excess.

Back to the question;

2 mol of CH3CHO produces 2 mol of CH3COOH  

0.454 mol would produce x

Solving for x;

x = 0.454 * 2 / 2 = 0.454 mol

Converting to mass;

Mass = number of moles* Molar mass

Mass = 0.454 mol *  60.052 g/mol = 27.26 grams

(b) how many grams of the excess reactant remain after the reaction is

complete

The excess reactant is O2

Number of moles left = Initial Number of moles - Number of moles that reacted

Number of moles left =  0.3125 mol - (0.454 mol / 2)

Number of moles left = 0.0855 mol

Converting to mass;

Mass = 0.0855 mol * 32 g/mol = 2.736 grams

6 0
3 years ago
What is the molarity if 2.00 liters containing 49.0 grams of sodium carbonate [Na2CO3)?
yKpoI14uk [10]

Answer: The molarity of solution is 0.231 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Molar mass of Na_2CO_3 = 2\times 22.99+1\times 12.01+3\times 16.00=105.99

moles of Na_2CO_3 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{49.0g}{105.99g/mol}=0.462mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.462mol}{2.00L}

Molarity=0.231M

Therefore, the molarity of solution is 0.231 M

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