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3 years ago

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A gas sample has an initial pressure of 547 mmHg and initial volume of 0.500 L. Pressure (in atm) when the volume of the sample

is decreased to 225 mL?

1 answer:

Alenkasestr [34]3 years ago

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What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is $O_{2}$ molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

First we have to calculate the moles of methane.

$\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}$ = $\frac{23g}{16.04g/mole}$ = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give $\frac{2moles\times 1.434moles}{1moles}$ moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is $O_{2}$ molecule.

6 0

3 years ago

I need someones help

valentinak56 [21]

Answer:

y or z one of those two i think its z but im so sorry if im wrong if not z then try y also I learned this last year.

3 0

2 years ago

What is the molarity of a solution with 2 moles of HCl in 2000 mL?

ikadub [295]

Answer:

"1 M" will be the right solution.

Explanation:

The given values are:

Number of moles,

= 2 moles

Volume of solution,

= 2000 mL

or,

= 2 L

Now,

The molarity of the solution will be:

= $\frac{Moles}{Liters}$

On substituting the values, we get

= $\frac{2}{2}$

= $1 \ M$

7 0

3 years ago

A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sampl

diamong [38]

<u>Answer:</u>

<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u> The rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

<u>Explanation:</u>

<u>For A:</u>

The average molecular speed of the gas is calculated by using the formula:

$V_{gas}=\sqrt{\frac{8RT}{\pi M}}$

OR

$V_{gas}\propto \sqrt{\frac{1}{M}}$

where, M is the molar mass of gas

Forming an equation for the two gases:

$\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}$ .....(1)

Given values:

$V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol$

Plugging values in equation 1:

$\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s$

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u>

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

$Rate\propto \frac{1}{\sqrt{M}}$

Where, M is the molar mass of the gas

Forming an equation for the two gases:

$\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}$ .....(2)

Given values:

$Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol$

Plugging values in equation 2:

$\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr$

Hence, the rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

8 0

2 years ago

Write the equation for the dissociation of KOH in water. (Include states-of-matter under TSS conditions in your answer. Use the

Genrish500 [490]

KOH(s) ----> K⁺(aq)+OH⁻ (aq)
s-solid,
aq - aqueous

6 0

3 years ago