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2 years ago

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A flask contains 0.25 mole of SO2(g), 0.50 mole of CH4(g), and 0.50 mole of O2(g). The total pressure of the gases in the flask

is 800 mm Hg. What is the partial pressure of the SO2(g) in the flask?

1 answer:

Gwar [14]2 years ago

4 0

<u>Answer:</u> The partial pressure of sulfur dioxide gas in the flask is 160 mmHg

<u>Explanation:</u>

We are given:

Moles of sulfur dioxide gas = 0.25 moles

Moles of methane gas = 0.50 moles

Moles of oxygen gas = 0.50 moles

To calculate the mole fraction of sulfur dioxide, we use the equation:

$\chi_{SO_2}=\frac{n_{SO_2}}{n_{SO_2}+n_{CH_4}+n_{O_2}}\\\\\chi_{SO_2}=\frac{0.25}{0.25+0.50+0.50}=0.2$

The partial pressure of a gas is given by Raoult's law, which is:

$p_{SO_2}=p_T\times \chi_{SO_2}$

where,

$p_{SO_2}$ = partial pressure of sulfur dioxide gas

$p_T$ = total pressure = 800 mmHg

$\chi_{SO_2}$ = mole fraction of sulfur dioxide = 0.2

Putting values in above equation, we get:

$p_{SO_2}=800\times 0.2=160mmHg$

Hence, the partial pressure of sulfur dioxide gas in the flask is 160 mmHg

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Explanation:

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<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

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To formulate the formula of the compound, we need to follow some steps:

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Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

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Total mass of sample = 100 g

Mass of carbon = 42.9 g

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To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

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<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

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$C_1:C_2=\frac{1}{2}:1=1:2$

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Therefore, mass of water = $volume\times density=998.8\times 0.9982=997\ g=0.997\ kg$

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First we have to write into the two half-reactions.

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Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

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Now balance the charge.

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