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2 years ago

A sample of gas contains 0.1200 mol of H2(g) and 0.1200 mol of O2(g) and occupies a volume of 11.5 L. The following reaction

Chemistry

1 answer:

Serjik [45]2 years ago

7 0

Answer:

5.75L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.

As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.

Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

<em>Where V is volume and n are moles of 1, initial state and 2, final state.</em>

Replacing:

V1 = 11.5L

n1 = 0.2400 moles

V2 = ?

n2 = 0.1200 moles

11.5L*0.1200 moles / 0.2400 moles = V2

V2 = 5.75L is the volume of the sample after the reaction

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Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is $2.9^0C$

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$M_2$ = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

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$T_b=49.1^0C$

Therefore, the change in boiling point is $(49.1-46.2)^oC=2.9^0C$

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