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mel-nik [20]
3 years ago
8

Helium gas is being pumped into a rigid container at a constant temperature. As a result, the pressure of helium in the containe

r is increasing. Select the one correct statement below.
a) As the pressure increases, helium atoms move faster, on average.
b) As the pressure increases, helium atoms move more slowly, on average.
c) As the pressure increases, helium atoms stay closer to the wall of the container, on average.
d) As the pressure increases, the volume of the container must decrease.
e) As the pressure increases, there are more collisions of helium atoms with the container wall.
Chemistry
1 answer:
icang [17]3 years ago
8 0

Answer: e. As the pressure increases, there are more collisions of helium atoms with the container wall.

Explanation:

We are given the information that helium gas is pumped into a rigid container at a constant temperature.

Due to this reason, there was a rise in the pressure of helium in the container. We should note that as the pressure rises, there are more collisions of helium atoms with the container wall.

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Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

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