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3 years ago

6

What would happen to the moon’s interia were to increase? A the moon will fly into space. B the moon will smash into Earth. C th

e moon will smash into pluto.

1 answer:

Montano1993 [528]3 years ago

5 0

Answer:

B

Explanation:

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A chemical reaction gives off 2100 kg of heat energy how many calories

dlinn [17]

The chemical reaction that gives off 2100 Kg of heat energy has 501.9126 calories

calculation

convert KJ to calories

that is

1 KJ = 0.239006 calories

2100KJ= ? calories

by close multiplication

= (2100 kj x 0.239006)÷ 1 kj =501.9126 calories

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3 years ago

In a decomposition reaction involving a gaseous product, what must be done for the reaction to reach equilibrium?

Vadim26 [7]

To prevent the gaseous product from escaping, the system must be sealed.

<h3>What are equilibrium and an example?</h3>

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Equilibrium can be classified as either stable, unstable, or neutral.

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2 years ago

1.) A balloon contains 0.126 mol of gas and has a volume of 2.74 L .

Ann [662]

Answer:

<u></u>

<u>Problem 1: 5.44L</u>

<u>Problem 2: 479mL</u>

Explanation:

<h2>Problem 1. </h2>

<u>a) Data:</u>

n₁ = 0.126 mol
V₁ = 2.74 L
n₂ = 0.126mol + 0.124mol
V₂ = ?
Conditions: constant temperature and pressure

<u>b) Physical principles:</u>

The volume occupied by a gas, when the temperature and pressure remain unchanged, is proportional to the number of moles of gas.

$\dfrac{n_1}{V_1}=\dfrac{n_2}{V_2}$

<u>c) Solution:</u>

Solve for V₂ , substitute and compute:

$V_2=n_2\times\dfrac{V_1}{n_1}\\ \\\\ V_2=(0.126mol+0.124 mol)\times \dfrac{2.74L}{0.126mol}\\ \\ \\ V_2=5.44L$

<h2>Problem 2.</h2>

<u>a) Data:</u>

n₁ = 0.547mol
V₁ = 295 mL
V₂ = ?
n₂ = 0.547mol + 0.342 mol
Conditions: constant temperature and pressure

<u>b) Physical principles:</u>

The same as above: the volume occupied by a gas, when the temperature and pressure remain unchanged, is proportional to the number of moles of gas.

$\dfrac{n_1}{V_1}=\dfrac{n_2}{V_2}$

<u>c) Solution:</u>

Solve for V₂ , substitute and compute:

$V_2=n_2\times\dfrac{V_1}{n_1}\\ \\\\ V_2=(0.547mol+0.342 mol)\times \dfrac{295mL}{0.547mol}\\ \\ \\ V_2=479mL$

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