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3 years ago

During which changes of state does the motion of the atoms decrease?

Chemistry

2 answers:

lesantik [10]3 years ago

4 0

freezing and condensation

Explanation:

The motion of the atoms decreases during freezing and condensation. These are phase changes that increases the molecular attraction between particles and forces them to come closer.

Freezing is a phase change from liquid to solid. It is usually achieved by removing heat from a substance and cooling it.

Condensation is a phase change from gases to liquid. .

Gases have the greatest motion because they have little to no attractive forces between them.

Phase changes from gas to solid to liquid generally reduces the motion of the atoms.

Learn more:

Phase change brainly.com/question/1875234

#learnwithBrainly

Thepotemich [5.8K]3 years ago

3 0

Answer:

FREEZING AND CONDENSATION

Explanation:

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The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

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3 years ago

Ill give brainlest to the one who answers right.

liraira [26]

The answer would be h2o

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Explanation:

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