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2 years ago

A sample of nitrogen occupies 10.0 liters at 25°c and 98.7 kpa. What would be the volume at 20°c and 102.7 kpa?

Chemistry

1 answer:

Fynjy0 [20]2 years ago

5 0

Answer:

9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

P₁ = 98.7 Kpa

T₂ = 20°C = 20 + 273.15 K = 293.15 K

P₂ = 102.7 KPa

V₂ = ?

2) Formula:

Used combined law of gases:

PV / T = constant

P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ = 9.4 liter ← answer

You can learn more about gas law problems reading this other answer on Brainly: brainly.com/question/12732788.

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Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

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m = mass of the substance

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$\Delta T$ = change in temperature

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Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

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Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

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Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

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We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

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Change in temperature = 34.04°C

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We are given:

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Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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