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Paladinen [302]
3 years ago
13

How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 to 100.0

Chemistry
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

Q = 19255.6 j

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of water = 46 g

change in temperature = ΔT = 100-0.0 = 100 °C

Heat absorbed by water = ?

Solution:

Specific heat capacity of water = 4.186 j/g. °C

Q = m.c. ΔT

Q = 46 g×4.186 j/g. °C×100 °C

Q = 19255.6 j

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A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sampl
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<u>Answer:</u>

<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u> The rate of effusion of SO_2 gas is 1.006\times 10^{-3}mol/hr

<u>Explanation:</u>

<u>For A:</u>

The average molecular speed of the gas is calculated by using the formula:

V_{gas}=\sqrt{\frac{8RT}{\pi M}}

     OR

V_{gas}\propto \sqrt{\frac{1}{M}}

where, M is the molar mass of gas

Forming an equation for the two gases:

\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}          .....(1)

Given values:

V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol

Plugging values in equation 1:

\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u>

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

Rate\propto \frac{1}{\sqrt{M}}

Where, M is the molar mass of the gas

Forming an equation for the two gases:

\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}          .....(2)

Given values:

Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol

Plugging values in equation 2:

\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr

Hence, the rate of effusion of SO_2 gas is 1.006\times 10^{-3}mol/hr

8 0
3 years ago
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