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Vitek1552 [10]
3 years ago
5

A sample of N2O3(g) has a pressure of 0.046 atm . The temperature (in K) is then doubled and the N2O3 undergoes complete decompo

sition to NO2(g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change. Enter your answer numerically, in terms of atm.
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

0.184 atm

Explanation:

The ideal gas equation is:

PV = nRT

Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.

So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.

The decomposition occurs:

N₂O₃(g) ⇄ NO₂(g) + NO(g)

So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.

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Answer:

850 Calories.

Explanation:

The following data were obtained from the question:

Mass (M) = 50 g

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Final temperature (T2) = 37 °C

Specific heat capacity (C) of water = 1 cal/g°C

Heat absorbed (Q) =..?

Next, we shall determine the change in temperature of water.

This can be obtained as follow:

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Final temperature (T2) = 37 °C

Change in temperature (ΔT) =..?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 37 – 20

Change in temperature (ΔT) = 17 °C

Finally, we shall determine the heat absorbed. This can be obtained as follow:

Mass (M) = 50 g

Specific heat capacity (C) of water = 1 cal/g°C.

Change in temperature (ΔT) = 17 °C

Heat absorbed (Q) =..?

Q = MCΔT

Q = 50 x 1 x 17

Q = 850 Calories

Therefore, the heat absorbed is 850 Calories.

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