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joja [24]
4 years ago
12

An important reaction sequence in the industrial production of nitric acid is the following:

Chemistry
1 answer:
Nataly_w [17]4 years ago
5 0

Answer:

25.0 mol O₂ are required in the second reaction

Explanation:

N₂ (g) + 3H₂ (g)  →  2NH₃ (g)

4NH₃ (g)  +  5O₂ (g)  →  4NO (g) + 6 H₂O(l)

Molar ratio in first reaction is 1:2

For every mol of N₂. I make 2 moles of ammonia. If I have 20 moles of N₂, i'm going to get, 40 moles of ammonia.

In the second reaction, molar ratio between products is 4:5.

If I obtained 40 moles of ammonia in first step, let's prepare the rule of three.

4 moles of ammonia react with 5 moles of O₂

40 moles of ammonia react with ( 40.5) /4 = 25moles

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it would have caused organisms to move in search of food and better conditions.

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g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water
Marina CMI [18]

<u>Answer:</u> The boiling point of water in Tibet is 69.9°C

<u>Explanation:</u>

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg      (Conversion factor:  1 atm = 760 mmHg)

P_2 = final pressure = 240. mmHg

\Delta H_{vap} = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature or normal boiling point of water = 100^oC=[100+273]K=373K

T_2 = final temperature = ?

Putting values in above equation, we get:

\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC

Hence, the boiling point of water in Tibet is 69.9°C

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3 years ago
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Answer:

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