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natta225 [31]
3 years ago
14

Which conversion factor is needed to solve the following problem?

Chemistry
1 answer:
USPshnik [31]3 years ago
3 0
Molar Volume is required to solve this problem. As we know that "1 mole of any gas at standard temperature and pressure occupies 22.4 L of volume". SO using this concept, we can calculate the volume of ammonia formed by reacting 54.1 L of Hydrogen gas as follow,

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
Name two example of oxygenase
Nataly_w [17]
Air sksksks and I oop-
4 0
3 years ago
Herbie left his house at 3:30 p.m. and arrived at Kit’s house at 5:00 p.m. The two boys live 7 miles apart. What was Herbie’s av
Kisachek [45]

Answer:

Step-by-step explanation:

Alright, lets get started.

Suppose they take t minutes to meet each other.

Distance covered by first friend in t minutes, = 0.2 *t=0.2∗t

Distance covered by second friend in t minutes , =0.15 *t=0.15∗t

Total distance is given as 7, so

0.2 t + 0.15 t = 70.2t+0.15t=7

0.35 t = 70.35t=7

t = 20t=20

means after 20minutes they will meet.

SO. the average speed is 10m: Answer

5 0
3 years ago
Calculate the internal energy of 2 moles of argon gas (assuming ideal behavior) at 298 K. Suggest two ways to increase its inter
Dominik [7]
I hope this helps you.

5 0
3 years ago
Answer the question below pls
Veseljchak [2.6K]

Answer:

The answer would be at 120°C

6 0
3 years ago
Read 2 more answers
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