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olganol [36]
3 years ago
8

A sample of nitrogen gas has a mass of 48.6 grams. How many N2 molecules are there in the sample? molecules Submit Answer &

Next
Chemistry
1 answer:
marysya [2.9K]3 years ago
6 0

Answer:

There are 1.05  x 10²⁴ molecules in 48.6 g N₂

Explanation:

1 mol of N₂ has a mass of (14 g * 2) 28 g.

Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.

Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be

(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05  x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).

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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}]

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

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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:

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

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