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uysha [10]
3 years ago
7

A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new

pressure is 841 mm hg and the temperature is now 26 °c. there are now ________ mol of gas in the flask.
Chemistry
1 answer:
garik1379 [7]3 years ago
8 0

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

 

1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) / 841 mm Hg

<span>n2 = 2.25 moles</span>

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1.113 kJ or 1,113 J.

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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
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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?

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<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}

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Because there are 14.01 grams per mile of nitrogen atoms.
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14.01 x 2.3= 32.223
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