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3 years ago

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A major textile dye manufacturer developed new yellow dye. The dye has a percent composition of 75.9 5% C, 17.72% N

1 answer:

AnnZ [28]3 years ago

5 0

Answer:

Such molecule must have molecular formula of C15N3H15

Explanation:

Mass of carbon in such molecule

$0.7595*240_{g/mol} =182.28_{g C/mol}$

The atomic mass of carbon is 12.01 g/mol, so in 182.28 g of carbon there is 15.18 mols of carbon.

Mass of Nitrogen in such molecule

$0.1772*240_{g/mol} =37.73_{g C/mol}$

The atomic mass of nitrogen is 14.01 g/mol, so in 42.53g of nitrogen there is 3.04 mols of nitrogen.

Mass of Hydrogen in such molecule

$0.0633*240_{g/mol} =15.19 {g C/mol}$

The atomic mass of Hydrogen is 1.00 g/mol, so in 15.19 g of Hydrogen there is 15.19 mols of Hydrogen.

Such molecule must have molecular formula of C15N3H15

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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 $61.29M^{-1}s^{-1}$

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Learn more about entropy at: brainly.com/question/21578229

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