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suter [353]
3 years ago
9

3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of

Chemistry
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen =  0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

                              no. of mole = mass in g / molar mass

Put value in above formula

                          no. of mole = 9.8 g/ 14 g/mol

                          no. of mole = 0.7

                           mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

                     no. of mole = mass in g / molar mass

Put value in above formula

                     no. of mole = 0.70 g/ 1 g/mol

                      no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

                       no. of mole = mass in g / molar mass

Put value in above formula

                       no. of mole = 33.6 g / 16 g/mol

                       no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

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Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

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Taking logarithm on both the sides, we get:

\ln k=-\frac{Ea}{RT}+\ln A             ............(1)

where,

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Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)

Ea=108318.365J/mol=108.318kJ/mol

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

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