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erma4kov [3.2K]
3 years ago
14

How many moles of N are in 0.203 g of N2O?

Chemistry
2 answers:
kondor19780726 [428]3 years ago
7 0
Molar mass N₂O = 14 x 2 + 16 => 44.0 g/mol

1 mole ---------- 44.0 g
moles ------------ 0.203 g

 0.203 / 44 => 0.004613 moles

1 mol N2O <span>contains 2 atoms N , therefore:

0.004613 x 2 => 0.009226 moles of N

hope this helps!</span>
Mandarinka [93]3 years ago
6 0
To calculate the number of moles of an element in the given mass of a compound, we need to convert mass into moles by using its molecular eight then relate the ratio of the elements.

0.203 g N2O (1 mol / 44.01 g N2O) ( 2 mol N / 1 mol N2O) = 9.23 x 10^-3 mol N
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The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
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Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

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T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

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Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

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q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

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