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3 years ago

What is the percentage composition of each element in dinitrogen monoxide, N2O?

Chemistry

2 answers:

Kitty [74]3 years ago

4 0

The correct answer is C :)

Lana71 [14]3 years ago

4 0

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

To calculate the percentage composition of an element in a compound, we use the formula:

$\text{Mass percent of element}=\frac{\text{Mass of element in compound}}{\text{Mass of the compound}}\times 100$

We are given:

Molar mass of $N_2O$ = 44.013 g/mol

Mass of nitrogen in compound = $[2\times 14.0067]=28.0134g/mol$

Mass of oxygen in compound = 15.999 g/mol

By putting and calculating the mass percentages of nitrogen and oxygen, we get:

$\text{Mass percent of nitrogen}=\frac{28.0134g/mol}{44.013g/mol}\times 100=63.64\%$

$\text{Mass percent of oxygen}=\frac{15.999g/mol}{44.013g/mol}\times 100=36.36\%$

Hence, the correct answer is Option c.

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$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

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Although we are not given any value about the present weight of $^{235}U$ .

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Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

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here;

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Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

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Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

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$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

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$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

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