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nordsb [41]
3 years ago
6

Which chemical equation demonstrates the law of conservation of mass!

Chemistry
1 answer:
zlopas [31]3 years ago
5 0

Answer:

F. 2NO + 02 —> 2NO

H. 4NH3 + 502 —> 4NO + 6H20

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.

2NO + 02 —> 2NO

From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.

4NH3 + 502 —> 4NO + 6H20

From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.

The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.

This is certified by the law of conservation of mass.

The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
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Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

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\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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