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3 years ago

A3B8 + C2 ---> AC2 +

Chemistry

1 answer:

AURORKA [14]3 years ago

5 0

Answer:

A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C

Explanation:

A₃B₈ + C₂ –> AC₂ + B₂C

The equation can be balance as illustrated below:

A₃B₈ + C₂ –> AC₂ + B₂C

There are 3 atoms of A on the left side and 1 atom on the right side. It can be balance by writing 3 before AC₂ as shown below:

A₃B₈ + C₂ –> 3AC₂ + B₂C

There are 8 atoms of B on the left side and 2 atoms on the right side. It can be balance by writing 4 before B₂C as shown below:

A₃B₈ + C₂ –> 3AC₂ + 4B₂C

There are 2 atoms of C on the left side and a total of 10 atoms on the right side. It can be balance by writing 5 before C₂ as shown below:

A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C

Now, the equation is balanced.

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3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

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3 years ago