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3 years ago

Which compound plays an important role in leather processing and electroplating?

Chemistry

2 answers:

Margaret [11]3 years ago

7 0

Answer;

Hydrochloric acid

Explanation;

Hydrochlorc acid plays a vital role in leather processing and electroplating.
Hydrochloric acid has a number of important applications in real life these includes; fertilizers, and dyes, textile, in electroplating, rubber and in the photographic industries.
Electroplating is the process that involves coating a metal using another metal for the purpose of improving their appearance or preventing rusting.
Hydrochloric acids is used in the processing of leather or the tanning of leather process, in which tanneries use the acid to stop the growth of bacteria and also maintain proper pH.

abruzzese [7]3 years ago

5 0

Answer:

hydrochloric acid is the correct answer.

Explanation:

hydrochloric acid compound plays an important role in leather processing and electroplating.

Electroplating is a process in which metal is coated with a thin layer of another metal.hydrochoric acid is used in electroplating as it enables oxidation of copper and removes the copper ions from the solution.

Hydrochloric acid is used in leather processing because tanneries use hydrochloric acid to stop the growth of bacteria and to control the pH level of the leather.

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3 years ago

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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]

Answer: The time taken by the reaction is 84.5 seconds

Explanation:

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

6 0

3 years ago

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Thepotemich [5.8K]

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

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3 years ago

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Explanation:

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3 years ago