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HACTEHA [7]
2 years ago
11

This lab question is: How can you distinguish a physical change from a chemical change?

Chemistry
1 answer:
geniusboy [140]2 years ago
4 0

Answer:b,and c

Explanation:

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The correct answer is A.) Volatile. please mark brainliest (:
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An example of a substance that can be decomposed by a chemical change is:
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What element is being oxidized in the following redox reaction?
DIA [1.3K]

Answer:

The element that has been oxidized is the N

Explanation:

Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)

See all the oxidation states:

Zn²⁺  → acts with +2

In ammonia, H acts with +1 and N with -3

Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.

Zn changed from 2+ to 0. The oxidation number, has decreased.

This element has been reduced.

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O acts with -2

The global charge is -1

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8 0
2 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

6 0
3 years ago
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