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shtirl [24]
3 years ago
15

Which best describes the law of conservation of mass? The coefficients in front of the chemicals in the reactants should be base

d on the physical state of the products. Products in the form of gases are not considered a part of the total mass change from reactants to products. When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not. The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Chemistry
2 answers:
devlian [24]3 years ago
5 0

Answer:

D. The mass of the reactants and products is equal and is not dependent on the physical state of the substances.

Explanation:

cestrela7 [59]3 years ago
3 0

Answer:

D) The mass of the reactants and products is equal and is not dependent on the physical state of the substances.

Explanation:

I took the quiz on Edgenuity

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An artifact contains one-fourth as much carbon-14 as the atmosphere. how old is the artifact
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<h3><u>Answer;</u></h3>

= 11,460 years

<h3><u>Explanation;</u></h3>
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<em>The initial amount of carbon-14 in this case was 1 whole; thus; </em>

<em>1 → 1/2 →1/4</em>

<em>To contain 1/4 of the value, 2 half-lives have passed. </em>

<em>But, 1 half life = 5,730 years</em>

<em>Therefore; The artifact is is therefore: 2 x 5,730 </em>

<em>          = 11,460 years </em>

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2 years ago
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First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Klio2033 [76]

Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

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t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

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t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

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To learn more about the specific heat capacity please click on the link brainly.com/question/16559442

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