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lorasvet [3.4K]
3 years ago
6

If you have 5.42 x 1024 aluminum atoms, approximately how many moles is that

Chemistry
1 answer:
sattari [20]3 years ago
4 0

Answer:

<h2>9.00 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.42 \times  {10}^{24} }{6.02 \times {10}^{23} }  \\  = 9.0033

We have the final answer as

<h3>9.00 moles</h3>

Hope this helps you

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How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
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Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

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ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

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Charles's law is an experimental gas law that shows the relationship between the temperature of a gas and its corresponding volu
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HNO3 + S --&gt; H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i
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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

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The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

HNO_3+S\rightarrow H_2SO_4+NO

<u>On the reactant side:</u>

Oxidation number of H = +1

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<u>On the product side:</u>

Oxidation number of H = +1

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As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

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Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

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