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3 years ago

3. How many moles are in 1.49 x 1023 molecules of iodine?

Chemistry

1 answer:

tino4ka555 [31]3 years ago

7 0

Answer:

<h3>The answer is 0.25 moles</h3>

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

$n = \frac{1.49 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{1.49}{6.02} \\ = 0.247508305...$

We have the final answer as

<h3>0.25 moles</h3>

Hope this helps you

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

<u>Calculating for first dilution:</u>

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

<u>Calculating for second dilution:</u>

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

2 years ago

_ an electron moved around the nucleus in circular paths

kifflom [539]

Answer: although it is convenient to think of the electron moving around the nucleus along circular paths

Explanation:

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3 years ago

Which of these contains generic material but is not classified as living?

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Viruses are not classified as being alive in because of the fact that they can't reproduce.

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2 years ago

When iron metal reacts with oxygen, the reaction can form Fe2O3. Write a balanced chemical equation for this reaction, and find

Art [367]

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3 years ago

Chemistry

madam [21]

Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1

Explanation:

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3 years ago