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3 years ago

Please help How many moles are there in 7.57 x 10^30 molecules of CO2?

Chemistry

1 answer:

Oxana [17]3 years ago

3 0

Answer:

The answer is

<h2>12,574,750.83 moles</h2>

Explanation:

In order to find the moles of CO2 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

N = 7.57 × 10^30 molecules of CO2

We have

$n = \frac{7.57 \times {10}^{30} }{6.02 \times {10}^{23} } \\ = 12574750.83056...$

We have the final answer as

<h3>12,574,750.83 moles</h3>

Hope this helps you

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Answer:

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Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

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This expression is substituted into the first equation and we solve for x.

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We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

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3 years ago

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2 years ago

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Answer:

Explanation:

Step 1:

The balanced equation for the reaction.

Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

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Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

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