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3 years ago

Grams needed to prepare the solution

Chemistry

1 answer:

erastovalidia [21]3 years ago

8 0

Answer:

grams of solute = 31.5 g

Explanation:

Given data:

Grams of solute = ?

%m/v = 3%

Volume = 1050 mL

Solution:

grams of solute = 3 g of NH₄Cl / 100 mL × 1050 mL

grams of solute = 0.03 g × 1050

grams of solute = 31.5 g

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Llana [10]

Answer:

I'm feeling nice today so heres the answer

Explanation:

In the portion of the cell membrane shown in the diagram, the arrow indicates the process of active transport.

Explanation:

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The other options presented are not correct, because

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2 years ago

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

$CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)$

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1\text{ and }n_1$ are the volume and number of moles of methane gas

$V_2\text{ and }n_2$ are the volume and number of moles of hydrogen chloride

We are given:

$V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol$

Putting values in above equation, we get:

$\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL$

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

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3 years ago

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3 years ago

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emmasim [6.3K]

Answer:

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