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pychu [463]
3 years ago
9

A container is filled with helium gas. It has a volume of 2.25 liters and contains 9.00 moles of helium. How many moles of heliu

m could be held in a 1.85 liter container at the same temperature and pressure?
Chemistry
1 answer:
Darina [25.2K]3 years ago
7 0

Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

Explanation:

Given: V_{1} = 2.25 L,     n_{1} = 9.0 mol

V_{2} = 1.85 L,            n_{2} = ?

Formula used to calculate the moles of helium are as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{2.25 L}{9.0 mol} = \frac{1.85 L}{n_{2}}\\n_{2} = \frac{1.85 L \times 9.0 mol}{2.25 L}\\= 7.4 mol

Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.

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What is the percent of nitrogen (N) by mass is a 55.0g sample of Fe(NO3)3? The molar mass of Fe(NO3)3 is 241.86g/mol.​
Anna35 [415]

Answer:

22.74%

Explanation:

The mass of nitrogen is 55g

The mass of Fe(NO3)3 is 241.86g/mol

%of nitrogen= mass of nitrogen/molar mass of Fe(NO3)3 ×100

= 55/241.86×100

=5500/241.86

=22.74%

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What is the gfm of CO2?
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4 0
3 years ago
1. Determine the number of atoms in 4.64 g gold.
liubo4ka [24]

Answer:

The answer to your question is below

Explanation:

1.-

a) Look for the mass number of Gold in the periodic table

Mass number = 197 g

Use proportions to answer this question

                          197 g of gold ------------- 6.023 x 10 ²³ atoms

                            4.64 g of gold ----------  x

                            x = (4.64 x 6.023 x 10²³) / 197

                           x = 1.42 x 10²² atoms

2.- mass of 4.12 x 10²⁴ molecules of glucose

Molecular mass = (6 x 12) + (12 X 1) + (6 x 16)

                          = 72 + 12 + 96

                          = 180 g

             180 g of glucose ------------------- 6.023 x 10 ²³ molecules

                x                        -------------------  4.12 x 10 ²⁴ molecules

                x = (4.12 x 10²⁴ x 180) / 6.023 x 10²³

               x = 1231.3 g of glucose                  

4 0
3 years ago
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