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3 years ago

How many liters of volume is one mole of gas at standard temperature and pressure?

Chemistry

1 answer:

horsena [70]3 years ago

3 0

22.7 liters

The molar volume of an ideal gas depends on the temperature and pressure. One mole of any ideal gas occupies 22.7 liters at 0 0C and 1 bar (STP).

Hope this helped

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Which one is the right answer?

mina [271]

Answer:

1,000

Explanation:

5 0

3 years ago

Writing net ionic equation: of cu with nitric acid

Lesechka [4]

The net ionic equation is

Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)

Molecular equation :

Cu(s) + 4HNO₃(aq) ⟶ Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(ℓ)

Ionic equation:

Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)

Net ionic equation

Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)

Note: The net ionic equation is the same as the ionic equation because there are no common ions to cancel on opposite sides of the arrow.

3 0

3 years ago

Given: There are 39.95 grams of Argon (39.95 g/1 mole) and one mole has a volume of 22.4 Liters (1 mole/22.4 L). What is the vol

marta [7]

Answer:

$V_2=19.23L$

Explanation:

Hello,

In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:

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

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

$n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol$

Thus, we find:

$V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L$

Best regards.

6 0

3 years ago

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

You should never heat chemicals in a closed system (closed bottle or flask) because

kondaur [170]

You should never heat chemicals in a closed system (closed bottle or flask) because it might explode.

4 0

3 years ago