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3 years ago

How many molecules are in 5.60 L of oxygen gas

Chemistry

2 answers:

Fynjy0 [20]3 years ago

5 0

Answer:

5.6L of O2 means we have 0.25 moles of O2. and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.

topjm [15]3 years ago

3 0

5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.

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Data:
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Answer:

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2 years ago

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

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