If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?

Answers

Answer 1

Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.

Explanation:

Given: [tex]n_{1}[/tex] = 4.00 moles,          [tex]V_{1}[/tex] = 5.0 L

[tex]n_{2}[/tex] = 3.00 moles,                 [tex]V_{2}[/tex] = ?

Formula used is as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]

Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.


Related Questions

a) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?

Answers

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

cuáles son las características de la luz y en qué consisten​

Answers

Answer:

Cuáles son las características de la luz y en qué consisten?

Explanation:

La luz es una radiación que se propaga en forma de ondas. Las ondas que se pueden propagar en el vacío se llaman ONDAS ELECTROMAGNÉTICAS. La luz es una radiación electromagnética

20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.

Answers

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

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