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3 years ago

If 8.00 mol of an ideal gas at stp were confined to a cube, what would be the length in cm of an edge of this cube?

1 answer:

6 0

To solve this problem it is fundamentally, just look for the volume of the gas and convert it to cm3. At STP 1 mole = 22.4 liters. 8.00 moles x 22.4 liters/mole = 179.2 liters = 179,200 cm^3 Then. get the cube root of 179,200 cm^3. This would be equal to 56.38 cm and thus would be the length of the edge of this cube.

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You unknowingly run over a small nail while driving and it punctures your tire. When you are getting in your car the next mornin

Norma-Jean [14]

The nails can puncture the tires of the vehicles making them deflated after a while. The process of tire deflation due to a puncture is due to the process called an effusion. Thus, option c is correct.

<h3>What is effusion?</h3>

Effusion is the process that defines the escape of the fluids or gases from a system through an outlet with a small diameter compared to the mean free path of the molecules.

When a tire is punctured then the gas from the tire starts to move out through the hole through effusion. This leads to deflation of the tire after some time as all the gas present inside had moved out completely.

Therefore, option c. a punctured tire deflates due to effusion.

Learn more about effusion here:

brainly.com/question/2823560

#SPJ1

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2 years ago

Which of these compounds is ionic?<br> A. KF<br> B. CO<br> C. CO2<br> D. SO2

sukhopar [10]

The answer is A) KF. The rest are covalent.

8 0

3 years ago

Read 2 more answers

What is an amu?<br> Will give Brainliest!

pantera1 [17]

Answer: Amu is an atomic mass unit.

8 0

2 years ago

50 ml of 0.2 M acetic acid is shaken with 10 g activated charcoal. The concentration of acetic acid is reduced to 0.5 times the

Olegator [25]

Answer:

$0.03 g_{acid}/g_{charcoal}$

Explanation:

The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.

So, we need to calculate the mass of the acetic acid as follows:

$m = n*M = C*V*M$

Where:

n: is the number of moles = C*V

M: is the molecular mass = 60.052 g/mol

C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L

V: is the volume = 50 ml = 0.050 L

$m_{acid} = C*V*M = 0.10 mol/L*0.050 L*60.052 g/mol = 0.30 g$

Now, the amount of solute adsorbed per gram of the adsorbent is:

$\frac{m_{acid}}{m_{charcoal}} = \frac{0.30 g}{10 g} = 0.03 g_{acid}/g_{charcoal}$

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.

I hope it helps you!

3 0

2 years ago

If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density

adelina 88 [10]

Explanation:

The given data is as follows.

Mass = 27.9 g/mol

As we know that according to Avogadro's number there are $6.023 \times 10^{26}$ atom present in 1 mole. Therefore, weight of 1 atom will be as follows.