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Sliva [168]
2 years ago
15

A 0.5 mol sample of He(g) and a 0.5 mol sample of Ne(g) are placed separately in two 10.0 L rigid containers at 25°C. Each conta

iner has a pinhole opening. Which of the gases, He(g) or Ne(g), will escape faster through the pinhole and why?
A) He because the He atoms are moving at a higher average speed than the Ne atoms.
B) Ne because its initial pressure in the container is higher
C) Ne because the Ne atoms have a higher average kinetic energy than the He atoms
D) Both gases will escape at the same rate because the atoms of both gases have the same average kinetic enery
Chemistry
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:

A) He because the He atoms are moving at a higher average speed than the Ne atoms.

Explanation:

The expression for the mean speed is:

C_{avg}=\sqrt {\dfrac{8RT}{\pi M}}

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

T is the temperature

Thus, as seen from the formula,  

The average speed of the gas is inversely proportional to the square root of the molar mass of the gas

So, Helium has lower molecular mass than neon and thus, The speed of the sound in helium is greater when compared to that of neon.

<u>Thus, He will escape faster through the pinhole.</u>

<u>Option A. is correct.</u>

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Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte.
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As you may know, ethyl alcohol, C2H5OH, can be produced by the fermentation of grains, which contain glucose, C6H12O6 → +2C2H5OH
fredd [130]

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆  →  2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

(5.55 .2)/1 = 11.1 moles.

Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

b. Let's determine the liters of ethyl alcohol we need.

1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

Density = Mass /Volume

0.79 g/mL = Mass / 111.1 mL

0.79 g/mL . 111.1 mL = 87.7 g

Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

Ratio is 2:1 so 2 moles of C₂H₅OH are produced by 1 mol of glucose

Therefore 1.91 mol would be produced by (1.91 .1)/2 = 0.954 moles

Finally we convert the moles of glucose to mass:

0.954 mol . 180 g/ 1mol = 171.7 grams.

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