All categories

2 years ago

What maximum fraction of the air in the room could be displaced by the gaseous nitrogen?

1 answer:

6 0

The atmosphere is made up of different gases. The air in any room can be likened to the air outside and so the same proportions of gases would apply. The air surrounding the earth is composed of 78% nitrogen, 21% oxygen, 1% argon and trace amounts of other gases including carbon dioxide. These values indicate therefore that the maximum fraction to be displaced is 78%.

You might be interested in

The pOH of a solution is 9.21. Calculate the hydrogen ion concentration of the solution. Be sure to report your answer to the co

11Alexandr11 [23.1K]

Answer:

[OH-] = 6.17 *10^-10

Explanation:

Step 1: Data given

pOH = 9.21

Step 2: Calculate [OH-]

pOH = -log [OH-] = 9.21

[OH-] = 10^-9.21

[OH-] = 6.17 *10^-10

Step 3: Check if it's correct

pOH + pH = 14

[H+]*[OH-] = 10^-14

pH = 14 - 9.21 = 4.79

[H+] = 10^-4.79

[H+] = 1.62 *10^-5

6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14

3 0

2 years ago

Which of these is a chemical property of a substance?

dlinn [17]

Answer:

Reactivity

Explanation:

8 0

2 years ago

What is the volume in liters of 321 g of a liquid with a density of 0.84 g/mL?

garri49 [273]

If we have 321 grams of a liquid, and the density is 0.84 g/mL, then we can easily find the volume of the liquid. We just need to take this 0.84 and multiply that by the number of grams. If we do 321 * 0.84, we get 269.64 mL. This is the volume that this liquid has.Remember this equation for future problems: V = D*M. V meaning volume, D meaning density, and M meaning mass. I hope this helps.

7 0

2 years ago

Read 2 more answers

A concentration cell consists of two zn/zn2+ electrodes. the electrolyte in compartment a is 0.10 m zn(no3)2 and in compartment

Dmitrij [34]

Answer:

E= -0.023V

Explanation:

find the solution below

7 0

3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

5 0

2 years ago