Which best describes how isotopes of the same element differ?

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

Answer: The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

2 years ago

When did jellyfish, bacteria and sponges thrive?

lina2011 [118]

The Cambrian period I think

3 0

2 years ago

If I have 7.5 L of a gas in a piston at a pressure of 1.75 atm and compress the gas until its volume is 3.7 L, what will the new

Sergio [31]

Answer:

3.55atm

Explanation:

We will apply Boyle's law formula in solving this problem.

P1V1 = P2V2

And with values given in the question

P1=initial pressure of gas = 1.75atm

V1=initial volume of gas =7.5L

P2=final pressure of gas inside new piston in atm

V2=final volume of gas = 3.7L

We need to find the final pressure

From the equation, P1V1 = P2V2,

We make P2 subject

P2 = (P1V1) / V2

P2 = (1.75×7.5)/3.7

P2=3.55atm

Therefore, the new pressure inside the piston is 3.55atm

7 0

3 years ago

What is the average mass of a single chlorine atom in grams

Archy [21]

If you look at the chlorine box, with the symbol Cl, you see the atomic mass is equal to 35.453 atomic mass units. This is the weighted average mass of chlorine, including its isotopes, as found in nature. This also means that one mole of chlorine atoms has a mass of 35.453 grams.

5 0

2 years ago

Calculate the molarity (M) of a 600 ml solution of 0.54 moles of H2SO4.

icang [17]

Molarity (concentration) can be calculated by the equation:
Concentration = moles / volume in L = 0.54 mol / 0.6 L = 0.9 M
Hope this helps!

8 0

3 years ago