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IrinaK [193]
3 years ago
13

10. Air is a homogenous mixture of gases. Oxygen is 20% by volume of air typically. What

Chemistry
1 answer:
raketka [301]3 years ago
3 0

Answer:

893 moles

Explanation:

An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are . 893 moles of Oxygen in 20.0 liters.

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Determine the mass in grams of 3.75 x 10^21 atoms of zinc. (the mass of one mole of zinc is 65.39 g)
Nastasia [14]

Answer: The mass in 3.75 \times 10^{21} atoms of zinc is 0.405 g.

Explanation:

Given: Atoms of zinc = 3.75 \times 10^{21}

It is known that 1 mole of every substance contains 6.022 \times 10^{23} atoms. So, the number of moles in given number of atoms is as follows.

Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol

As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.

Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g

Thus, we can conclude that the mass in 3.75 \times 10^{21} atoms of zinc is 0.405 g.

6 0
3 years ago
If a 22.4 L volume of a sample of gas has a density of 0.900 grams/L at 1.00 atm and 0.00°C. Given
disa [49]

Answer:

Formula Weight of gas sample = 20.1 g/mole => Neon (Ne)

Explanation:

Use Ideal Gas Law formula to determine formula weight and compare to formula weights of answer choices.

PV = nRT = (mass/fwt)RT => fwt = (mass/Volume)RT = Density x R x T

Density = 0.900 grams/L

R = 0.08206 L·atm/mole·K

T = 0.00°C = 273Kfwt = (0.900g/L)(0.08206L·atm/mole·K )(273K)

= 20.1 g/mol => Neon (Ne)

4 0
3 years ago
Land breezes near bodies of water occur because
Aleksandr [31]

B) land cools off much faster than water (creating a air region of high density  above land and lower density above sea)

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4 years ago
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4 years ago
How many atoms are there in total in a molecule of sulfur trioxide SO3??​
MAXImum [283]

Answer:

I got 1.14 mol So3 as the answer

hopefully if this could help

7 0
3 years ago
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