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ycow [4]
3 years ago
14

Composition of Dry AirGas Percent by Volume (%)Nitrogen (N2) 78Oxygen (O2) 21Argon (Ar) 0.9Carbon dioxide (CO2) 0.04 What volume

of air contains 12.3 g of oxygen gas at 273 K and 1.00 atm
Chemistry
1 answer:
Minchanka [31]3 years ago
5 0

Answer:

40.95 L

Explanation:

We'll begin by calculating the number of mole in 12.3 g of O₂. This can be obtained as follow:

Mass of O₂ = 12.3 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mole of O₂ =?

Mole = mass / Molar mass

Mole of O₂ = 12.3 / 32

Mole of O₂ = 0.384 mole

Next, we shall determine the volume occupied by 0.384 mole of O₂. This can be obtained as follow:

Number of mole (n) of O₂ = 0.384 mole

Pressure (P) = 1 atm

Temperature (T) = 273 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of O₂ =?

PV = nRT

1 × V = 0.384 × 0.0821 × 273

V = 0.384 × 0.0821 × 273

V = 8.6 L

Thus, the volume of O₂ is 8.6 L

Finally, we shall determine the volume of air that contains 8.6 L of O₂. This can be obtained as follow:

Volume of O₂ = 8.6 L

Percentage of O₂ in air = 21%

Volume of air =?

Percentage of O₂ = Vol of O₂ / Vol of air × 100

21% = 8.6 / Vol of air

21 / 100 = 8.6 / Vol of air

Cross multiply

21 × Vol of air = 100 × 8.6

21 × Vol of air = 860

Divide both side by 21

Volume of air = 860 / 21

Volume of air = 40.95 L

Therefore, the volume of air is 40.95 L.

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Lynna [10]

Answer:

<em>Hi Todoroki here!!! </em>

Explanation:

Chlorine has the electron configuration [Ne]3s 2 3p 5, with the seven electrons in the third and outermost shell acting as its valence electrons. Like all halogens, it is thus one electron short of a full octet, and is hence a strong oxidising agent, reacting with many elements in order to complete its outer shell.

<em>Your welcome!!</em>

7 0
3 years ago
How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?
GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

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3 years ago
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3 years ago
1. The pressure and temperature of a gas are held constant. Which of the following is true for the volume of the gas?
katovenus [111]
<h3>1.</h3>

C) The volume of the gas is proportional to the number of moles of gas particles.

The Avogadro's law applies to ideal gases with constant pressure and temperature. By that law, the volume of an ideal gas is proportional to the number of moles of particles in that gas.

<h3>2.</h3>

B) The gas now occupies less volume, and the piston will move downward.

Boyle's Law applies to ideal gases with a constant temperature. The volume of an ideal gas is inversely related to its pressure. A high pressure drives gas particles together, such that they occupy less volume. The gas trapped inside the piston has a smaller volume. As a result, the the piston will move downward.

Alternatively, consider the forces acting on the piston. Both the atmosphere and gravity are dragging the piston down. In order for it to stay in place, the gas below it must exert a pressure to balance the two forces. Now the pressure from outside has increased. The gas inside needs to increase its pressure. It needs a smaller volume to create that extra pressure. As a result, its volume will decrease, and the piston will move downwards.

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Draw the alkyl bromide(s) you would use in a malonic ester synthesis of ethyl 3-methylbutanoate.
Gennadij [26K]

The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate.

Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.

The alkylation of diethyl malonate or a related ester of malonic acid at the carbon alpha (immediately next) to both carbonyl groups, followed by conversion to a substituted acetic acid, characterizes the chemical reaction known as the malonic ester synthesis.

As a result, it is evident from the structure of ethyl 2-methyl-4-pentenoate that ethyl and methyl bromides are the alkyl bromides employed.

To learn more about Malonic ester synthesis refer here:

brainly.com/question/17237043

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