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umka21 [38]
3 years ago
13

What is the mass of 4.32 mol GaI3? 455 g 1,230 g 1,720 g 1,950 g

Chemistry
1 answer:
denis23 [38]3 years ago
8 0

Answer:

D. 1,950 g

Explanation:

I completed the test and it is the correct answer!

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Which of the following is the best explanation for why it is important to follow lab safety guidelines
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Answer:

B

Explanation:

The main aim of lab rules is to minimize the no of accidents

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3 years ago
78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a
saul85 [17]

1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>2.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}P_{O_2}=1.57\text{ }atm

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2

<em>3.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>3.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}P_{F_2}=1.13\text{ }atm

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

P_T=P_A+P_B

4.2- Plug in the known quantities.

P_T=1.57\text{ }atm+1.13\text{ }atmP_T=2.7\text{ }atm

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

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