39122.4.33 molar mass 23.4 O. 12=mplar mass of cl₂ = 40.90° d. If 464mL of an ideal gas is at 27°C and 1.05atm, what will be new volume at STP? Gas Law: Solve for the Unknown:

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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please help with 2(d)
752.348mmHg/
c. kPa
0.4899 atm=0.9899atmx 101.325 kPa
latm
760mmHg
=100.3kfa
2. With the given information: i) State the gas law that can be used and ii) Solve for the
unknown.
a. If a 425mL of He gas is cooled from 76°C to 22°C, what is the new volume of the
gas?
Gas Law: Charle's law
V₁ V₂
T₁ T₂
Solve for the Unknown:
T₁=76°C=76+273K=349K
T2-22°c=22+273K = 295k
v=425ML
4251245
V2=359 24mL
b. The new pressure of a gas when it expands fo 1.25L when originally 235mL and
5.34atm.
Gas Law: Boyle's law
P₁V₁ = P₂V₂
-0.1011 atm
V₁ = √₂
1₁ 1₂
c. If 23.4grams of Argon is at a volume if1.35L, what volume will the same mass of Cl2
gas take up given that all other conditions are the same? (hint for this one-compare
moles)
Gas Law:
Avogadro's law
b. Pressure of the Helium gas in atm
Solve for the Unknown:
235x5.34 =1250x₂ 1.004atm
P₂=1.004 atm
Solve for the Unknown:
V₁=1.35L
n. 23.49
molar mass of Ar 39.948.
13.4
12=mplar mass of cl₂ = 40.90²
d. If 464mL of an ideal gas is at 27°C and 1.05atm, what will be new volume at STP?
Gas Law:
Solve for the Unknown:
359.24mL
c. Total Pressure of the container using parts a and b.
121.4.33
1.35V2 2.76L
0.76L
23.4 20.5857 0.5857 0.33
3. Suppose 2.94g of H2 and 2.34g of He are trapped in 2.65L container at standard
temperature. Determine the....
a. Pressure of the Hydrogen gas in atm
Transcribed Image Text:752.348mmHg/ c. kPa 0.4899 atm=0.9899atmx 101.325 kPa latm 760mmHg =100.3kfa 2. With the given information: i) State the gas law that can be used and ii) Solve for the unknown. a. If a 425mL of He gas is cooled from 76°C to 22°C, what is the new volume of the gas? Gas Law: Charle's law V₁ V₂ T₁ T₂ Solve for the Unknown: T₁=76°C=76+273K=349K T2-22°c=22+273K = 295k v=425ML 4251245 V2=359 24mL b. The new pressure of a gas when it expands fo 1.25L when originally 235mL and 5.34atm. Gas Law: Boyle's law P₁V₁ = P₂V₂ -0.1011 atm V₁ = √₂ 1₁ 1₂ c. If 23.4grams of Argon is at a volume if1.35L, what volume will the same mass of Cl2 gas take up given that all other conditions are the same? (hint for this one-compare moles) Gas Law: Avogadro's law b. Pressure of the Helium gas in atm Solve for the Unknown: 235x5.34 =1250x₂ 1.004atm P₂=1.004 atm Solve for the Unknown: V₁=1.35L n. 23.49 molar mass of Ar 39.948. 13.4 12=mplar mass of cl₂ = 40.90² d. If 464mL of an ideal gas is at 27°C and 1.05atm, what will be new volume at STP? Gas Law: Solve for the Unknown: 359.24mL c. Total Pressure of the container using parts a and b. 121.4.33 1.35V2 2.76L 0.76L 23.4 20.5857 0.5857 0.33 3. Suppose 2.94g of H2 and 2.34g of He are trapped in 2.65L container at standard temperature. Determine the.... a. Pressure of the Hydrogen gas in atm
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