Answer:
Explanation:
a) 1 troy ounce is equal to 31.103 g
2.41 troy ounce is equal to 2.41 x 31.103 g
= 74.958 g
b) 1 ounce = 1 / 16 lb = 1/16 x 453.6 g = 28.35 g.
1 troy ounce = 31.103 g
so 1 troy ounce is heavier than 1 ounce.
The volume of the gas at a temperature of 405.0 K would be 607.5 mL. Making option D the right answer to the question.
What is the volume of the gas?
To find the volume of the gas, the equation to be used would have to be combine gas law.
Combine gas law as the name suggest uses the combination of Charles law which measures Volume against temperature, and Gay-Lussac's law which measures Pressure/Temperature, and Boyle's law which measures pressure X volume where k is constant.
Using the combine law to find the volume, we have:
P₁V₁/T₁=P₂V₂/T₂
Where P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
P₁ = 2.25atm
V₁ = 450.0 mL
T₁ = 300 K
T₂ = 405.0 K
V₂ = ?
D) 607.5 mL
= [2.25(450)]÷300=[2.25(V₂]÷405
Making V₂ the subject
3.375=2.25 V₂ ÷ 405
V₂ = 3.375 x 405 ÷ 2.25
V₂ = 607.5 mL
In summary, a gas with an initial pressure of 2.25atm, an initial pressure of 450.0 mL and an initial temperature of 300 K would have a final volume of 607.5 mL if the temperature is increased to 405.0 K.
Learn more about Combine gas law here: brainly.com/question/13538773
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Explanation:
a)phenotype= 100% big
b) genotype=1:2:1 (FF-25%, Ff-50%, ff-25%)
phenotype=3:1 (big=75%, small= 25%)
Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where 
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
