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
#SPJ1
Answer:
electrons located outside the nucleus
Explanation:
electrons are said to be located outside the nucleus of an atom and are in orbit around the nucleus.
according to Bohr in 1913.
7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x ]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
Answer:
Yes.
Explanation:
Because distillation separates substances based on their different boiling points.
