Answer : The vapor pressure of propane at is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at
= ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane =
= normal boiling point of propane =
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
Answer:
B) 0.6M
Explanation:
I apologize in advance if it is not correct :l
The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.
(M1*V1= M2*V2)
so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.
M1= 6M (M1*V1= M2*V2)
V1= 5mL
M2= ?
V2= 50mL
now we plug in what we know into the equation to find the unknown (M2)
(6M*5mL= M2*50mL)
now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!
We are going to put the 50mL under the (6M*5mL) for division.
This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M
honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)
Good Luck!