V1 = 30 mL
P1 = 760 torr
P2 = 1520 torr
V2 = ?
applying Boyle's Law
P1*V1 = P2*V2
760 torr * 30 mL = 1520 torr * V2
V2 = 760 torr * 30 mL / 1520 torr
( C ) is correct
The correct answer is B. balance
I think because its the only one to be liquid at normal temperatures.
Answer:
52.45g
Explanation:
The computation of the mass of pure acetic acid in 125mL of this solution is shown below:
The percentage of mass would be equivalent to the g of solute in each 100g of water
As we know that
density = mass ÷ volume
So,
Volume = mass ÷ density
V = 100g / 1.049 (g / ml)
V = 95.328 mL
Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2
i.e.
each 125ml of C_2H_4O_2 there are 52.45g
SO,
x = 40g. 125ml ÷ 95.328
x = 52.45g
The Kj of heat that are needed to completely vaporize 1.30 moles of H2O if the heat of vaporization for water is 40.6 Kj/mole is calculated as below
Q(heat) = moles x heat of vaporization)
=1.30 mol x40.6 kj/mol= 52.78 Kj is needed