The density of benzene at 15 ∘c is 0.8787 g/ml. calculate the mass of 0.1000 l of benzene at this temperature.
2 answers:
Answer: The mass of benzene is 87.87 grams
Explanation:
To calculate mass of a substance, we use the equation:
We are given:
Density of benzene =
mass of benzene = ?
Volume of benzene = 0.1000L = 100ml
Putting values in above equation, we get:
Mass of benzene = 87.87g
Hence, the mass of benzene is 87.87 grams
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Answer:
Final volumeof the gas = 2.84 L
Explanation:
The formular to be used here is the general gas equation. the formular is being used because it gives the relationship between the three gas parameters (volume, temperature and pressure) mentioned.
The general gas equation is given as;
where;
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature
From the question,
P1 = 1.00 atm
P2 = 0.85atm
T1 = 25C + 273 = 298K (Converting to kelvin)
T2 = 15C + 273 = 288K (Converting to kelvin)
V1 = 2.5L
V2 = ?
from the equation, making V2 subject of formula we have;
V2 = (1*2.5*288)/(298*0.85) = 2.84 L.
Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:
From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton,
n = 0.146 moles
for chlorine,
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.
= 32.94 g of
So, 32.94 g of the product will form.