Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
Answer:
C₂ = 0.334 M
Explanation:
Given data:
Volume of HCl = 0.0780 L
Concentration of HCl = 0.12 M
Volume of LiOH = 0.0280 L
Concentration of LiOH = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Concentration of HCl
V₁ = Volume of HCl
C₂ = Concentration of LiOH
V₂ = Volume of LiOH
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.12 M × 0.0780 L = C₂ × 0.0280 L
0.00936 M.L = C₂ × 0.0280 L
C₂ = 0.00936 M.L/0.0280 L
C₂ = 0.334 M
Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
The reaction between boron sulfide and carbon is given as:
2B2S3 + 3C → 4B + 3CS2
As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.
Given data:
Mass of C = 2.1 * 10^ 4 g
Mass of B = 3.11*10^4 g
Mass of CS2 = 1.47*10^5
Mass of B2S3 = ?
Now based on the law of conservation of mass:
Mass of B2S3 + mass C = mass of B + mass of CS2
Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5
Mass of B2S3 = 15.7 * 10^4 g
We are tasked to solve for the volume of the gas that occupies when pressure and temperature changes to 400 Torr and 200 Kelvin from Torr and 400 Kelvin. We can use ideal gas law assuming constant gas composition and close system. The solution is shown below:
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1P2
V2 = 800*72*200 / 400*400
V2 = 72 ml
The answer for the volume is 72 ml.
