Answer:
30.12 g/m is the molar mass of the compound
Explanation:
Freezing point depression to solve this. The formula for the colligative property is:
ΔT = Kf . m
ΔT = T° freezing pure solvent - T° freezing solution
Kf = Cryoscopic constant
m = molality (mol/kg)
T° freezing pure benzene: 5.5°C
(5.5°C - 1.02°C) = 5.12 °C/m . m
4.48°C = 5.12 °C/m . m
4.48°C / 5.12 °C/m = m → 0.875 mol/kg
Mol = mass / molar mas
Molality = mol /kg
Let's find out the molar mass, with this equation:
(6.59 g / Molar mass) / 0.250 kg = 0.875 mol/kg
6.59 g / molar mass = 0.875 mol/kg . 0.250 kg
6.59 g / molar mass = 0.21875 mol
6.59 g / 0.21875 mol = molar mass → 30.12 g/m
Answer:

Explanation:
We know that:
1 atm = 760 mm of Hg (Standardly)
Multiply both sides by 2.27
2.27 atm = 760 * 2.27 mm of Hg
2.27 atm = 1,725 mm of Hg
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Hope this helped!
<h3>~AH1807</h3>
Answer:
The correct answer is: K'= 0.033.
Explanation:
The formation of HI from H₂ and I₂ is given by:
H₂ + I₂ → 2 HI K= 29.9
The decomposition of HI is the reverse reaction of the formation of HI:
2 HI → H₂ + I₂ K'
Thus, K' is the equilibrium constant for the reverse reaction of formation of HI. It is calculated as the reciprocal of the equilibrium constant of the forward reaction (K):
K' = 1/K = 1/(29.9)= 0.033
Therefore, the equilibrium constant for the decomposition of HI is K'= 0.033
The grams of the sugar in 125 g of the drink is calculated as below
%M/m) = mass of the solute (sugar)/ mass of the solvent(drink) x100
let the mass of the solute(sugar) be represented by y
convert % into fraction by dividing by 100 = 10.5/100
10.5/100 = y/125
by cross multiplication
100y =1312.5
divide both side by 100
y=13.125 grams