The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer:
The % yield is 56.6 %
Explanation:
This is the reaction:
C₂H₄ + Cl₂ → C₂H₄Cl₂
Molar mass of ethylene gas: 28 g/m
Mol = mass / molar mass
140 g / 28 g/m = 5 moles
Ratio is 1:1, so 5 moles of ethylene produce 5 moles of dichloro ethane.
Molar mass of C₂H₄Cl₂ = 98.9 g/m
Mass of C₂H₄Cl₂ produced = 98.9 g/m . 5 m → 494.5 g
% yield reaction
(280 g / 494.5 g ) . 100 = 56.6%
I think it would be B
hope this works
2 are formed :) hope this helped!
Answer:
The partial pressure of nitrogen is 0.402 atm.
Explanation:
Given data:
Number of moles of helium = 1 mol
Number of moles of nitrogen = 2 mol
Total pressure of mixture = 0.60 atm
Partial pressure of nitrogen = ?
Solution:
First of all we will calculate the mole fraction of nitrogen.
mole fraction of nitrogen = moles of nitrogen / total number of moles
mole fraction of nitrogen = 2 mol / 3 mol = 0.67
Partial pressure of nitrogen:
P₁ = [ n₁ /n(t)] × Pt
P₁ = 0.67 × 0.60 atm
P₁ = 0.402 atm
