0.528344 gallons in a 2 liter
The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
Answer:
20.3 % NaCl
Explanation:
Given data:
Mass of solute = 45.09 g
Mass of solvent = 174.9 g
Mass percent of solution = ?
Solution:
Mass of solution = 45.09 g + 174.9 g
Mass of solution = 220 g
The solute in 220 g is 45.09 g
220 g = 2.22 × 45.09
In 100 g solution amount of solute:
45.09 g/2.22 = 20.3 g
Thus m/m% = 20.3 % NaCl
