Answer:
= 61.25 g
= 88.75 g
Explanation:
= 
= 50 g
⇒ 
= 
= 1.25 (moles)
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
2 : 1 : 1 : 2
1.25 (moles)
⇒ 
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ 
= 0.625 × 98 = 61.25 g
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ = 0.625 × 142 = 88.75 g
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.
Relatively few hydrogen atoms
Answer:
V2 = 35.967cm^3
Explanation:
Given data:
P1 = 0.2atm
P2 = 1.4atm
V1 = 250cm^3
V2 = ?
T1 = 10°C + 273 = 283K
T2 = 12°C + 273 = 285K
Apply combined law:
P1xV1/T1 = P2xV2/T2 ...eq1
Substituting values:
0.2 x 250/283 = 1.4 x V2/285
Solve for V2:
V2 = 14250/396.2
V2 = 35.967cm^3
