When m for water is only 5 % so, m for the acid is (m%) = 95%
and D = 1.84 g/ml, assume 1 L of solution therefore,
1- D= m / v
m = D x v = 1.84 g/ml x 1 L x 10^3 ml /L
= 1840 g solution
2- m% = m acid / m sol x 100 %
m acid = 1840 g H2So4 x 0.95
= 1750 g H2So4
3- n = 1750 g H2So4 x 1 mol H2So4 x 98.09 H2So4
= 17.8 mol H2So4
4- So the molarity of { H2So4} = n / v = 17.8 mol H2So4 / 1 L
= 17.8 M
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
This is an example of Charles’ Law problems, the basic equation is: V1/T1 = V2/T2. One vital thing to recall for all gas law problems is that the temperature must be in Kelvin (not Celsius).
So our given is 10.0 C = 283 K. So
V1/T1 = V2/T2
733/283 = 950/T2
T2 = 367 K
Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
