Answer:
P(mixture) = 1.92 atm
Explanation:
Given data:
Mass of H₂ = 0.200 g
Mass of N₂ = 1.00 g
Mass of Ar = 0.820 g
Volume = 2 L
Temperature = 20°C
Pressure of mixture = ?
Solution:
Pressure of hydrogen:
Number of moles of hydrogen = mass / molar mass
Number of moles of hydrogen = 0.200 g / 2 g/mol
Number of moles of hydrogen = 0.1 mol
P = nRT / V
P = 0.1 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 2.41 atm. L /2 L
P = 1.2 atm
Pressure of nitrogen:
Number of moles of nitrogen = mass / molar mass
Number of moles of nitrogen = 1 g / 28 g/mol
Number of moles of nitrogen = 0.04 mol
P = nRT / V
P = 0.04 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 0.96 atm. L /2 L
P = 0.48 atm
Pressure of argon:
Number of moles of argon = mass / molar mass
Number of moles of argon = 0.820 g / 40 g/mol
Number of moles of argon = 0.02 mol
P = nRT / V
P = 0.02 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 0.48 atm. L /2 L
P = 0.24 atm
Total pressure of mixture:
P(mixture) = pressure of hydrogen + pressure of nitrogen + pressure of argon
P(mixture) = 1.2 atm + 0.48 atm + 0.24 atm
P(mixture) = 1.92 atm
The mass of water is calculated as follows
find the moles of each reagent
that is moles = mass/molar mass
for H2s = 84.7/ 34= 2.485 moles
O2 = 78.4 / 32 = 2.45 moles
since 2 moles of H2S react with 3 moles of O2 therefore 2.45 moles of oxygen will be used up therefore O2 is the limiting reagent and H2S is in excess
2H2S + 3O2 ----->2So2 + 2H2O
by use of mole ratio between O2 and H20 which is 3:2 the moles of H2O is therefore = 2.45 x2/3= 1.63 moles of H2O
mass of H2O = moles x molar mass
= 1.63 g x 18g/mol = 29.4 g
Answer:
I can´t see, it isnt clear
Explanation:
Answer:a reddish- or yellowish-brown flaky coating of iron oxide that is formed on iron or steel by oxidation, especially in the presence of moisture.
Explanation: so basically rusting happens when a metal comes in contact with moisture and air, at a brown, dirty metal coat forms on it.
