Answer:
New volume = 8.00 Liters
Explanation:
Boyles Law => P₁V₁ = P₂V₂
P₁ = 1.0 Atm
V₁ = 20.0 L
P₂ = 2.5 Atm
V₂ = ?
P₁V₁ = P₂V₂ = V₂ = V₁(P₁/P₂) = 20.0L(1.0Atm/2.5Atm) = 8.00L
The answer to your question is "Participant Observation".
Answer:
The molar mass of the metal is 54.9 g/mol.
Explanation:
When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.
Patm = Pwater + PH₂
PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar
The pressure of H₂ is:
The absolute temperature is:
K = °C + 273 = 25°C + 273 = 298 K
We can calculate the moles of H₂ using the ideal gas equation.
Let's consider the following balanced equation.
M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)
The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:
Answer:
4.79 g of water
Explanation:
From the reaction equation;
C2H6(g) + 7/2O2(g) ----> 2CO2(g) + 3H2O(g)
Next we convert the given masses of reactants to moles of reactants.
For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles
For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles
Next we determine the limiting reactant, the limiting reactant yields the least amount of product.
For ethane;
From the reaction equation,
1 mole of ethane yields 3 moles of water
0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water
For oxygen;
3.5 moles of oxygen yields 3 moles of water
0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water
Hence oxygen is the limiting reactant.
Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water
