In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
Answer:
Concentration AgBr at saturation = 7.07 x 10⁻⁷M
Explanation:
Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]
I --- 0 0
C --- +x +x
E --- x x
[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M
Answer is: 10 moles of water will be produced.
Balanced chemical reaction of formation of water:
2H₂ + O₂ → 2H₂O.
n(H₂) = 10 mol; amount of hydrogen gas.
From balanced chemical reaction: n(H₂) : n(H₂O) = 2 : 2 (1 : 1).
n(H₂O) = n(H₂).
n(H₂O) = 10 mol; amount of water.
