Solving this chemistry is a little bit hard because the question didn't give some important detailed.
So first, there are a couple problems with your question.
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be:
Kc = [C] / [A] [B]^2
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.
Answer: pH = 14
Explanation: Please see the attachments below
Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol
3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction 3 mol 2 mol
mass from reaction 3mol* 111.1g/mol 2 mol*133.5g/mol
333.3 g 267.0 g
mass from problem 45.7 g x g
Proportion:
333.3 g CaCl2 ------- 267.0 g AlCl3
45.7 g CaCl2 -------- x g AlCl3
x=45.7*267.0/333.3= 36.6 g AlCl3
Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
