a) when Kc = concentration of products / concentration of reactants
So according to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)
∴ Kc =[BrCl] ^2 / [Br2][Cl2]
b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m
[Br2] = 1 m
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9
- and we can see that q > Kc
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.
C) by using ICE table:
Br2(g) + Cl2(g) → 2BrCl (g)
initial 1 1 3
change -X -X +X
Equ (1-X) (1-X) (3+X)
when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X) by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m
[BrCl] = 3+0.215 = 3.215 m
Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL
Explanation: in 200mL , 0.1mole
mw NaOH = 40g/mol —> 4g in 0.1 mole
4g in 200mL so 2g in 100mL
density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water
Balanced chemical reaction:
2Na₃PO₄(aq) + 3CaCl₂(aq) → 6NaCl(aq) + Ca₃(PO₄)₂(s).
Ionic reaction:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ca²⁺(aq) + 6Cl⁻(aq) → 6Na⁺(aq) + 6Cl⁻(aq) + Ca₃(PO₄)₂(s).
Net ionic reaction: 2PO₄³⁻(aq) + 3Ca²⁺(aq) → Ca₃(PO₄)₂(s).
<span>(aq) means that
substances are dissociated on cations and anions in water.
</span>(s) means solid.
Answer:
0.0164 g
Explanation:
Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.
Ag⁺(aq) + 1 e⁻ → Ag(s)
We can establish the following relations.
- 1 A = 1 C/s
- The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
- 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
- The molar mass of silver is 107.87 g/mol
The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:
Answer:
I think its might be 1 because the ionic numbers for CA is +2 and for P its +3
