D. 18 as the elements in this group are the noble gases which exist as monoatomic elements with a full outer shell
We have that you can you prepare 450 ml of 1 X buffer A by adding 45ml of 10X buffer A with 405ml(450-45) of water
From the question we are told
If you have 100 ml of 10X Buffer A, how could you prepare 450 ml of 1 X buffer A
Generally the equation for the Concentration Volume relationship is mathematically given as
Therefore
You can you prepare 450 ml of 1 X buffer A by adding 45ml of 10X buffer A with 405ml(450-45) of water
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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
