# data[].attributes.body
Kstab = [CuCl4 2-] / [Cu(H2O)6 2+][Cl -]^4

Kstab >1, hence eq lies on the RHS

The olive-green soln turns pale blue (depending on the conc, "olive-green" could be yellow-brown.)

# data[].attributes.body
Kstab = [CuCl4 2-] / [Cu(H2O)6 2+] [Cl-]^4
K stab is large, therefore equilibrium lies on the RHS
changing temperature of solution
changing concentrations of reactants or products

# data[].attributes.body
Kstab = [CuCl4 2-] / [ Cu(H2O)6 2+] [ Cl-]^4
Large Kstab means equilibrium lies on the right hand side
Kstab can be changed if temperature changes
Position of equilibrium can be changed if the concentration of reactants/products are changed

# data[].attributes.body
Kstab = [CuCl4 2- (aq)] / [Cu(H2O)6 2+ (aq)] [Cl- (aq)]^4

Kstab is large, so the equilibrium lies to right.

Change the temperature.

Change concentration of reactants. Increase: equilibrium shifts right. Decrease: equilibrium shifts left.

# data[].attributes.body
1.Kstab = [CuCl4 2-]/ [Cu(H2O)6 2+][Cl-]^4

2.Since 4x10^5 is large, reaction lies to the right.

3.change temperature

4. change concentration of water

# data[].attributes.body
Kstab = [[CuCl.4]^2- (aq)]/ [[Cu(H2O)6 2+] [Cl-(aq)]^4
Kstab > 1 therefore equilibrium lies on the RHS
Change temperature
Change conc

# data[].attributes.body
Kstab = [CuCl4 ²⁻] / [Cu(H2O)6 ²⁺][Cl-]⁴

Since 4x10⁵ is greater than 1, it means that the concentration of [CuCl4 ]

# data[].attributes.body
Kstab= 4x10^5 means equilibrium lies to the right since it is greater than 1, ie at equilibrium the  (multiplication) product of product concentrations is 4x10^5 greater than that of the reactants. Thus [Cu(H2O)6 2+] more stable than [CuCl42-]

Change value of Kstab by increasing temp (if forward reaction exothermic, this causes Kstab to decrease) and vv.

Change position of eq by increasing concentration of Cl- (shifts left) and [CuCl4 2-] which shifts eq to RHS

# data[].attributes.body
Kstab = [CuCl42-]/[Cu(H2O)6 2+] [Cl-]^4

FOr 4x10^5, it will lie on the the right hand side

To change the value by changing the concentration of the reactants/products. Change the temperarture.

# data[].attributes.body
(1)
Kstab=[CuCl4,2-]/[Cu(H20)6,2+][Cl-]4
(2)
Equilibrium lies to RHS due to kstab>1
(3)
Altering concentration of one of the reactants
Changing temperature
(4)
Same as 3

# data[].attributes.body
Kstab = [CuCl4 2-]/ [Cu(H2O)6 2+] [Cl-]4
Equilibrium lies on the right hand side of the equation.
Change the conc of reactants and products
Add water

# data[].attributes.body
[CuCl42-] / [Cu(H20)62+] x [Cl-]4

Since Kstab is positive, it means that eqm lies to the RHS

By changing the concentration of water, temperature  you can shift equilibrium

# data[].attributes.body
Kstab:

[CuCl42-]/[cl-]^4 X [Cu(H2O)62+]

Lies  to the rhs

Change this value by changing concentration of reactants/ products

Change temperature  or pressure

# data[].attributes.body
Kstab = [CuCl4 2-] / [Cu(H2O)6 2+] [Cl- 4]

If it is 4x10^5, it means it lies to the RHS as it is a positive value. There is more CuCl4 than Cu(H2O)6

To change the value, you shift the position of the equilibrium which you can achieve by changing the concentration of CuCl4 or Cu(H2O)6. Changing the temperature can also shift it like how temperature affects Kc

# data[].attributes.body
Kstab = [CuCl42-]/[Cl]^4 [Cu(H2O)6 2+]

The equilibrium lies to the right hand side.
Increase concentration of CuCl4 so equilibrium shifts left/decrease concentration of CL- or the hexaqua chloride.

# data[].attributes.body
1. Kstab: [CuCl4]2- / [Cu(H2O)62+][Cl-]4

2. Large Ksp value so equilibrium lies to the right

3. Change the concentration and temperature