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Kaylis [27]
3 years ago
9

Solid Mg(OH)2 is mixed with water and the following reversible reaction occurs:________.

Chemistry
1 answer:
hram777 [196]3 years ago
7 0

Answer:

Mg^2+ and OH- are the chemical species present at the equilibrium. Mg(OH)2 will not affect the equilibrium.

Explanation:

Step 1: data given

Reactants are Solid Mg(OH)2 and H2O(l)

Kc1 = 1.8 * 10^-11

Step 2: The balanced equation

Mg(OH)2(s)  ⇄ Mg2+(aq) + 2OH-(aq)

Step 3: Define the equilibrium constant Kc

Kc = [OH-]²[Mg^2+]

Pure solids and liquids do not have any effect or influence on the equilibrium in the reaction. So they are not included in the equilibrium constant expression.

This means Mg^2+ and OH- are the chemical species present at the equilibrium. Mg(OH)2 will not affect the equilibrium.

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Paha777 [63]

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6 0
4 years ago
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0.5 moles of sodium chloride is dissolved to make 0.05litres of solution. find the molar concentration
Gala2k [10]
M = n / V

M = 0.5 / 0.05

M = 10 mol/L-1

Hope this helps!
7 0
3 years ago
When you mix copper sulphate solution and steel wool, what is the chemical property that can be observed.
bazaltina [42]

Answer:

Explanation:

depending on the activity series there will probably be a single replacement reaction  possibly heat or color change and the copper precipitate out of solution

6 0
3 years ago
Calculate how much Kool-aid you will need to make 0.5 L of each of the following solutions: 0.1M, 0.4M, & 0.7M.
icang [17]

Answer:

the 4th option

Explanation:

because i said so

8 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
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