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PIT_PIT [208]
3 years ago
9

The ionic equation for MgCl2 and KOH.

Chemistry
2 answers:
saw5 [17]3 years ago
5 0
The reaction of Mg Cl2 and KOH can be described as a double substitution type of reaction. This means the cations of the reactants are exchanged in places when the products are formed. In this case, the balanced reaction is expressed
MgCl2 (s) + 2KOH (aq) = Mhg (OH)2 (aq) + 2KCl (s)
natta225 [31]3 years ago
4 0

solution:

the ionic equation of MgCl2 and KOH are as follows:

MgCl2+KOH

for the we have to multiply by 2 in KOH.

Now,

MgCl2+2KOH --> Mg(OH)2 + 2KCL

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Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

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Log (N₀/N) = 2.592 / 2.303

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N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

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