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kirill [66]
3 years ago
14

Suppose you have a mixture of solid mg(oh)2 and baso4, how do you separate the mixture

Chemistry
1 answer:
Step2247 [10]3 years ago
5 0
This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.

Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O

The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.

MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄

Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
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Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9
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Answer : The current passing between the electrodes is, 1.056\times 10^{-2}A

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First we have to calculate the charge of sodium ion.

q=ne

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e = charge on electron = 1.6\times 10^{-19}C

Now put all the given values in the above formula, we get:

q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C

Now we have to calculate the charge of chlorine ion.

q'=ne

where,

q' = charge of chlorine ion

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Now put all the given values in the above formula, we get:

q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C

Now we have to calculate the current passing between the electrodes.

I=\frac{q}{t}+\frac{q'}{t}

I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}

I=1.056\times 10^{-2}A

Thus, the current passing between the electrodes is, 1.056\times 10^{-2}A

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