Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.

1 answer:

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When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220( 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-

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The solubility of in water at a certain temperature is 35.7 g /100. g . Suppose that you have 330.0 g of . What is the minimum v

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The question is incomplete, here is the complete question:

The solubility of substance X in water at a certain temperature is 35.7 g /100. g. Suppose that you have 330.0 g of substance X. What is the minimum volume of water you would need to dissolve it all? (Assume that the density of water is 1.00 g/mL.)

Answer: The minimum volume of water that would be needed is 940.17 mL

Explanation:

We are given:

Solubility of substance X in water = 35.1 g/100 g

This means that 35.1 grams of substance X is dissolved in 100 grams of water

Applying unitary method:

If 35.1 grams of substance X is dissolved in 100 grams of water

So, 330.0 grams of substance X will be dissolved in = $\frac{100}{35.1}\times 330=940.17g$ of water