Help with Solubility questions, please!

[From: ] [author: ] [Date: 11-08-24] [Hit: ]

:http://answers.yahoo.com/question/index;…Suppose you had bought 10kg of NaHCO3 10H2O for $72 by mistake, instead of 10kg of NaHCO3 for $80. (a) How much water did you buy and how much did you pay per liter of mass? (The mass of I L of water is 1kg.......

(c) barium and mercury(I) ions; #3 add NaCl to precipitate BaCl2
Same as (a)

(d) silver and zinc ions #3 add NaCl to precipitate AgCl
Same as (a)

I appreciate any and all help!
Also, if you have a chance, if you wouldn't mind looking at my other chem question involving Avogadro's constant and moles and whatnot. Please and thank you! :
http://answers.yahoo.com/question/index;…

Suppose you had bought 10kg of NaHCO3 10H2O for $72 by mistake, instead of 10kg of NaHCO3 for $80. (a) How much water did you buy and how much did you pay per liter of mass? (The mass of I L of water is 1kg.)

$72 for 10kg = $7.20 per kg of each component

1 mole of NaHCO3 * 10 H2O contains 10 moles of H2O

Mass of 1 mole of NaHCO3 = 23 + 1 + 12 + 48 = 84 g
Mass of 10 moles of H2O = 10 * (2 + 16) = 180 g
Mass of 1 mole of NaHCO3 * 10 H2O = 84 + 180 = 264 g

What fraction of NaHCO3 * 10 H2O is water? 180/264

What is the mass of water in 10 kg of NaHCO3 * 10 H2O? 180/264 * 10 kg = 6.818 kg = 6.818 liters

how much did you pay per liter of mass? $7.20 per kg = $7.20 per liter

(b) What would have been a fair price for the hydrated compound, valuing water at zero cost?

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Think about the solubility rules:

a) When I see Pb, I immediately think of halides (Cl-, Br- and I- are soluble except for Pb, Ag and Hg), but I don't see Cu here, so if I add HCl (as a source of Cl- ) to a solution of Pb^2+ and Cu^2+, the Pb will precipitate Pb^2+ + Cl^- --> PbCl2(s)

b) Oh, good - all ammonium salts are soluble, so all I need is something that will precipitate Mg^2+. Hydroxide comes to mind Mg^2+ + OH^- --> Mg(OH)2(s)

c) Back to the halides of (a) - Cl- will precipitate Hg(I) but not Ba Hg2^2+ + Cl^- --> Hg2Cl2(s)

d) We get a lot of mileage out of (a) Cl- will precipitate Ag, but not Zn Ag^+ + Cl^- --> AgCl(s)

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