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uranmaximum [27]
3 years ago
6

A solution containing a mixture of metal cations was treated with dilute hcl and a precipitate formed. the solution was filtered

and h2s was bubbled through the acidic solution. a precipitate again formed and was filtered off. then, the ph was raised to about 8 and h2s was again bubbled through the solution. this time, no precipitate formed. finally, the solution was treated with a sodium carbonate solution, which resulted in formation of a precipitate. which metal ions were definitely present, which were definitely absent, and which may or may not have been present in the original mixture?
Chemistry
1 answer:
likoan [24]3 years ago
6 0
If you really keep an eye on the flow chart, the only ions you can consider as being "Definitely not present" are: Cr3+, Fe3+, and Zn2+. The rest of the ions should be considered under "Possibly present", as we cannot conclude if any of the ions are "Definitely present". 
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What should i write on a essay about glue and liquid corn starch
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Hmm... interesting topic you're writing about here!
Well, for this essay, there must be specific criteria, correct? I'll make some suggestions, but of course you don't have to go by them if you don't like 'em. So... here they are!:

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8 0
3 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
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