# chemistry

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a large 1.5 x 10^6liter tank contains water polluted with lead (II) nitrate at a molarity concentraion of
2.5 x 10^-6 M

to the tank, you add 59.7 liters of 0.075 M potasium chloride solution.

did you add enough potassium chloride to remove all the lead(II) ions out of the solution??

• chemistry -

I didn't work the problem but the answer is no. PbCl2 is relatively soluble and a saturated soln of PbCl2 still contains fairly large concns of Pb.
I think, however, that you are supposed to write and balance the equation, then determine mols Pb in the tank. That will be M x L = ?
Then determine mols KCl you added, go through the stoichiometry and see if there is any Pb remaining.

• chemistry -

You are a chemist requested to review and analyse certain debris found on thebottom of aboot found in a crime scene. The debris contains metal fragments. If you identify the metal fragments---it will help the detctive identify the locations traveled by the person who did the crime.

from the bottom of the boot---a sample of unknow metal, X(s) and react it with Oxygen gas according to the following equation: 2x(s) + O2 >>> 2XO(s)

if you react 397.2 grams of X(s) with 100.0 grams of O2(g) to produce 497.2 grams of XO(s), then what is the identity of metal X???

which is it based on the items below

iron
copper
gold
silver

• chemistry -

You are a chemist requested to review and analyse certain debris found on thebottom of aboot found in a crime scene. The debris contains metal fragments. If you identify the metal fragments---it will help the detctive identify the locations traveled by the person who did the crime.

from the bottom of the boot---a sample of unknow metal, X(s) and react it with Oxygen gas according to the following equation: 2x(s) + O2 >>> 2XO(s)

if you react 397.2 grams of X(s) with 100.0 grams of O2(g) to produce 497.2 grams of XO(s), then what is the identity of metal X???

which is it based on the items below

iron
copper
gold
silver

• chemistry -

........2X + O2 ==> 2XO
.....397.2..100.0....497.2
Note that the grams add to the same on both side; therefore, you know neither X nor O2 was in excess. Thus it is not a limiting reagent problem.
How many moles O2 did you use. That is 100/32 = 3.125
According to the equation, how many mols X were used. That must be 2 x 3.126 = 6.250.
Then since moles = grams/atomic mass we can rearrange to atomic mass = grams/moles.
Substitute and look up the atomic mass on the periodic chart. I vote for Cu.

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