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Chemistry

1 answer:

EastWind [94]3 years ago

4 0

These could all go either way, hardness and other special properties are what I'm guessing would be the most accurate in determining the kind of material.
luster, cleavage, streak, and color can all be affected by other factors. but I guess cleavage would also be accurate. so I guess hardness special properties and cleavage would be the most reliable.

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The iodide ion concentration in a solution may be determined by the precipitation of lead iodide. Pb2 (aq) 2I-(aq) PbI2(s) A stu

MatroZZZ [7]

Answer:

$M_{I^-}=0.6841M$

Explanation:

Hello.

In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

$Pb(NO_3)_2+2I^-\rightarrow PbI_2+2NO_3^-$

In such a way, since 15.71 mL of a 0.5770-M solution of lead (II) nitrate precipitates out lead (II) iodide, we can first compute the moles of lead (II) nitrate in the solution:

$n_{Pb(NO_3)_2}=0.5570\frac{molPb(NO_3)_2}{L}*0.01571L=0.01393molPb(NO_3)_2$

Next, since there is a 1:2 mole ratio between lead (II) nitrate and iodide ions, we compute the moles of those ions:

$n_{I^-}=0.01393molPb(NO_3)_2*\frac{2molI^-}{1molPb(NO_3)_2} =0.02785molI^-$

Finally, since the mixing of the two solutions produce a final volume of 40.71 mL (0.04071 L), the resulting concentration (molarity) of the iodide ions in the student's unknown turns out:

$M_{I^-}=\frac{0.02785molI^-}{0.04071L}\\\\ M_{I^-}=0.6841M$