Chemistry, 09.10.2019 19:10 ohartshorn1599

Heavy metal ions like lead(ii) can be precipitated from laboratory wastewater by adding sodium sulfide, na2s. will all the lead be removed from 11.2 ml of 7.10×10-3 m pb(no3)2 upon addition of 12.4 ml of 0.0117 m na2s? if all the lead is removed, how many moles of lead is this? if not, how many moles of pb remain?

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Answer from: mariap3504

The moles of lead removed is $7.95\times 10^{-5}mol$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For sodium sulfide:

Molarity of sodium sulfide solution = 0.0117 M

Volume of solution = 12.4 mL = 0.0124 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.0117M=\frac{\text{Moles of }Na_2S}{0.0124L}\\\\\text{Moles of }Na_2S=(0.0117mol/L\times 0.0124L)=1.451\times 10^{-4}mol$

For lead nitrate:

Molarity of lead nitrate solution = $7.10\times 10^{-3}$ M

Volume of solution = 11.2 mL = 0.0112 L

Putting values in equation 1, we get:

$7.10\times 10^{-3}M=\frac{\text{Moles of }Pb(NO_3)_2}{0.0112L}\\\\\text{Moles of }Pb(NO_3)_2=(7.10\times 10^{-3}mol/L\times 0.0112L)=7.95\times 10^{-5}mol$

The chemical equation for the reaction of sodium sulfide and lead nitrate follows:

$Pb(NO_3)_2+Na_2S\rightarrow PbS+2NaNO_3$

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 1 mole of sodium sulfide.

So, $7.95\times 10^{-5}mol$ will react with = $\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol$ of sodium sulfide.

As, given amount of sodium sulfide is more than the required amount. So, it is considered as an excess reagent.

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

As, all the lead from lead nitrate is getting converted to lead sulfide. So, all the lead (II) is removed from the solution.

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead sulfide.

So, $7.95\times 10^{-5}mol$ will produce = $\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol$ of lead sulfide.

Hence, the moles of lead removed is $7.95\times 10^{-5}mol$

Answer from: Quest

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.

Answer from: Quest

Butterfly wingspan is 7.7

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