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3 years ago

This formula equation is unbalanced.

Chemistry

2 answers:

Afina-wow [57]3 years ago

6 0

The balanced equation will read:

Pb(NO3)2 (aq) + LiSO4 (aq) ➡️ PbSO4 (s) + 2️⃣LiNO3 (aq)

have a great day!!

Alex_Xolod [135]3 years ago

3 0

<u>Answer:</u> The coefficient '2' must be added in front of $LiNO_3$ to form a balanced equation.

<u>Explanation:</u>

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. The total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the given chemical equation:

$Pb(NO_3)_2(aq.)+Li_2SO_4(aq.)\rightarrow PbSO_4(s)+2LiNO_3(aq.)$

The coefficient '2' must be added in front of $LiNO_3$ to balance the equation.

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Answer:

Hi !

Here is your answer !

Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place.

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2 years ago

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2 years ago

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$