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3 years ago

Options are upper left upper right bottom left bottom right for both problems please help!

Chemistry

1 answer:

polet [3.4K]3 years ago

6 0

Answer:

(upper right) corner of the periodic table to the bottom left corner

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

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<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

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

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

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$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

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