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Nata [24]
3 years ago
15

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

NH4I(aq)⟶PbI2(s)+2NH4NO3(aq) What volume of a 0.250 M NH4I solution is required to react with 121 mL of a 0.360 M Pb(NO3)2 solution? volume: mL How many moles of PbI2 are formed from this reaction? moles: mol PbI2
Chemistry
1 answer:
Setler79 [48]3 years ago
7 0

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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Answer:

Explanation:

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B= A metal that lost one electrons is Y⁺

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C=A nonmetal that gained one electrons is X⁻

When non metals gain the electrons anions are formed and negative charge is created on atom due to addition of electron. So when non metal gained one electron -1 charge is created.

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4 years ago
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nata0808 [166]
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7 0
3 years ago
Calculate the osmotic pressure of a magnesium citrate laxative solution containing 25.5 g of magnesium citrate in 244 mL of solu
Vera_Pavlovna [14]

Answer:

The answer to your question is π = 12.47 atm

Explanation:

Data

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volume of solution = 244 ml

temperature = 37°C

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Process

1.- Calculate the moles of magnesium citrate

                         214 grams ----------------- 1 mol

                          25.5 grams ---------------  x

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