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3 years ago

Number of representative particles in 3 moles Sn

Chemistry

1 answer:

AURORKA [14]3 years ago

5 0

Answer:

18.066 × 10²³ particles

Explanation:

Given data:

Number of moles of Sn = 3 mol

Number of representative particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 3 mole of Sn:

3 × 6.022 × 10²³ particles

18.066 × 10²³ particles

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Answer:

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Explanation:

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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

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Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

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The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

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Therefore, the percent yield is, 93.4%

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Answer:

Explanation:

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The molar solubility of pbi2 is 1.5 103 m.

Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

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[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

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