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2 years ago

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In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury

1 answer:

Nuetrik [128]2 years ago

5 0

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211 = 0.84789 g

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7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s

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Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$ .

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$

Let's assume both $CO_{2}$ and $NH_{3}$ behaves ideally.

Therefore partial pressure of $NH_{3}$ , $P_{NH_{3}}= x_{NH_{3}}.P_{total}$ and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, $P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm$

$P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm$

So, $K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888$

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Explanation:

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Answer:

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Explanation:

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