Chemistry, 03.07.2021 01:00 daeshawnc14

What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorine and the rest is bromine

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Answer from: Richwave17

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we calculate the number of moles of each element, using their respective molar masses:

82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we divide those numbers of moles by the lowest number among them:

0.9594 mol Mo / 0.9594 = 11.290 mol Cl / 0.9594 = 1.34 ≅ 12.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

Crust,mantle,core,earth

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What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of mol...