Chemistry, 17.07.2019 17:30 CoreyHammond6247

How many fluorine atoms bond with calcium to form calcium fluoride? one two three four five

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Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+. Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.

Ca²⁺ F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.

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

the answer to this question is 3.

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