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3 years ago

The process of fractionation is based on which property of the components of a solution ?

Chemistry

1 answer:

Leto [7]3 years ago

4 0

Boiling points, in this sense they are talking about fractional distillation.

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Please help me!!!!

miv72 [106K]

Answer:

0.8g/ml

Explanation:

d=m/v

d=20/25

d=0.8

7 0

3 years ago

Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ

JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

$\frac{1}{2}$ Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is $\frac{1134.9}{4}$ = 283.725 kJ ⋅ mol − 1

4 0

3 years ago

Consider the reaction:

goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0

3 years ago

12.

Veronika [31]

Answer:

Nonbonding pairs of electrons.

Explanation:

Both oxygen atoms in the diatomic molecule have two nonbonding pairs. This results in the oxygen molecule having a planar geometric shape. This is because nonbonding pairs repel each other are significant in determining the shape of a molecule.

3 0

3 years ago

Read 2 more answers

What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use

ch4aika [34]

Explanation:

The given data is as follows.

$M_{1}$ = 10 mM = $10 \times 10^{-3}$ M

$V_{1}$ = 750 ml, $V_{2}$ = 5 ml

$M_{2}$ = ?

Therefore, calculate the molarity of given NaCl stock as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml$

$M_{2}$ = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.

7 0

2 years ago