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timofeeve [1]
2 years ago
14

If your unknow solid has an impurity that is insoluble in cyclohexane, will this impurity result in the molar mass of the unknow

n solid being recorded as high, low, or unaffected? Explain.
Chemistry
1 answer:
ser-zykov [4K]2 years ago
5 0

Answer:

High

Explanation:

When the unknown compound contains an impurity that is insoluble in cyclohexane, the solute will not dissolve in the solvent (cyclohexane) completely. ∆T of the solution would be smaller than it is supposed to be, when compared to a compound without such insoluble impurity. Molecular weight determination won't be accurate because the molecular weight obtained will be higher as a result of the fact that the mass of the solute would include the actual solute that is changing the temperature and the excess mass of the impurity.

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DaniilM [7]

Answer:you observe the light?

Explanation:

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In a Lewis structure for carbon tetrachloride (CCl4), which of the following is the central atom?
Sonbull [250]
Carbon is the answer it is the main molecule in the atom
3 0
3 years ago
What is the pH of a<br> 2.5 x 10-5 M solution<br> of HCI?
aalyn [17]

Answer:

The pH of the solution is 4.60.

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

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7 0
2 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
2 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
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