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rodikova [14]
3 years ago
6

Select the correct answer. What is the enthalpy for the reaction represented in the following energy diagram? A. +200 kJ B. +350

kJ C. +150 kJ D. +250 kJ

Chemistry
1 answer:
Stella [2.4K]3 years ago
6 0

Answer:

Option C. +150KJ

Explanation:

Data obtained from the question include:

Heat of reactant (Hr) = 200KJ

Heat of product (Hp) = 350KJ

Change in enthalphy (ΔH) =..?

The enthalphy of the reaction can be obtained as follow:

Change in enthalphy (ΔH) = Heat of reactant (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

ΔH = 350 – 200

ΔH = +150KJ

Therefore, the enthalphy for the reaction above is +150KJ

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7 0
3 years ago
A solution is prepared by dissolving 4.66 g of KCl in enough distilled water to give 250 mL of solution. KCl is a strong electro
Leokris [45]

Answer:

Depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.

Explanation:

On adding any non volatile solute to a solvent its boiling point increases and its freezing point decreases [these are two of the four colligative properties].

The depression in freezing point is related to molality of solution as:

ΔTf=iK_{f}Xmolality

where

ΔTf= depression in freezing point

Kf= cryoscopic constant of water = 1.853 K. kg/mol.

i = Van't Hoff factor = 2 ( for KCl)

molality = \frac{molesofsolute}{massofsolvent(Kg)}

moles of solute = mass / molarmass = 4.66 / 74.55 =0.0625

mass of solvent = mass of solution (almost)

considering the density of solution to be 1g/mL

mass of solvent = 250 grams = 0.250 Kg

molality = \frac{0.0625}{0.25}= 0.25

Putting values

depression in freezing point = 2 X 1.853 X 0.25 = 0.9625

Thus this will be the difference between the freezing point of pure water and the solution.

7 0
4 years ago
Read 2 more answers
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