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3 years ago

If 980 Kj of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be?

Chemistry

1 answer:

jekas [21]3 years ago

4 0

Answer:

T(final) = 328.78 K

Explanation:

Given data:

Energy added = 980 kj

Mass of water = 6.2 L

Initial temperature = 291 K

Final temperature = ?

Solution:

Mass of water = 6.2 L = 6.2 Kg = 6200 g

Specific heat of water = 4.184 j/g.K

Energy added = 980 kj = 980 × 1000 = 980,000 J

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

980,000 J = 6200 g × 4.184 j/g.K × T(final) - 291 K

980,000 J = 25940.8 j/K × T(final) - 291 K

980,000 J / 25940.8 j/K = T(final) - 291 K

T(final) - 291 K = 37.78 K

T(final) = 37.78 K + 291 K

T(final) = 328.78 K

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Answer: The value of $K_p$ for the equation is $2.5\times 10^3$

Explanation:

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The chemical equation for which the equilibrium constant is to be calculated follows:

$H_2(g)+C_2N_2(g)\rightleftharpoons 2HCN(g);K_p$

As, the equation is the result of the reverse of given reaction. So, the equilibrium constant for the equation will be the inverse of equilibrium constant for the given reaction.

The value of equilibrium constant for the equation is:

$K_p=\frac{1}{K_1}$

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$K_1=4.00\times 10^{-4}$

Putting values in above equation, we get:

$K_p=\frac{1}{4.00\times 10^{-4}}=2.5\times 10^3$

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3 years ago

Write a balanced half-cell equation for the reaction occurring at the anode. ignore phases in the reaction.

r-ruslan [8.4K]

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Explanation:

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grandymaker [24]

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