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3 years ago

Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.

Chemistry

1 answer:

shtirl [24]3 years ago

8 0

Answer:

Q = 246 kJ

Explanation:

It is given that,

Mass of water, m = 200 g

Let initial temperature, $T_i=5^{\circ}$

Final temperature of water, $T_f=300^{\circ} C$

We know that the specific heat capacity of water, $c=4.18\ J/g-^{\circ} C$

So, the heat energy needed to raise the temperature is given by :

$Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J$

Q = 246 kJ

So, the heat energy of 246 kJ is needed.

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

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What is the oh- in a solution with a poh of 5.71

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Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

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