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S_A_V [24]
3 years ago
14

20g of water at 30°C was mixed with 40g of water at 75°C. What is

Chemistry
1 answer:
Kipish [7]3 years ago
4 0

Answer:

The temperature of the mixture is 60°C

Explanation:

We can write the energy of water as follows:

E = m×C×T

<em>Where E is energy in Joules, m is mass of water, C is specific heat of water = 4.184J/g°C and T is temperature</em>

<em />

Replacing fot both samples:

E = 20g×4.184J/g°C×30°C

E = 2510.4J

E = 40g×4.184J/g°C×75°C

E = 12552J

The total energy of the mixture is 12552J + 2510.4J = 15062.4J

Mass = 60g:

15062.4J = 60g×4.184J/g°C×T

60°C = T

<h3>The temperature of the mixture is 60°C</h3>

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2 years ago
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Right Side:
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This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
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meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

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Answer:

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Answer:

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Part B: It releases energy because when the compounds are added together they are immediately releasing it by bubbling and turning cold.

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