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Svetllana [295]

3 years ago

15

During a change of state, the temperature of a substance _____.

2 answers:

Natali5045456 [20]3 years ago

4 0

Hello,

Your questions states:

During a change of state, the temperature of a substance _____?

In which you gave us some choices:

A. decreases if the arrangement of particles in the substance changes.
B. remains constant until the change of state is complete.
C. increases if the kinetic energy of the particles in the substance increases.
D. increases during melting and vaporization and decreases during freezing and condensation.

Your answer would be:

B. remains constant until the change of state is complete.

Your explanation/Reasoning:

It absorbs the energy, then after the phase changes it then increases the temperature all over again.

Have a nice day:)

Hope this helps!

~Rendorforestmusic

BabaBlast [244]3 years ago

3 0

B. Remains constant until the change of state is complete.

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A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon

ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

$q_1=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_1=675J/^oC\times 29.62^oC=19993.5J$

To calculate the heat absorbed by water, we use the equation:

$q_2=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[19993.5+114690.12]J=134683.62J=134.7kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

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n = number of moles of hydrocarbon = 0.500 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol$

Hence, the enthalpy of the reaction is 269.4 kJ/mol

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