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3 years ago

What is s phisical change

Chemistry

2 answers:

cestrela7 [59]3 years ago

3 0

Physical change is where u change physically or on the "outside" hope this helps, have a great day!

Mama L [17]3 years ago

3 0

*scrolls through phone in confuzion* .-.

*the whole time me wifi disconnected* - windy outside..

-sigh impaitently ......

soo basically a "physical change is a change.. and i guess it can affect; *it affects* chemical substances...."

i quoted :)

example: boiling water

- hope this helps you and have a wonderful day/evening.

sorry if this joke was annoying or corny, just wanted to add nonsense humor to make someone sth... or find this relatable.. ://

*gives cookie*

-Wolfs c:

*remember to smile and say oof*

-and ofc eat cookie

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Answer:

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3 years ago

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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,

q = amount of heat absorbed = 134.7 kJ

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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4 years ago