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lora16 [44]
3 years ago
5

What would be the greatest difference in using an open ceramic coffee mug rather than an insulated mug with a lid as a calorimet

er? increased evaporation of the water loss of gaseous reactants inability to contain liquids exchange of energy with the surroundings
Chemistry
2 answers:
STALIN [3.7K]3 years ago
6 0

Answer:

Hi. Increased evaporation would be the greatest difference.

Explanation:

Have you ever had a drink in a mug with a lid? When you would remove the lid, all of the liquid on the lid is the liquid that would evaporate if there was no lid.

Anarel [89]3 years ago
6 0

Answer:

D

Explanation:

took the test and got it right

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When a glow stick is crushed and shaken, the two liquids mix and the stick glows.
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Answer:

A chemical change has occurred, with energy being given off.

Explanation:

The liquids mix and the stick gives off energy in light form

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3 years ago
Can a chemical reaction happen with only one substance? For example, if you select water as the first substance but you don't ad
KATRIN_1 [288]

Explanation:

Yes, a chemical reaction can happen with only one substance.

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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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