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erica [24]
2 years ago
5

A gas is at an original pressure of 412 torr and a temperature of 79°C. What is the new temperature of the gas (in K) if the pre

ssure changes to 341 torr?
Chemistry
1 answer:
olga_2 [115]2 years ago
7 0

Answer:

Whenever you are doing calculations with gases and temperature, you must convert the temperature from C into Kelvin. Then, just plug in your values and do the calculation. So, for the first one,

32C + 273 = 305 K

63C + 273 = 336 K

Then, P1/T1 = P2/T2

575 torr / 305 K = P2 / 336 K

P2 = 633 torr

For the second one:

P1/T1 = P2/T2

575/305 = 5750 / T2

T2 = 3050 K

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Which of the following lists elements in order of increasing malleability?
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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
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<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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