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sp2606 [1]
3 years ago
10

2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
5 0
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
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Answer:

Option C. By increasing the temperature

Explanation:

From the graphical illustration above, we see clearly that the volume and temperature of the gas are directly proportional. This implies that as the temperature increases, the volume will also increase and as the temperature decreases, the volume will also decrease. This can further be explained by using the ideal gas equation as shown below:

PV = nRT

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V is the volume.

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Divide both side by P

V = nRT/P

Since n and P are constant, the equation above becomes:

V & T

V = KT

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The above equation i.e V = KT implies that:

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Considering the question given above,

The volume of the gas can be increased if the temperature is increased.

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