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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:
Answer : The volume of gas will be 29.6 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 12 atm
= final pressure of gas = 14 atm
= initial volume of gas = 23 L
= final volume of gas = ?
= initial temperature of gas = 200K
= final temperature of gas = 300K
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the new volume of gas will be 29.6 L
Answer:
vigorous
Explanation:
As you go down group one of the periodic table, the reactions become and more vigorous.
We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
