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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<u>Answer:</u> The moles of water produced are 1.54 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of ethane = 15.42 g
Molar mass of ethane = 30.07 g/mol
Putting values in above equation, we get:
The chemical equation for the combustion of ethane follows:
By Stoichiometry of the reaction:
2 moles of ethane produces 6 moles of water
So, 0.513 moles of ethane will produce = 
of water
Hence, the moles of water produced are 1.54 moles.
Answer:
Increasing its mass to twice its original value
Explanation:
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Answer:
12.78
Explanation:
pOH= -log(0.0600) = 1.22184875
pH= 14-1.22184875 = 12.78
