Answer:
D
Explanation:
Hopefully this helps you!
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:
100% find the gfm of both sides then divide
Answer
For this we use ideal gas equation which is:
P1V1 = P2V2
P1 = 1.10 atm
V1 = 326 ml
P2 = 1.90
V2 = ?
By rearranging the ideal gas equation:
V2 = P1V1 ÷ P2
V2 = 1.10 × 326 ÷1.90
V2 = 358.6 ÷ 1.90
V2 = 188.7 ml
