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:
A.
Explanation:
hope this helped sorry if its wrong!
Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm
1109, 5.9 * 10 squared, -0.041, -4.2 times 10 to the -3, and -7.6 times 10 to the -5
Hope this helps
