When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:
According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.
Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
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Answer:
234.35 °C
Explanation:
Given data:
Volume of balloon = 125000 mL
Moles of oxygen = 3 mol
Pressure = 1 atm
Temperature = ?
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Volume of balloon = 125000 mL × 1 L /1000 mL
Volume of balloon = 125 L
Now we will put the values:
Ideal gas constant = R = 0.0821 atm.L/mol.K
PV = nRT
T = PV/nR
T = 1 atm × 125 L/ 0.0821 atm.L/mol.K × 3 mol
T= 125 /0.2463 /K
T = 507.5 K
K to °C
507.5 K - 273.15 = 234.35 °C
Answer:
88 gram is the answer in 2 significant digits
Answer:
1—methylethyl ethanoate
Explanation:
The compound is named as as follow:
1. Name the group after the O, starting the numbering from the carbon bonding to the O. The name is:
1—methylethyl
2. Name the group before the O, ending the name with —oate. The name name is ethanoate since there are 2 carbon atoms.
Now we combine the names together, starting with the name after the O as illustrated below:
1—methylethyl ethanoate
Answer:
3rd option
Explanation:
They are all organic compounds, that is, they contain the element carbon. Carbohydrates and lipids both contain carbon (C), hydrogen (H), and oxygen (0)
