A copper rod that has a mass of 200.0 g has an initial temperature of 20.0°C and is heated to 40.0°C. If 1,540 J of heat are nee
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Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = Heat gained = ?
c = Specific heat =
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
Now we have to calculate molar enthalpy of combustion of this substance :
where,
= enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane =
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Hi there!
We can use the following conversions to solve:
Total mass --> amount of mols --> amount of atoms (Avogadro's number)
Begin by calculating the amount of boron trifluoride in 3.61 grams:
3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃
Use avogadro's number to convert:
0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms