Ethene is double and ethane is single
In order to answer this, we mus know the data for the heat of combustion of propane. This is an empirical data that you can search online. The heat of combustion is -2220 kJ/mol. The molar mass of propane of 44.1 g/mol. The solution is as follows:
ΔH = -2220 kJ/mol (1 mol/44.1 g)(1000g/1kg)(20 kg)
<em>ΔH = -1006802.721 kJ or -1 GJ</em>
Answer:
Explanation:
<u>1) Data:</u>
a) Solute: acetone
b) Solvent: benzene
c) n (solute) = 0.863 mol
d) mass of solvent = 500 g = 0.500 Kg
e) ΔTf = 8.84 °C
f) Kf = ?
<u>2) Formulae:</u>
- Depression of freezing point: ΔTf = m . Kf
- Molality: m = n (solute) / kg solvent
<u>3) Solution:</u>
a) m = 0.863 mol / 0.500 kg = 1.726 m
b) Kf = ΔTf / m = 8.84°C / 1.726 m = 5.12 °C/m ← answer
Answer:
6.73g
Explanation:
T½ = 5.2days
No = 80g
N = ?
T = 20.8days
We'll have to find the disintegration constant first so that we can plug it into the equation that will help us find the mass of the sample after 20.8 days
T½ = In2 / λ
T½ = half life
λ = disintegration constant
λ = In2 / T½
λ = 0.693 / 5.8
λ = 0.119
In(N / No) = -λt
N = final mass of the radioactive sample
No = initial mass of the sample
λ = disintegration constant
t = time for the radioactive decay
In(N/No) = -λt
N / No = e^-λt
N = No(e^-λt)
N = 80 × e^-(0.119 × 20.8)
N = 80 × e^-2.4752
N = 80 × 0.0841
N = 6.728g
The mass of the sample after 20.8 days is approximately 6.73g
Answer:
2CaO + 2CO2 → 2CaCO2 + O2 is right
