They are directly proportional to each other, in other words, when temperature of an object increases, the motion of it's particles also increases
Hope this helps!
Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol
C( n)H(2n) is the general formula, eg: Cyclohexane is C6H12
Answer:
Volume of container = 0.0012 m³ or 1.2 L or 1200 ml
Explanation:
Volume of butane = 5.0 ml
density = 0.60 g/ml
Room temperature (T) = 293.15 K
Normal pressure (P) = 1 atm = 101,325 pa
Ideal gas constant (R) = 8.3145 J/mole.K)
volume of container V = ?
Solution
To find out the volume of container we use ideal gas equation
PV = nRT
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First we find out number of moles
<em>As Mass = density × volume</em>
mass of butane = 0.60 g/ml ×5.0 ml
mass of butane = 3 g
now find out number of moles (n)
n = mass / molar mass
n = 3 g / 58.12 g/mol
n = 0.05 mol
Now put all values in ideal gas equation
<em>PV = nRt</em>
<em>V = nRT/P</em>
V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa
V = 121.87 ÷ 101,325 pa
V = 0.0012 m³ OR 1.2 L OR 1200 ml
As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
