Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
Answer:
Explanation:
wavelength λ = 12.4 x 10⁻² m .
energy of one photon = h c / λ
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 12.4 x 10⁻²
= 1.6 x 10⁻²⁴ J .
Let density of coffee be equal to density of water .
mass of coffee = 255 x 1 = 255 g
heat required to heat up coffee = mass x specific heat x rise in temp
= 255 x 4.18 x ( 62-25 )
= 39438.3 J .
No of photons required = heat energy required / energy of one photon
= 39438.3 / 1.6 x 10⁻²⁴
= 24649 x 10²⁴
= 24.65 x 10²⁷ .
An acidic solution is 0.1M in HCl and 0.2 H2so4. volume is equal to no of moles divided by molarity.
number of moles of HCl is 450ml x 0.1 divided by 1000 which is equal to 0.045 moles
volume of HCl is therefore 0.45 divided by 0.16 which is 2.81 litres
Number of moles of H2so4 is 450ml x 0.2 divided by 1000 which is equal to 0.09 moles
volume of H2SO4 IS 0.09 divided by 0.16 which is equal to 0.56 litres
