Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s
Answer:
A.
Explanation:
Using the ideal gas equation, we can calculate the number of moles present. I.e
PV = nRT
Since all the parameters are equal for both gases, we can simply deduce that both has the same number of moles of gases.
The relationship between the mass of each sample and the number of moles can be seen in the relation below :
mass in grammes = molar mass in g/mol × number of moles.
Now , we have established that both have the same number of moles. For them to have the same mass, they must have the same molar masses which is not possible.
Hence option A is wrong
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
It’s 5% because without the 5% you wouldn’t make it to 100 equally
