Answer:
Explanation:
Assume the reaction is the combustion of propane.
Word equation: propane plus oxygen produces carbon dioxide and water
Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)
Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)
(a) Table of enthalpies of formation of reactants and products
(b)Total enthalpies of reactants and products
ΔᵣH° is negative, so the reaction is exothermic.
Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
=
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
=
P₂ = 15.8atm
Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :
Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).
= 58%