Answer:
2.5×10⁶ s
Explanation:
From the question given above, the following data were obtained:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
The half-life of a first order reaction is given by:
Half-life (t½) = 0.693 / Rate constant (K)
t½ = 0.693 / K
With the above formula, we can obtain the half-life of the reaction as follow:
Rate constant (K) = 2.8×10¯⁷ s¯¹
Half-life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 2.8×10¯⁷
t½ = 2.5×10⁶ s
Therefore, the half-life of the reaction is 2.5×10⁶ s
<em>This statement is false.</em> When you say full octet, it means that an element opts to have eight electrons in its valence shell to become stable. However, since a hydrogen atom only has 1 electron when neutral, it is impossible to reach a full octet. That is why Hydrogen is one of the exceptions to this octet rule.
Answer:
1. C4H8 + 6O2 -----> 4CO2 + 4H20
2. 3836.77 kcal
Explanation:
1. Balanced equation for the complete combustion of cyclobutane:
C4H8 + 6O2 -----> 4CO2 + 4H20
2. Heat of combustion of cyclobutane = 650.3 kcal/mol
Molecular weight of cyclobutane, C4H8 = 56.1 g/mol
Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane
Mole of C4H8 = 331/56.1 = 5.9 mol
Energy released during combustion = 5.9 mol × 650.3 kcal/mol = 3836.77kcal
Therefore the energythat is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal
