The chemical formula for acido perbromico is HBrO4 or perbromic
acid or bromate. It is an inorganic compound and an oxoacid of bromine. When an
H+, the –ate ion is –ic acid: one less O is –ous acid, ttwo less is hypo- -ous
acid and one more is per- -ic acid.
The charge of Br changed from –1 to 0, therefore it is the
element which is oxidized. Since it is oxidized then Br is also the reducing
agent.
The charge of Mn changed from +4 to +2 therefore it is the
element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing
agent.
The half –reactions are:
Br: 2Br --> Br2 + 2e-
Mn: MnO2 --> Mn2+
First balance oxygen by adding H2O:
MnO2 --> Mn2+ + 2H2O
Then balance hydrogen by adding H+ ions:
4H+ + MnO2 --> Mn2 + 2H2O
Then the appropriate electrons:
4e- + 4H+ + MnO2 --> Mn2 + 2H2O
Multiply the half-reaction of Br by 2 because the half-reaction
of Mn has 4 electrons.
4Br --> 2Br2 + 4e-
Combine the two half reactions and cancel common factors:
4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O
When compounds form, the atoms that bonded get a stable arrangement of electrons.
Compounds form because their atoms get a more stable arrangement than they had in the reactants.
A stable arrangement is a <em>complete octet</em> of eight electrons in the valence shell
.
Answer:
Option C. +150KJ
Explanation:
Data obtained from the question include:
Heat of reactant (Hr) = 200KJ
Heat of product (Hp) = 350KJ
Change in enthalphy (ΔH) =..?
The enthalphy of the reaction can be obtained as follow:
Change in enthalphy (ΔH) = Heat of reactant (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
ΔH = 350 – 200
ΔH = +150KJ
Therefore, the enthalphy for the reaction above is +150KJ
