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
Answer: Final temperature = 206∘C
Explanation:
Heat Energy is given as
q= mCΔT
ehere
q= Heat energy = 87.4J
m= mass=1.25g
C=specific heat c= 0.386Jg∘C) ,
ΔT = Change in temperate of which the final temperature= 25.0∘C
q= mCΔT
ΔT = q/mC
ΔT = 87.4/ 1.25 X 0.386=181.14∘C
But,
T final- T initial = ΔT
T final = T initial + ΔT
T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C
Answer:
B) 0.32 %
Explanation:
Given that:
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)
Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
Answer:
Rubidium
Explanation:
Rubidium is an alkali metal that has 37 protons 48 neutrons and 1 valence electron
