(I posted the picture for a ray diagram and the first law of reflection.) The rays show the first law of reflection since everything gets reflected. Look at the picture for more:
Molarity = moles / liter
a) M = 2/4 = 0.5 M
b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M
c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)