Na_2O_2 rightarrow Na_2O+ 1/2 O_2 +99 KJ -99KJ +198 KJ cannot determine without
ID: 999169 • Letter: N
Question
Na_2O_2 rightarrow Na_2O+ 1/2 O_2 +99 KJ -99KJ +198 KJ cannot determine without delta H_f for O_2 Chlorine will oxidize I^- as shown in this equation: Cl_2 + 2l^- rightarrow l_2 + 2Cl^- Use reactivity trends to predict which can be used to prepare Br_2 from Br^-. F + Br^- Cl_2 + Br^- l_2 + Br^- III only I and II only II and III only I, II, and III A buried metal tank will not corrode if it is wired to a metal rod that corrodes more easily, Which combination protects the tank from corrosion? Mg rod and S_n tank Fe rod and Z_n tank Fe rod and Al tank Cu rod and Zn tank Cd^2+(aq) + 2e^- rightarrow Cd E'' = - 0.402 Ni^2+(aq) + 2e^- rightarrow Ni E'' = -0.236 -0.638 v What happen during this reaction? Cd + NiO_2 rightarrow Cd(OH)_2 + Ni(OH) Cadmium metal is oxidized. Cadmium metal is reduced The O atom in H_2O is reduced. The Ni atom is NiO_2 oxidized. The solubility of a salt is water usually increase increasing temperature because the process is exothermic so dissolution occurs faster. exothermic so dissolution occurs faster. exothermic to the higher temperature favors dissolution. exothermic to the higher temperatures favor dissolution. Which reagent, when added to BaCl_2(aq) and filtered, can produce a solution that is essentially HCl(aq)? HC_2H_3O_2 H_2SO_4Explanation / Answer
16) Ho = Hof of products - Hof of reactants
= -416 - [ -515 + 0]
= + 99 kJ
17) B - I,II only
F, Cl are more reactive than Br.
I is less reactive than Br.
18) Mg rod and Zn tank
Because, Mg corrodes easily.
21) Cd anode , Ni cathode
Eo = cathode reduction potential - anode reduction potential
= -0.236 - (-0.402)
= + 0.166 V
Hence, ans = c
22) Oxidation number of Cd is increased from 0 to +2.
Therefore,Cadmium metal is oxidized.
Hence,
ans = a
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