Solution for Balance the following reactions: Are the reactions oxidation-reduction reactions? a) Ag(s) + CN (aq) + O2(g) + H2O1) --> Ag(CN)2 (aq) + OH'(aq) Question: A) Ag(s) + CN-(aq) + O2(g) ----> Ag(CN)2-(aq)b) NO2-(aq) + Al(s) ----> NH3(g) + AlO2-(aq)I'd Appreciate If You Go Deeply Into Detail For These.the silver is with the use of the cyanide ion CN- through the following reaction in basic solution: Ag(s) + 2 CN-(aq) + O 2 (g) Ag(CN) 2 - (aq)
Solved: A) Ag(s) + CN-(aq) + O2(g) ----> Ag(CN)2-(aq)b) NO... - Chegg
A balanced chemical equation provides a great deal of information in a very to react with 0.429 mol of Al according to the following equation (see Figure 1)? molecules of Zn(CN)2 are produced by the reaction of 35.27 g of K[Ag(CN)2]? Balance: __ HNO3 (aq) + __ H3AsO3 (aq) ---> __ NO (g) + __ H3AsO4 (aq) + __ H2O (l). 2 HNO3 Ag (s) + CN[-] (aq) + O2 ---> Ag(CN)2 [-] (aq). 4 Ag (s) +8 Ag(s) + CN–(aq) + O2(g) → Ag(CN)2–(aq). FREE Expert Solution. 79% (133 ratings). Our videos prepare you to succeed in your college classes. Let us help you simplify your studying. If you are having trouble with Chemistry, Organic, Physics, Calculus, or Statistics, we got your back! Our videos will help you understand concepts, solve your homework, and do great on your exams.
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I triedbut this is wrong2H2O+ AG + CN{-} + O2 = AG(CN)2{-}+4OH{-} I am supposed to do half reactions. Balance the following redox equation occurring in a basic solution: AG + CN{-} + O2 = AG(CN)2{-} I triedbut this is wrong2H2O+ AG + CN{-} + O2 = AG(CN)2{-}+4OH{-} I am supposed to do half reactionshow do I spit the product AG(CN)2{-}?????I got 1.0×10−8M . The way I see this is that the KCN and AgNO3 react to form an initial concentration of Ag(CN)−2 . The formation of . I got 1.0 xx 10^(-8)"M". The way I see this is that the "KCN" and "AgNO"_3 react to form an initial concentration of "Ag"("CN")_2^(-). The formation of "Ag"("CN")_2^(-) is extremely favorable; we know that K_f = 1/(4.0 xx 10^(-10)) = 2.5 xx 10^9. So, this is first a limiting reactant problem, and then it becomes a common-ion effect problem with leftover excess reactant. We assume that these solutions are those concentrations after they were mixed, so that the solution volume can be arbitrarily "1 L". 2"KCN"(aq) + "AgNO"_3(aq) -> "K"["Ag"("CN")_2] (aq) + "KNO"_3(aq) I assume you can realize that "AgNO"_3 is the limiting reactant. We need twice as much "KCN", so we use "0.08 M" of the "KCN" to convert all "0.04 M" of the "Ag"^(+) to "Ag"("CN")_2^(-) at first, and are left with "0.04 M KCN". So, our ICE table begins with "0.04 M Ag"("CN")_2^(-) and "0.04 M CN"^(-). "Ag"("CN")_2^(-)(aq) rightleftharpoons "Ag"^(+)(aq) + 2"CN"^(-)(aq) "I"" ""0.04 M"" "" "" "" ""0.00 M"" "" ""0.04 M" "C"" "-x" "" "" "" "" "+x" "" ""4 Ag(s) + 8 CN-(aq) + O2(g) + 4 H+(aq) ® 4 Ag(CN)2-(aq) + 2 H2O(l) do you expect to occur spontaneously in the forward direction (for standard conditions)?Ag(s) + CN- (aq) + O2(g) → Ag(CN), (aq) 2 Ag(s) + 4CN (aq) +202 + H2O(l)(9) + 2 (aq) + 402 + 2 H20(1)(g) + 4 Ag(CN)2 (aq) +8 OH(aq) 4 Ag(s) + 2 CN (aq) + O2 + (aq) → N (9)+Cu (aq) 4 Cu2+ (aq) N2H4 (aq) + 40H(aq)-> N2(g) + Cut (aq) + following redox reaction in acidic solution H2O2(aq) + Cr20-2-(aq) - O2(g) +.
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