Answer to Question #154072 in Inorganic Chemistry for Mani




Calculate between free radical and cationic process from

  


1. Propagation (RP)

  


2. DP
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setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v 1.5 kg = 1.5 × 1000 g = 1500 g∴ ratio of 1.5 kg to 2500 g= 1500 g : 2500 g = `1500/2500` = `3/5` = 3 : 5 the q value for the nernst equation can be determined by  the overall reaction of the cell notation. recall that q is the reaction quotient which can be calculated as:determine the overall reaction using the mnemonic for cell notation, abc:step 1. determine the overall reaction and find the anode and cathode. the anode (one being oxidized) and cathode (one being reduced). using the mnemonic, it appears that the anode is zn(s) while the cathode is cu(s).step 2. write the equation where the anode, zn (s), is being oxidized to to zn2+ (from zn(no3)2) and the cathode, cu (s), is produced by reduction of cu2+ (from cu(no3)2zn (s) - anode; cu (s) - cathode zn (s) + cu2+ (aq) → cu (s) + zn2+ (aq)step 3. setup the q expression using the concentration of zn2+ and cu2+ provided. note that (s) and (l) are not included in q or keq expressionsstep 4. plugin q to lnq ln(q) = ln(1) = 0 answer is 0 a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v we are going to do the following steps:step 1: determine the classification of the compoundstep 2: write the formula of the compound[readmore]step 1: determine the classification of the compoundrecall the definition of the following:ionic compound - usually made up of a metal and a non-metalcovalent compound - composed of more than one nonmetal togetheracid - substance whose molecules yield hydrogen ions (h+) when dissolved in waterhydrate - substance that contains water or its constituent elementsfor cobaltic iodidecobaltic – metaliodide – nonmetalmetal + nonmetal = ionicstep 2: write the formula of the compoundthe rules for naming ionic compounds are as follows:1. the metal keeps its name and is named and written first.– for transition metals, we use roman numerals to denote its charge.2. the nonmetal keeps its base name but has its ending changed to –ide.for cobaltic iodidecobaltic  → transition metal → +2 charge → co2+iodide → group 7a → -1 charge → i-criss-cross the charges and make them the subscript of the other ioncobaltic iodide → coi2answer: cobaltic iodide is an ionic compound with a formula of coi2 kclo3(s) — heat→ kcl (s) + o2(g)      reactant        productk        1                   1cl       1                   1o       3                    2[readmore]2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)      reactant        productk        2                   2cl       2                   2o       6                    6the balanced equation for the reaction is 2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g) on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2-