Planck\'s constant h = 6.626 times 10^-34 J middot s. Avogadro\'s Number = 6.023

Question

Planck's constant h = 6.626 times 10^-34 J middot s. Avogadro's Number = 6.023 times 10^23 particles per mol. The Rydberg constant R = 0.01097 nm^-1 Classify each of the following reactions as precipitation, acid-base neutralization, or oxidation-reduction reaction: a) NiCl_2(aq) + Na_2S(aq) rightarrow NiS(s) + 2NaCl(aq) b) Ca(OH)(aq) + 2HNO_3(aq) rightarrow Ca(NO_3)_2(aq) + 2H_2O(l) c) MnO_2(s) + 4HCl(aq) rightarrow MnCl_2(aq) + Cl_2(g) + 2H_2O(l) d) Pb(NO_3)_2(aq) + 2HCl(aq) = PbCl_2(s) + 2HNO_3(aq) Write net ionic equations for reactions a) b) and d) only above. Complete and balance the following reactions and write a net balanced equation for each. a) Fe(OH)_3(s) + HNO3(aq) rightarrow ? b) HClO_3(aq) + NaOH(ag) rightarrow ? a) If the following reaction will take place, complete and balance it. If there is no reaction, so state: AgNO_3(aq) + Cu(s) rightarrow b) Balance the following half reactions: i) VO^2+(aq) rightarrow V^3+(aq)(acidic) ii) Ni(OH)_2(s) rightarrow Ni_2O_3 (s) (basic) Balance the following redox reaction then tell which ion is the oxidizing agent and which is the reducing agent: b) MnO_4^-(aq) + C_2O_4^2- (aq) + H^+ (aq) rightarrow Mn^2+ (aq) + CO_2(g) + H_2O(l) Calculate the energy in kJ of one mole of red light with a wavelength 653.6 nm. What is the wavelength of the line in the Balmer series in the spectrum of the hydrogen atom with n = 3? In which region of the electromagnetic spectrum is this line found? 5.51 M&F; 7th edn

Explanation / Answer

Q1.-

precipitation --> must form solids

acid-base = must be OH- and H+ neutralization + slat formation

redox = reduction/oxidation formation, electrno flow

a)

NiCl2 + Na2S = NiS + 2NaCl

there is precipitation, since NiS(s) forms

b)

This is an acid / base reaction,

OH- from Ca(OH)2 and h+ from HNO3 will form water + respective salt Ca(NO3)2

c)

This is a redox reaction

since

Mn goes from +4 to +2

and Cl from -1 to 0

therefore, electron flow is present

d)

This is precipitation since PbCl(s) forms

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