South Carolina State University Department of Biological and Physical Sciences N

Question

South Carolina State University Department of Biological and Physical Sciences Name 52. VERSION BV2. Fall 2017 t your name on your seantron page. Mark the correct answer on your scantros pe Which one of the following species could not react as a A) HCN Bronsted-Lowry acin IHiS C) D) CaO CH3COoH 2 In the equiation HF + H2O-> H,O" + F"? A) 1B) H:O is a base, and HF is its conjugate acid. H:O is an acid, and HF is the conjugate base. HF is an acid, and F is its conjugate base. HF is a base, and F is its conjugate acid. D) In the following reaction, which species is oxidized? 3AgS(s) + 8H,(aq) + 2No, (aq) 6Ag.(aq) + 3S(s) +2NO(g) + 411,0 A) B) C) D) NO Ag S Ag' No What is the oxidation number of arsenic in HAsO32 A) B) C) D) 3 4 Which of the following would have the highest freezing point? 0.14m Na CO 0.11m Znl 0.36m C&H120; Pure water loyr mush beat is absorbed in the complete reaction of 3.00 grams

Explanation / Answer

Q1

First, let us define Bronsted Lowry acid/base:

Bronsted Lowry acid: any species that will donate H+ (protons) in solution, and makes pH lower (i.e HCl)

Bronsted Lowry base: any species that will accept H+ (protons) in solution, and makes pH higher (NH3 will accept H+ to form NH4+)

Typically, acid/bases are shown in the left (reactants)

when we write the products:

Bronsted Lowery conjugate base = the base formed when the B.L. acid donates its H+ proton ( i.e. HCl -> Cl-

Bronsted Lowery conjugate acid = the acid formed when the B.L. base accept its H+ proton ( i.e. NH4+ has accept H+ proton)

Note that, typically conjugate bases/acids are shown in the right (product) side

So, from your options

HCN = can donate H+

H2S = can donate H+

CaO = can't donate H+

CH#COOH = Can donate H+

therefore, CaO cant donate H+, it cant be Bronste Lowry acid

Q2

HF is donating H+, then, acid

H2O is accepting H+, then th ebase

so

Hf is an acid, F- its conjguate (C)

Q3

oxidation = loss of electrons

S goes from -2 in Ag2S, to 0 in S(s), therefore, loses 2 electrons,

Ag2S has the species which is oxidixed

Q4

HAsO3 = -2

+1 + As + 3(-2) = -2

As = -2-1+6 = +3

choose

Q5

highest freezing point --> must be the pure sample, since all other will decrease/depress the freezing point due to colligative properties

choose D

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