( ) SENIOR 4 CHEMISTRY FINAL PRACTICE REVIEW TEST VALUE: TOTAL 100 MARKS. Multiple Choice. Ca (PO ) 3Ca + 2PO. Name Student Number
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1 SENIOR 4 CHEMISTRY FINAL PRACTICE REVIEW TEST Name Student Number Attending Phone Number Address NonAttending ANSWER EY VALUE: TOTAL 100 MARS PART A Multiple Choice 1. (c) Using the solubility chart i) Solution I: Cannot be Ag, Pb, Hg, Cu ii) Solution SO : could be Ag, Pb, Ca, Ba, Sr 4 iii) Cation could be Ca, Ba, Sr iv) Solution OH : could not be group 1A, NH 4, Sr, Ba From (iii) and (iv) only Ca remains choice (c). (c) Pb OH Pb OH g/l 41 g/mol M M M. (b) Ca (PO ) Ca PO 4 (s) (aq) 4(aq) = Ca PO (aq) 4(aq) 4. (c) Solubilities of compounds given are as follows CuI; χ = = AgI; χ = = CaCO ; χ = = BaSO ; χ = = (a) Both CuCl and ( NH ) NH Cl CuSO CuCl NH SO SO are soluble compounds. See solubility chart, page 1 of the exam. 6. (d) ZnS Zn s ( aq) S( aq) Since the solution is saturated, equilibrium exists between the solid ZnS and its ions and there would be no change in their concentrations. Therefore, the correct answer is choice (d).
2 Final Practice Review Test ey FTR 00 Senior 4 Chemistry 7. (b) The sudden rise in Ag concentration ( AgNO Ag NO) ( s) (aq) ( aq) must be from the addition of the AgNO. This is confirmed by the CO concentration decreasing, which it should because of the shift in equilibrium to the left to precipitate so AgCO to maintain equilibrium of the silver carbonate equilibrium, that is Ag CO Ag CO 8. (c) SrF Sr F ( s) ( aq) ( aq) S S S where S = solubility = Sr F = S S = 4S S = 4 9. (c) Select the largest which is AgCl; = 1.8 CO (b) From the acide chart on page 41, Module, Lesson 5, we see that H SP is the strongest acid (highest up on the lefthand side). 11. (b) Normal rain has a ph of about 6, due to dissolved CO from hydrocarbon combustion. Acid rain is rain with a ph lower that 6, due to SO dissolved in water. 1. (c) Choice (a) is a mass unit, choice (b) is a quantity unit, and choice (d) is a rateof reaction unit. 1. (a) NiC l Ni s ( aq ) Cl ( aq 1 ) 0.60 M 0.60M Ni = = 0.0 M 14. (c) See solubility chart on page 1 of the exam. S would cause both Zn and Sr to precipitate Cl both Zn and Sr are soluble CO both Zn and Sr are soluble SO ZnSO is soluble but SrSO is not 15. (a) PbS s Pb S aq ( aq 1 ) = Pb S =.9 10 = (c) According to equilibrium equation the Ag and Cl concentrations would have to be equal and only graph (b) shows this to be the case. 17. (c) An oxidizing agent undergoes reduction and therefore must gain (accept) electrons.
3 Final Practice Review Test ey FTR 00 Senior 4 Chemistry 18. (d) A reducing agent itself is, undergoes, oxidation (that is, loses electrons). This means that its oxidation number must increase. 1 Mn IO H 0 MnO I H 4 1 5, 1, 7, 0 oxidization number increases 19. (a) Use the reduction potential chart in your notes. Select the substance located on the lefthand side of your chart, which is highest up. Only I gains; Li, Au, and Hg occur on the RHS. 0. (b) HCl is a strong acid and strong acids in equation format should always be represented as H or more correctly H O. Only answer (b) does this. 1. (a) CO MnO Mn CO 4, 4, 4, oxidation numbers C has increased by 1. (a) Using the standard reduction potential chart, select the substance from the RHS of the equilibrium arrows that is higher up than Pb (s).. (b) Any equation which shows changes in oxidation numbers. CuS H H S Cu, 0 0 1, 4. (c) The zinc electrode is the anode and the copper electrode is the cathode. Therefore, copper ions migrate to the copper plate. Cations migrate to the copper plate. Copper ions migrate to the copper plate and gain electrons to become copper metal, and stick to the plate and thus the mass increases. Anions migrate to the anolde which is the zinc electrode. 5. (c) 0 0 all Zn Cu E = E E = 0.76 V 0.4 V = 1.10 V 6. (d) Both cations and anions would have to migrate to maintain electrical neutrality within the electrolytic cell. 7. (a) To electroplate anything (such as an iron oon), it would have to be the cathode (i.e., reduction would have to occur). The only answer which shows the reduction of silver is choice (a). 8. (b) Using Le Chatelier s Principle, any change which causes a shift in equilibrium to the left would work. Choice (a) shifts it right as do (c) and (d). 9. (b) MgO and Mg(OH) are solids and should not apear in the mass actions expression, i.e., the RHS of the expression. Therefore, choice (b). 0. (a) The larger the eq the more complete the reaction is (that is, favours products.) Therefore we selected the smallest eq. 1. (b) Reducing the volume would increase all concentrations but would not affect equilibrium because we have an equal number of moles on both sides of the equation.
4 4 Final Practice Review Test ey FTR 00 Senior 4 Chemistry. (a) NO4 NO Initial 0.00 M M React = M M Equilibrium 0.10 M M. (c) Use solubility chart on page 1 of the exam. a) NO both soluble b) SO 4 both precipitate c) CO both precipitate 4. (b) Both Sr(OH) and H SO 4 are strong and should be written in ionic format. Also SrSO 4 is a precipitate and H 0 is a molecular substance and should be written as such. This all occurs in choice (d). 5. (b) The vertical rise in the forward reaction graph indicates addition of HI and immediate increase in its rate followed by a decrease in rate. Correondingly, the reverse reaction rate would increase as more product molecules are formed. 6. (a) Acid strength depends on degree of ionization. Strong means essentially 100% ionization. That means your concentration and ph values must agree. For example, a) 0.01 M = 10 = ph of b) 0.01 M does not equal ph of (c) Addition of HCl, which is a strong acid and therefore ionizes 100%. HCl H O H O Cl Therefore, the H O concentration would increase greatly and the ph value would decrease. 8. (a) 9. (c) CO 4 (x) 4() = x8 = x = 6 x = As O 4NO 7H O4H = 6H AsO 4NO ,,, 5, 1,5, Oxidizing agents are reduced (i.e., there oxidization numbers should decrease or gain electrons). 40. (b) WO WO 5 5, 4, reduced 41. (b) Reaction #1: Te is a stronger oxidizing agent than V Reaction #: U 4 is a stronger oxidizing agent than Te Therefore (decending order): U 4, Te, V 4. (a) Since the nickel electrode supplies the electrons, it must be the anode (ie, Ni (s) Ni e ) 4. (b) Cell negative ions (NO ) would migrate to the Ni electrode and all positive ions (, Pd, Ni ) would
5 5 Final Practice Review Test ey FTR 00 Senior 4 Chemistry migrate to the palladium electrode. 44. (c) Ni Ni e 0 E = 0.5V Pd e Pd E = x 0 Ni Pd Ni Pd Ecell = 1.1V 0.5 x = 1.1; x=0.96v (d) The substance N O 4 is formed in the first step but used up in the second step and is therefore a reaction intermediate. 46. (c) The formation if iron rust must have both H O and O present, but not N. 47. (d) Reaction occurring at the cathode is always reduction. Only choice (a) or (d) are reduction reactions but choice (d) has the more positive value of the two and would occur before choice (a). o o Choice (a), E =.71; choice (d) E = 0.8. PART B Written Reonse (1) 1. (a) Ag IO AgIO (aq) (aq) (aq) () (b) 15.0 Ag = x 0.50= IO x 0.50=0.5 (excess IO = [ 0.0 ]) 50.0 (Trial) = Ag IO = [ 0.15][ 0.5] = S ince ( ; a precipitate of AgIO will form (Trial) 8 (AgIO ) =.1 x 10 ) (). (). 1 ( (Ag CO ) = 8. x 10 ) AgCO (s) Ag(aq) CO(aq) AgNO (s) Ag NO (aq) (aq) x x x = Ag CO ; 8. x 10 = 1. x 10 CO 1 CO = 8. x 10 = 4.9 x x H 8e ClO Cl 4H O x 4 Al Al e x 8 4H ClO 8Al Cl 8Al 1H O 4 4 (1) 4. (a) Cu (s) Ag (aq) Cu (aq) Ag (s) () 4. (b) AgCH COO (s) Ag (aq)ch COO (aq)
6 6 Final Practice Review Test ey FTR 00 Senior 4 Chemistry = Ag CH COO (aq) (aq).00g Cu = =.15 x g/mol A g = Cu = (.15 x 10 ) = 6.0 x x 10 mol = = mol 1.00L L 6.0 x 10 = = = Ag 6.0 x 10 Ag CH COO 6.0 x L = 6.0 x 10 =.97 x 10 () 5. (a) (1) (b) i) on diagram (electrolytes) () ii) Anode: Pb (s) Pb (aq) e E = 0.1V Cathode: Ag (aq) e Ag (s) E = 0.80V () iii) Add equations above Pb (s) Ag (aq) Pb (aq) Ag (s) E = 0.9V (1) iv) on diagram (1) v) Pb, Ag left to right : NO right to left cations anion (1) 6. (a) Anything below Fe on the Reduction Potential Chart ie Cr, Zn, Al (1) (b) Cr metal will undergo oxidation: Cr (s) Cr e and therefore iron cannot undergo oxidization (i.e., rust), Fe Fe e ( s) (1) 7. (a) I I e
7 7 Final Practice Review Test ey FTR 00 Senior 4 Chemistry () (b) i) H (hydrogen) () 8. (a) ii) Because reduction of water occurs at electrode B (cathode) producing OH (base) ions: H O e OH H (1) (b) Equilibrium will shift to the right (product) side & therefore the concentration of Cl decreases. (1) (c) Equilibrium shifts right & therefore eq increases. (4) g g/mol moles Ca OH = = 1.68 x 10 mol 1.68 x 10 Ca ( OH ) = = 1.68 x L Ca OH Ca OH 1.68 x x x 10 = Ca OH 6 = 1.68 x 10.6 x 10 = x 10 () 10. Ba (aq) SO 4 (aq) BaSO 4(s) = 1.90 x 10 5 (6) 11. Nicotinic acid, HC 6 H 4 NO, is a weak acid found in vitamin B. Calculate the ph of M HC 6 H 4 NO ( a = 1.4 x 10 5 ).
8 8 Final Practice Review Test ey FTR 00 Senior 4 Chemistry 6 4 H C6H4NO HC H NO Solution A H C6H4NO = HC H NO H C6H4NO HC H NO x x x ( x)( x) x 10 = "x" can be dropped from the denominator as it is very small composed to the value x 5 x = ( 1.4 x 10 )( 0.100) = 1.4 x 10 6 x = 1.4 x 10 = 1. x 10 6 H = x = 1. x 10 ph = log 1. x 10 =.9 () 1. (a) None of the atoms change oxidation number and therefore (a) is not a Redox Reaction. (b) H changes oxidation number from O 1 and N changes oxidation number from O therefore (b) is a Redox Reaction () 1. e (g) (aq) (s) Cl Cl E = 1.6V (aq) Ni Ni e E = 0.5V (g) (s) ( aq) cell Cl Ni Cl E = 1.61V () 14. Mg e Mg It.00 x 4.00 x 60 x 60 N e = = = 0.98 mol e 96,500 96,500 Mg e Mg mol 1 mol mol x = mol 1 mol 0.98 mol x = = Mg (s) Mass Mg = mol x 4.1 g/mol m =.6 g (4) 15. Cu (aq ) e Cu(s) mol 1 mol x x mol = mol 1 mol = 0.00 mol e
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