GRADE 12 CHEMISTRY MID TERM PRACTICE REVIEW TEST

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1 GRDE 1 CHEMISTRY MID TERM PRCTICE REVIEW TEST Name Student Number ttending Phone Number ddress Non-ttending NSWER KEY VLUE: TOTL 100 MRKS PRT 1. (d) From the equation, we note that the coefficient for NO is 4 whereas for N O 5 is. Therefore the NO is forming at 4/ or (twice) the rate at which N O 5 is decomposing. Therefore (.5 x 10-6 ) 5.0 x 10-6 mol/5.. (d) Just a straight recall question from your notes.. (b) ctivation energy for reverse reaction ie from products to reactants 170kJ 0kJ 140 kj CHEMISTRY 40S 1 MID TERM PRCTICE REVIEW KEY

2 4. Step 1 Cl (g) + O (g) ClO (g) + O (g) Step O (g) + ClO (g) Cl (g) + O (g) Final Reaction Cl + O + O Cl + O (d) Reaction intermediate is the substance(s) that would not appear in the final reaction given here. The final reaction is determined by adding step 1 & step reactions and canceling out what is common to both sides. In the is case it is the ClO (g). 5. (c) Straight definition question from the notes. 6. (c) The forward reaction rate decreases immediately and continues to decrease as the reactant molecules are being used up causing reactant molecules to decrease. The converse is true for the reverse reaction rate. 7. (d) Increasing the volume of the system would decrease the concentration of all species (NO included) therefore using Le Chateliers Principle the reaction that produces more particles (NO ) would be favored since this would again increase the concentration. 8. (c) ny substance which is a liquid or a solid does not appear in the expression, since its concentration is constant and is thus incorporated into the Keq. 9. (b) If the equilibrium shifts left then the H appears on the RHS of the equation, meaning forward reaction is exothermic. If exothermic the H is always written with a negative sign in front of it ie H -? kj 10. (b) Equation can be rewritten as follows: PCl 5(g) kj PCl (g) + Cl (g) using Le Chateliers Principle, decreasing the temperature (energy) would favor the reaction which produces energy & that is the reverse reaction. Therefore your PCl and Cl concentrations decrease & your PCl 5 concentration would increase which makes your Keq value to decrease since Keq [ PCl][ Cl] [ PCl ] (a) [ ][ ] [ ][ ] ( )( ) ( )[ ] CO H K eq ; 5.7 CH H O 0.40 H O 4 [ ] ( 0.0)( 0.80) ( 0.40)( 5.7) H O (b) Since K TRIL is greater than Keq the reverse reaction is forward since it decreases the concentration of O and increases the concentration of O. K TRIL O K eq ; 55 ; 95 O [ ] [ ] [ ] [ ] 1. (d) Rate description must have a unit of time in its measurement & only choice d) has this. CHEMISTRY 40S MID TERM PRCTICE REVIEW KEY

3 14. (c) n increase in T increases the number of collisions per unit of time and also increases the number of successful collisions per unit of time (ie particles with the correct amount of activation energy sufficient KE) 15. (b) Using Le Chateliers Principle: Choice b) would shift equilibrium to the right ie increase decomposition 16. (a) Brönsted-Lowry base} cid proton donor Base proton acceptor 17. (b) ll acids have the same concentration & ie 10M. Therefore we have to look at the degree of dissociation of the acid & that we find in the cid Chart in your notes & this happens to be H SO 4 ie it dissociates to the greatest degree & therefore the greatest electrical conductor of the 4 choices. 18. (c) H + H + HCHO + HIO HCHO +IO B B (d) From the acid strength chart all we need to do is compare HNO with HSO - 4 & we see from the chart that HSO - 4 is higher up on the chart than HNO & therefore the reactants are favoured. 0. (a) HCl + H O H O + + Cl - ddition of HCl essentially increases the [H O+] & using Le Chateliers Principle again, this would shift the equilibrium to the left. Therefore your [OH - ] decreases but your [H O + ] is still higher than before because you added a strong acid HCI 1. (a) x 10 H O x x 10 or 6.7 x x 10. (a) poh 14 ph OH anti log of ( 7.48) Using a calculator i) punch in 7.48 ii) then key in nd fcn key & log key x10. (b) NaOH Na + + OH - The OH - from the base would combine with the H O + in the reaction given & reduce its concentration & causing equilibrium to shift to the light 4. (c) This is a weak acid (HNO ) and a strong base (NaOH) reaction HNO + OH - NO - + H O. Weak acids are always written in molecular form then & that s why we write HNO. Strong bases are just written showing the OH (a) Basic solutions always have metals in their formulas & therefore we need only look for an oxide having a metal in it & that is choice (a) 6. (a) Brönsted-Lowry acid ie lose an H + ; Brönsted-Lowry base ie gain an H + HClO 4 H + + ClO - 4 not HClO 4 + H + + H ClO 4 7. (d) Strength of acids depend on degree of ionization 8. (d) Since reactants are favored, the reverse reaction was favored over the forward reaction this means that 8 CHEMISTRY 40S MID TERM PRCTICE REVIEW KEY

4 H X was stronger than HZ - 9. (d) NaOH is a strong base, which means it dissociates complete ie NaOH Na + + OH M 0.05 M 0.05 M poh - log 0.05 M 1.60 ph (b) Formula would be identical to that in Module, Lesson 4, Page 0, except it would be solved in terms of OH (a) VxC VxC B B C VxC B B V 5.0 x or5.00x10 x 5.0. (c) HCl + NaOH NaCl + HOH V x C V B x C B 10.0 ml x 0.10 m 5.0 ml x m + - moles of H moles of OH There is no excess of H or OH & solution is neutral ph (b) The principal quantum number n always indicates the number of sublevels 4. (c) Ionization energy increases from left to right in the periodic table & from the choices given choice d) is the furthest to the right 5. (c) 8 c.00 x 10 m/s f -7 λ 6.0 x 10 m 5.0 x Hz x 10 Hz -4 8 hc 6.6 x 10 J/Hz x.00 x 10 m/s (c) -19 E. x 10 λ 6. x 10 m 7. (b) Correct choice is where the superscripts add up to 0, choice (b), exactly equal to the atomic number for calcium. 8. (c) Valence electrons are the electrons in the outermost orbits subshells s and p Valence shell S P 4 } equals 6e s. 9. (c) Only the 1,, and 8 are significant digits. The zeroes are not significant because they are neither in between nor following non-zero digits. 40. (c) nswer should have significant figures. See Module, page, Rule - When multiplying or dividing a value (1.4 g) with uncertainty in order to use a different SI prefix (kg), the original number of sig digits () are retained identically. CHEMISTRY 40S 4 MID TERM PRCTICE REVIEW KEY

5 41. (d) nswer should have significant figures. See Module, page, Rule Multiply or divide and then round off to the least number of significant digits found in the question. 5. m m m 4 m S.F. S.F. answer should have S.F. 4. (c) g 46.9 g 4. d) electronegativity S B covalent with slight polarity 44. (d) The superscripts add up to 5 which is the number of e s in the atom and therefore also the atomic #5manganese and from its electron configuration (4S ) we see it has e s. 45. (a) When ionization jumps by 10 0 times original value (ie E 1, as it does to E 5 ) it means that electrons are being removed from a different energy level. Therefore E 1, E, E, E 4 must be the removal of electrons from the valence or outer shell. 46. (b) l has a valence of, i.e., l + S has a valence of -, i.e., S - l S } This is the simplest ratio & the empirical formula. Other formulas possible are l 4 S 6, l 6 S 9 & so on but these are not empirical formulas 47. (c) n (4) PRT B Written Response 1. a) Reaction is exothermic since the heat of the reaction (94kJ) occurs on the product side. Therefore it should be written as H -94kJ b) Only the sfe e s can collide with the O c) Greater sfe area in dust form results in more collisions/unit time & thus a faster reaction.. H + I HI Initial?? 0 React Equilibrium χ χ HI K eq ; 64 H I χ χ χ H EQUIL H - H H INITIL RECT EQUIL INITIL ( ) H CHEMISTRY 40S 5 MID TERM PRCTICE REVIEW KEY

6 . a) Indicates a 4 particle collision (NO and H ) and this is next to impossible. Most collisions are particle collisions. b) i) NO + H HO + N final reaction ( NO + NO+ H N+ N + HO ) sum of reactions 1 and NO NO N NO+ N NO ii) Reaction intermediates are the species in the reaction mechanism (step 1, & ) that do not appear in the final reaction. They are N + N O 4. a) Left to right activation energy kj b) H f kJ Since products are at a lower level than reactants the H must be negative H -10 kj c).e. (5 or 4) 10 5 or 4 kj d) Drawn must be lower than 5kJ (peak of the curve) but beginning and ending levels must be the same. 5. Keq [ I ] [ I ] [ HI ] [ ][ ] ( 0.40) ( 0.40) ( 0.01)( ) x ; H I I ( 0.01)( 7.1x10 ) CHEMISTRY 40S 6 MID TERM PRCTICE REVIEW KEY

7 6. i) V 1 xc 1 C xc 40.0mL x mL x C C [KOH] ii) KOH K + + OH iii) x 10 H O 6.5 x CH COOH CH COO +H - + K + H CHCOO CH COOH 8. a) weak Brönsted-Lowry base is a poor acceptor of protons. -5 x x 10 ; x 1.8 x 10 x x x x 7. x 10.7 x 10 H phz - log.7 x 10.6 b) ny species high up on RHS of the cid Strength chart C10 4 -, I -, Br - etc C H OCOOH C H OCOO + H phof.95 H 1.1x10 K K + + CHOCOO 5 H [ C H OCOOH ] 5 1.5x ( 1.1x10 )( 1.1x10 ) 1.00x NaCN (s) Na + (aq) + CN - (aq) ; CN - (aq) + HOH HCN + OH - (aq) 11. a) at # of ( ) Vanadium b) at # of ( ) 40 Zirconium 1. a) Tl 1S S P 6 S P 6 4S D 10 4P 6 5S 4D 10 5P 6 6S 4F 14 5D 10 6P 1 b) Fe ls S P 6 S P 6 4S D 6 CHEMISTRY 40S 7 MID TERM PRCTICE REVIEW KEY

8 1. Total # of electrons: 1 x S 1 x x 0 4 x 6 4 charge e - [ ] c.00 x 10 m/s 14 f 4.9 x 10 Hz -7 λ 6.10 x 110 m 15. a) b) λ +8 c.00 x 10 m/s -5.1 x 10 m 1 f 9.6 x 10 Hz -4 1 E hf 6.6 x 10 J/Hz x 9.6 x 10 Hz -1 E 6.41 x 10 J 16. c) Molecule in i) is polar covalent ii) is nonpolar covalent CHEMISTRY 40S 8 MID TERM PRCTICE REVIEW KEY