Notes
a) three points
[NH4+] = [OH¯] = x
[NH3] = 0.150 mol/L - x
Kb = ([NH4+][OH¯]) ÷ [NH3]
1.8 x 10¯5 = [(x) (x)] ÷ (0.150 - x)
approximately equals x2 ÷ 0.150
x = [OH¯] = 1.6 x 10¯3 mol/L
% diss = [(1.6 x 10¯3) / (0.150)] x 100% = 1.1%
b) three points
[NH4+] = 0.0500 mol / 0.100 L = 0.500 mol/L NH4+
[NH3] = 0.150 mol/L
OR
mol NH4+ = 0.0500 mol NH4+
mol NH3 = 0.150 mol/L x 0.100 L = 0.0150 mol
THEN
1.8 x 10¯5 = [(0.500) (x)] ÷ (0.150)
OR
pOH = 4.74 + log (0.500 / 0.150)
THEN
x = [OH¯] = 5.4 x 10¯6 mol/L
pOH = 5.27
pH = 14.00 - 5.27 = 8.73
c) three points
Mg(OH)2(s) <===> Mg2+ + OH¯
[Mg2+] = 0.0800 mol / 0.100 L = 0.800 mol/L (0.800 ± x = no credit)
([OH¯] = 5.4 x 10¯6 mol/L from b.)
Q= [Mg2+][OH¯]2 = (0.800) (5.4 x 10¯6)2
Q = 2.3 x 10¯11
Ksp = 1.5 x 10¯11
since Q > Ksp, Mg(OH)2 precipitates
(Q must be defined in the same way as Ksp)
OR
Ksp = 1.5 x 10¯11 = (0.800)[OH¯]2
[OH¯] = 4.3 x 10¯6 mol/L
since 5.4 x 10¯6 > 4.3 x 10¯6 mol/L, then Mg(OH)2 precipitates
a) three points; one each for form of rate law, HgCl2 exponent, C2O42¯ exponent
Rate = k [HgCl2][C2O42¯]2
b) three points; two for correct substitution and arithmetic, one for units
0.52 x 10¯4 M/min = k [0.0836][0.202]2
k = 1.52 x 10¯2 M¯2 min¯1
(note: sometimes this unit is written as L2 mol¯2 min¯1)
c) one point
+ d[Cl¯] / dt = 0.52 x 10¯4 x (1 C2O42¯ / 2 Cl¯) =2.6 x 10¯5 M/min
no credit for inverting ratio
d) two points
1.27 x 10¯4 M/min = (1.52 x 10¯2 M¯2 min¯1) (0.0316)[C2O42¯]2
[C2O42¯]2 = 0.264 M2
[C2O42¯] = 0.514 M
a) three points
1.518 g KHC8H4O4 ÷ (204.2 g / mole) = 7.434 x 10¯3 moles KHP
7.434 x 10¯3 mol KHP = 7.434 x 10¯3 mol NaOH
7.434 millimole NaOH ÷ 26.90 mL NaOH = 0.2764 M NaOH
OR
7.434 x 10¯3 mol NaOH / 0.02690 L = 0.2764 M NaOH
b) three points
(28.35 mL NaOH / 25.00 mL HNO3) x (0.2764 moles NaOH / L) x (1 mole HNO3 / 1 mole NaOH) = 0.3134 M HNO3 solution
0.3134 M x 500 ml dilute soln = x (10.00 mL conc. HNO3 soln)
x = (0.3134) (500) ÷ (10.00) = 15.67 M HNO3
c) three points
% = (g HNO3 ÷ g soln) x 100
to get numerator:
(15.67 HNO3 / L soln) x (63.02 g HNO3 / mole) = 987.5 g HNO3
(in a L)
to get denominator:
(1.42 g soln / mL soln) x (1000 mL / L) = 1,420 g HNO3 soln (in a L)
then:
% = (987.5 g HNO3 / 1,420 g soln) x 100 = 69.5% HNO3
a) Ca + H2O ---> Ca(OH)2 + H2
Ca2+ + OH¯ earns one point
b) MgO + CO2 ---> MgCO3
c) H2S + Ni2+ ---> NiS + H+
d) OH¯ + Al(OH)3 ---> Al(OH)4¯
other acceptable answers:
Al(OH)63¯
AlO2¯
Al(OH)4(H2O)2¯
e) Ag + H+ + NO3¯ ---> Ag+ + NO (or NO2) + H2O
1 product = 1 pt
2 products = 1 pt
3 products = 2 pts
f) OH¯ + H2PO4¯ ---> PO43¯ + H2O
g) H2O2 + Fe2+ ---> Fe3+ + H2O
Fe(OH)3 only as a product earns one point (Note: the scoring standard on this question has a two next to the formula, but in context of below, a one seems more appropriate)
h) C3H7OH + O2 ---> CO2 + H2O
spurious products: minus 1 pt.
2 pts for correct molecular equation where ionic equation is apropriate
5) Special rules for this question:
An ambiguous 2 pt. response receives only 1 pt
An incorrect statement in an otherwise correct 2 pt response will result in a score of 1 pt
The answers labeled (i) below received two points; (ii) received one point.
a) two points
The radii of the alkali metal ions increase with increasing atomic number because:
(i) the principle quantum number (or shell or energy level) increases
(ii) there is an increase in shielding (or the number of orbitals increases)
b) two points
The chloride ion is larger than the chlorine atom because:
(i) electron- electron repulsion increases (or shielding increases or the electron-proton ratio increases or the effective nuclear charge decreases)
(ii) an extra electron generally increases the size
c) two points
The ionization energy of Mg is greater than that for Al because:
(i) the 3p orbital is at a higher energy than the 3s orbital (or the electron in Al is shielded from the nucleus more completely by the 3s electron than the 3s electrons shield one another from the nucleus)
(ii) a 3p electron is easier to remove than a 3s electron
d) two points
The much greater difference between the 2nd and 3rd ionization energies in Mg (relative to the difference between the 1st and 2nd) is due to the 3rd electron being removed from the 2p subshell after the first 2 were removed from the 3s subshell.
6) in part (b) and (c) there is a maximun of 3 points for electrode reactions and one point each part for correct designation of electrodes
a) one point
Na+ is not reduced as easily as H2O (or H+, or OH¯)
OR
if Na(s) were formed is would immediately react with the water to reform Na+ and H2(g)
b)
ANODE:
2 Cl¯ ---> O2 + 4H+ + 4e¯
CATHODE:
2e¯ + H2O ---> H2 + 2OH¯
or
2H+ + 2e¯ ---> H2
c)
ANODE:
2Cl¯ ---> Cl2 + 2e¯
CATHODE:
2e¯ + 2H2O ---> H2 + 2 OH¯
or
2H+ + 2e¯ ---> H2
d) any two; one point each for a total of two points
H2 - "pop" with a glowing splint or other suitable test
O2 - ignite a glowing splint of other suitable test
Cl2 - yellowish-green color or other suitable test
a) two points
The freezing point depression (or any other colligative effect) that occurs when a mole of a salt is dissolved is greater than when a mole of a non-dissociating substance is dissoved.
b) two points
When a salt is dissolved in water the solution conducts electricity.
c) two points
neutralization between a strong monoprotic acid and a strong base involves the reaction H+(aq) + OH¯(aq) ---> H2O regardless of the acid or base used, because both acid and base are completely dissociated. Spectator ions have no appreciable effect.
d) two points
water, because of its polar natur, is capable of solvating the ions which results from the dissociation, whereas the non-polar benzene interacts very weakly with these ions
OR
water, because of its greater dielectic constant, is more capable of separating the ions.
a) six points, two for each numbered entry
(i) the enthalpy increases ([delta]H is positive) since the reaction absorbs heat as shown by the decrease in temperature
(ii) the entropy increases ([delta]S is positive) since solid reactants are converted to gases and liquids possessing a much higher degree of disorder
(iii) The free energy decreases ([delta]G is negative) since the reaction occurs (is spontaneous)
(Correct use of the equation [delta]G = [delta]H - T [delta]S to deduce the sign of any one of the above was accepted for full credit)
b) two points
(i) The water on the wood froze because the endothermic reaction lowered its temperature below its freezing point (0 °C)
(ii) the solution in the beaker did not freeze because the presence of ions and dissolved gases lowered its freezing point below -15 °C. The freezing point depression is given by the equation [delta]T = (Kf) (m) where m is the molality of the solution, and Kf is the molal freezing point depression constant.
a) three points
It is impossible to determine (or measure) bothe the position and momentum of any particle (or object, or body) simutaneously.
OR
The more exactly the position of a particle is known, the less exactly the position or velocity of the particle is known.
OR
([delta] x) ([delta] p) [greater than or equal to] h-bar (or h / 4pi)
h = Planck's constant
[delta] x = uncertainty in position
[delta] p = uncertainty in momentum
Notes:
1 point is given for the notion of simultaneous determination. (A number of students give the first sentence but omit the word simultaneously. (They got 2 out of 3.) If the second or third versions of the answer are given, simultaneity is understood.
If they give the equation, they must have a > sign, not just an = sign, or they lose 1 pt.
A student who gives a correct answer but adds erroneous material gets one point deducted.
b) five points
Bohr postulated that the electron in a H atom travels about the nucleus in a circular orbitand has a fixed angular momentum. With a fixed radius of orbit and a fixed momentum (or energy),
([delta] x) ([delta] p) < h/4pi
and violates the uncertainty principle.
Students receive 2 pts total for the above. If they say only "the electron travels in circular orbits", they get 1 point. They also get credit for saying elliptical orbits. To receive full credit, they must describe an aspect of the Bohr theory.
The following portion of the answer is worth three points.
The wavelength of a particle is given by the deBroglie relation:
[gamma] = h/mv
For masses of macroscopic objects, h/m is so small for any v that [gamma] is so small as to be undetectable. For an electron, m is so small that h/mv yields a detectable [gamma].
OR
They may say the product of the uncertainties in postion and velocity depends on h/m and since h is so small (h = 6.63 x 10¯34 J s) unless m is very small, as for an electron, the product of the uncertainties is too small to be detected.
Students may discuss the fact that measuring the position and momentum requires having a photon strike the particle. The photon has an energy comparable to that of an electon but small compared to that of a macroscopic object. They must stress mass rather than size as the important distinction to get full credit.
A student who says [gamma] or ([delta] x) ([delta] p) depends on "size" and not mass gets 2 points if discussion is otherwise okay.
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Educational Testing Service. All rights reserved.