ÉRETTSÉGI VIZSGA ● 2013. október 22.
Kémia angol nyelven
középszint
Javítási-értékelési útmutató 1313
KÉMIA
ANGOL NYELVEN
KÖZÉPSZINTŰ ÍRÁSBELI
ÉRETTSÉGI VIZSGA
JAVÍTÁSI-ÉRTÉKELÉSI
ÚTMUTATÓ
EMBERI ERŐFORRÁSOK
MINISZTÉRIUMA
Javítási-értékelési útmutató
Kémia angol nyelven — középszint
Basic guidelines to the evaluation of written tasks
The evaluation of the written test papers follows the distributed instructions of correction.
Evaluation of the theoretical questions
•
No deviation from the correction instruction is allowed.
•
½ points cannot be given; the questions can only be evaluated according to the
allowed partial points in the correction key.
Evaluation of the calculation problems
•
Solutions in the paper which follow the method of the correction key, must be
evaluated according to the partial points given in the key.
• Besides being objective, the correction has to be bona fide. During the evaluation,
punishment with a pedagogical intention cannot be applied!
• In a given correct solution no points can be subtracted because of the lack of not
required subresults (which are given in the correction key). (Those subresults help
only the evaluation of partial solutions.)
• Approaches differing from the correction key – if correct – get maximum points or
partial points according to the nodes of the correction key.
• For a bare result without any derivation or explanation only 1-2 points can be given
as a maximum according to the points of that result in the correction key!
• A calculation gets maximum points even if it contains a theoretically incorrect
reaction equation if it is not necessary to the solution (and the question was not asked
to be written).
• In the case of a problem containing several subproblems, partial points for a certain
subproblem can be given even if the candidate makes the calculation using an
incorrect result of the previous subproblem – if the solution does not lead to a
contradiction.
• Relations, which can be regarded as trivial, can be used without any derivation in the
calculation problems of the maturity examination, and they get maximum points –
even without detailed explanation. For example:
• conversion of mass, number of moles, volume and number of particles,
• trivial facts following from Avogadro’s law (equal stoichiometric ratios or
volume ratios in the case of gases under the same conditions, etc.),
• using the equation of mixing (dilution), etc.
• For each calculation error maximum 1-2 points can be subtracted (if the candidate
continues the calculation correctly with the incorrect subresult, he or she should get all
other partial points for the further part of the calculation).
•
In the case of a smaller theoretical error, the candidate does not get points for the
incorrect part of the calculation, but the following steps of the calculation using
incorrect data get the corresponding points. A smaller error is for example:
• incorrect use of density in the conversion of volume and mass,
• other incorrect but simple mathematical procedure,
• incorrectly balanced equation, which does not lead to an obviously unrealistic
result.
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• In the case of a gross error, the candidate does not get further points for the specific
part of the calculation given in the correction key, even if he continues the calculation
correctly with the incorrect subresult. A gross error is for example:
• a calculation based on an incorrect (e.g. not occurring) reaction equation,
• if the result estimated from the data is obviously unrealistic (for example if the
mass of the solution calculated from the mass of the solute is smaller than the
mass of the solute dissolved in it, etc.).
(Naturally, the solution of further subcalculations, which can be regarded as
independent calculation units, can be evaluated according to the previously discussed
principles. Points can be given – if calculating correctly with incorrect subresults – if
the calculation doesn’t lead to unrealistic results.)
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1. . Simple choice (5 points)
For each correct answer one point can be given.
1. B
2. E
3. A
4. D
5. B
2. Case study (13 points)
a) 0
b) Reducing agent
c) The total surface of the iron particles is approximately hundred times
larger than that of the iron filings, thus its reactivity is approximately
hundred times as well
d) compound A: tetrachloroethene
compound B: trichloroethene
compound E: ethene
e) Geometrical isomerism / cis-trans isomerism
Cl
1 point
1 point
1 point
1 point
H
C
H
1 point
1 point
1 point
1 point
C
1 point
1 point
1 point
1 point
1 point
Cl
trans-1,2-dichloroethene
f) 4 Fe + 3 O2 = 2 Fe2O3
g) Hydrogen gas
h) Fe + 2 H2O = Fe(OH)2 + H2
3. Multiple choice (8 points)
For each correct answer one point can be given.
1. D
2. A
3. C
4. B
5. B
6. A
7. D
8. C
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4. Panel question (15 points)
1. Water (also name or correct formula can be accepted)
2. Acidic
3. CH3COOH + H2O
CH3COO– + H3O+
4. Acetate ion
5. For example, sodium hydrogen carbonate (or NaHCO3)
6. CH3COOH + NaHCO3 = CH3COONa + H2O + CO2
7. Sodium acetate
8. For example, ethanol (or the proper formula)
9. Ester
CH3CO–O–CH2CH3 + H2O
10. CH3COOH + CH3CH2OH
(If not double way arrow was used in the answer, only 1 point can be given.)
11. Ethyl acetate (ethyl ethanoate)
12. For example, sodium
13. 2 CH3COOH + 2 Na = 2 CH3COONa + H2
14. Sodium acetate
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
2 points
1 point
1 point
1 point
1 point
5. Alternative question
A) Panel questions and analytical task (14 points)
1. Benzene
2. Liquid
3. Substitution
4.
5.
6.
7.
1 point
1 point
1 point
Fe
C6H6 + Br2
C6H5Br + HBr (even without catalyst)
Toluene
Liquid
C6H5–CH3 + 9 O2 = 7 CO2+ 4 H2O
(If the coefficient of oxygen is incorrect, 1 point can be given.)
8. Phenol
9. Solid
10. Acidic
11. Vinyl group / CH2=CH– group
12. Plastics production / production of polystyrene
2 points
1 point
1 point
2 points
1 point
1 point
1 point
1 point
1 point
B) Calculation (14 points)
a)
Let’s take 1000.0 cm3 of solution
V(solution) = 1000.0 cm3, m(solution) = 1000.0 cm3×1.051 g/cm3 = 1051g
mass %: 7.30
100 (or the use of the concept)
1 point
1 point
1 point
m(NaCl) = 0.073×1051 g = 76.7 g
n(NaCl) = 76.7 g / 58.5 g/mol = 1.31 mol
c(solution) = n(NaCl) / V(solution) (or the use of the concept)
c(solution) = 1.31 mol/dm3
(Second solution: let’s take 100.0 g solution:
1 point
1 point
1 point
1 point
1 point
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100 (or the use of the concept)
1 point
m(solution) = 100.0 g, m(NaCl) = 7.30 g
V(solution) = 100.0 g / 1.051 g/cm3 = 95.15 cm3
n(NaCl) = 7.30 g / 58.5 g/mol = 0.125 mol
c(solution) = n(NaCl) / V(solution) (or the use of the concept)
c(NaCl) = 0.125 mol / 0.09515 dm3 = 1.31 mol/dm3)
V(ice) = 1.00 m2 × 1 cm = 10000 cm2 × 1 cm = 10000 cm3
m(ice) = m(solvent)
m(solvent) = 10000 cm3×0.917 g/cm3
m(solvent) = 9170 g
the addition of x g NaCl:
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
mass %: 7.30
b)
100
2 points
7.30
9170
x = 722.1 g, (at least) 722 g of salt has to be used
(Any other correct method of solution is worth maximum points.)
1 point
6. Panel question (15 points)
H
H C
1.
O H
H
or
H3C
1 point
OH
O
2. H
3. O
H
C
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
O
4. Liquid
5. Liquid
6. Gas
7. 2 H2 + O2 = 2 H2O
8. C + O2 = CO2
9. CH3OH + 1.5 O2 = CO2 + 2 H2O
10. ΔrH = ΔfH(products) – ΔfH(reactants) (or the use of it)
ΔrH = 2×ΔfH(H2O) + ΔfH(CO2) – ΔfH(CH3OH) =
= (–286)×2 + (–394) + 239 = –727 kJ/mol
11. m = n×M = n×18.0 g/mol (or the use of it)
n(H2O) = 4 mol, m = 72.0 g
12. V = n×VM = n×24.5 dm3/mol (or the use of it)
n(CO2) = 2 mol, V = 49.0 dm3
1 point
1 point
1 point
1 point
1 point
7. Experiment evaluation and calculation (15 points)
a)
b)
The solution becomes pink/red/purple/fuchsia,
because phenolphthalein is pink/red/purple/fuchsia in basic medium
pH = – lg[H3O+], [H3O+] = 0.100 mol/dm3
1 point
1 point
c(HCl) = 0.100 mol/dm3
1 point
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c)
d)
e)
pOH = – lg[OH–] (pH + pOH = 14), [OH–] = 0.0100 mol/dm3
c(NaOH) = 0.0100 mol/dm3
n = c×V
in 100.0 cm3 NaOH of solution n(NaOH) = 0.00100 mol = 1.00×10–3 mol
V(HCl) = 100.0 cm3, n(HCl) = 0.010 mol
As n(HCl) > n(NaOH), HCl is in excess after mixing,
the mixture becomes acidic and phenolphthalein is colourless in acidic
medium, thus the solution becomes colourless.
– In the case of lime stone, the solid substance dissolves and bubble
formation is observed.
Lime stone is CaCO3, which reacts with hydrochloric acid while carbon
dioxide is formed.
(CaCO3 + 2 HCl = CaCl2 + H2O + CO2)
The point is given without the equation.
The point is given for a correct equation without explanation.
– In the case of quicklime, the solid substance dissolves (without bubble
formation).
(The answer “nothing can be observed” cannot be accepted.)
Quicklime is CaO, which reacts with hydrochloric acid while calcium
chloride is formed.
(CaO + 2 HCl = CaCl2 + H2O)
The point is given without the equation.
The point is given for a correct equation without explanation.
– Yes, it can be decided: the solid which shows bubble formation is lime
stone, the other (which does not) is quicklime.
No change is observed.
Copper does not dissolve in hydrochloric acid / does not react with
hydrochloric acid / cannot reduce the hydrogen ions of hydrochloric acid,
because it has a positive standard potential.
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
8. Calculation (15 points)
a)
b)
m(water) = 40.0 g, and also m(water) = 60.0 g
m(96% alcohol) = 1010.0 cm3 × 0.802 g/cm3 = 810.02 g
m(solution) = m(KI) + m(I2) + m(water) + m(alcohol)
m(solution) = 1000 g
The method of calculating the mass % in any step
.
⁄ %
mass % KI:
100%
.
mass % I2:
.
.
.
100%
.
m(ethanol) = 0.96×810.0 g = 777.6 g
.
mass % ethanol:
100%
.
c)
⁄
1 point
%
1 point
.
⁄
%
n(Na2S2O3) = c×V (or the use of the formula)
n(Na2S2O3) = 0.0126 dm3 × 0.281 mol/dm3 = 0.00354 mol
n(I2) = 0.5× n(Na2S2O3) = 0.00177 mol
m(I2) =0.00177 mol × 253.8 g/mol = 0.449 g
m(solution) = 0.449 g / 0.05 = 8.98 g
d(solution) = 8.98 g / 10.0 cm3 = 0.898 g/cm3
(Any other correct solution is worth maximum points.)
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2013. október 22.