MOCKS 1
NAME ______________________________________ INDEX NO _____________________
DATE ____________________________ SIGN ______________________________
233/1
CHEMISTRY
PAPER 1
TIME 2HRS
K ______________________ (1mk)
M ______________________ (1mk)
K ____________________________ (1/2 mk)
M ___________________________ (1/2mk)
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Chemistry Topic By Topic Questions And answers (All Topics)
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State and explain the observations made (3mks)
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(ii) What is the purpose of (ii) above
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Experiment I
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Experiment II
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Mass of evaporating dish = 26.2g
Mass of evaporating dish + saturated solution = 42.4g
Mass of evaporating dish + dry solid y = 30.4g
Using the information, determine the solubility of solid Y at 30oc in grams per 100g of water (2mks)
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by using an energy cycle diagram, determine the molar heat of combustion of carbon(ii)oxide (3mks)
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21D + 31T abY + 10n
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Cr2O72-(aq) + ZOH–(aq) 2Cro2-4 + H2O (l)
Orange Yellow
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Explain the effect of yield AB by
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Name the process (3mks)
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B ______________________________
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MOCKS 1 2023
NAME…………………………..………………DATE ………………………………………
INDEX NO.……….……….………………………..… SIGNATURE ……………..………
233/2
CHEMISTRY
(THEORY)
PAPER 2
THEORY
| A | ||||||||
| I | B | C | D | E | ||||
| F | G | H | ||||||
………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
(b) What is the nature of the oxide formed by element C? 1mark
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Above 95.50C
Rhombic sulphur Monoclinic sulphur
Below 95.50C
Give a name to the temperature 95.50C 1mark
…………………………………………………………………………………………………
X
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Y
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Z
……………………………………………………………….
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I – The dust precipitator and drying power
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Step 2:
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Step 3:
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Step 4:
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Step II
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Step III
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(ii) Draw the structure of 1, 2 – dibromopropane 1mark
4.
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The following data was obtained from the above experiment.
Mass of burner + methanol before burning = 62.74g
Mass of burner + methanol after burning = 62.36gFinal temperature of water = 38.50C
Initial temperature of water = 23.50C
Volume of water used = 100cm3
(Density of water =1g/cm3, C = 12, O=16, H= 1.0)
Specific heat capacity of solution 4.2Kj K-1g K-1)
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C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔH = – 2209 KJmol-1
H2(g) + ½ O2(g) H2O(l) ΔH = -286KJmol-1
C(s) + O2(g) CO2(g) ΔH = -406KJmol-1
Use this information to find the heat of formation of propane. (3marks)
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5.
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| Indicator | Colour in | |
| Acidic solution | Alkaline solution | |
| Phenolphthalein | __________ | Pink |
| Methyl Orange | Pink | __________ |
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Changes in mass were noted at one minute intervals and were as follows;
| Time (Min) | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Loss in mass (g) | 0.26 | 0.46 | 0.60 | 0.69 | 0.73 | 0.73 | 0.73 |
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a.Write the equation for the reaction between anhydrous sodium carbonate and dilute nitric v acid (1mk)
c.Calculate the mass of anhydrous sodium carbonate reacted (3mks)
d.Calculate the mass of water in 0.715g of hydrated sodium carbonate (1mk)
MOCKS 1 2023
Name _________________________________________ Index No. _____________________________
Candidate’s Signature ___________________
Date _________________________________
233/3
CHEMISTRY
PAPER 3
PRACTICAL
2 ¼ HOURS
INSTRUCTIONS TO CANDIDATES
FOR EXAMINER’SUSE ONLY
| Question | Maximum Score | Candidate’s Score |
| 1 | 19 | |
| 2 | 10 | |
| 3 | 11 | |
| TOTAL SCORE | 40 |
This paper consists of 8 printed pages
Turn Over
– Solution A, a mixture of two bases sodium hydroxide and sodium carbonate solids
dissolved in a 1 litre solution.
– Solution B, 0.2M hydrochloric acid.
– Phenolphthalein and methylorange indicators.
– Solution C, barium chloride solution.
You are required to determine the concentration of each of the reactants in the mixture.
Procedure 1
Pipette 25.0cm3 of solution A into a conical flask.
Add two drops of methylorange indicator.
Titrate solution A with B until the yellow colour just changes to pink.
Record your results in the table below.
Repeat the procedure to obtain two more readings.
Table I (3 marks )
| 1 | 2 | 3 | |
| Final burette reading (cm3) | |||
| Initial burette reading (cm3) | |||
| Volume of solution B used (cm3) |
(a) Calculate the average volume (V1) of solution B used. (1 mark )
(b) Calculate the number of moles of hydrochloric acid that reacted. (1 ½ marks )
Procedure II
Pipette 25.0cm3 of solution A into a conical flask. Measure 15.0cm3 of barium chloride
solution (solution C) with clean measuring cylinder.
Add it to the solution A in the conical flask. Shake it gently and add three drops of phenolphthalein indicator.
Titrate solution B into the conical flask until the pink colour just changes to colourless.
NB: The white precipitate should remain in the flask.
Repeat the procedure to obtain two more readings.
Table II ( 3 marks )
| 1 | 2 | 3 | |
| Final burette readings (cm3 | |||
| Initial burette reading (cm3) | |||
| Volume of solution B used (cm3) |
(c ) Calculate the average volume (V2) of solution B used. (1 mark )
(d) The equation for the formation of white precipitate
Na2CO3 (aq) + BaCl2 (aq) BaCO3 (s) + 2NaCl (aq)
During titration II the white precipitate formed after adding barium chloride does
not take part in the titration but all the hydroxide ions (OH–) in the solution are neutralized.
(i) Calculate the moles of the acid (solution B) reacting in titration II. ( 1 ½ marks )
(ii) Calculate moles of sodium hydroxide (OH–) reacting during the titration. ( 1 ½ marks )
(e) Calculate number of moles of acid that reacted with sodium carbonate in the mixture. ( 1 ½ marks )
(f) Calculate the concentration of solution A in terms of sodium hydroxide in
moles per litre. ( 1 ½ marks )
(g) Write an ionic equation for the reaction of the acid with sodium carbonate. ( 1 mark)
(h) (i) Calculate the number of moles of sodium carbonate in the mixture. (1 mark )
(ii) Calculate concentration of solution A in terms of sodium carbonate in
moles per litre. (1 ½ marks )
and inferences.
(a) Using a clean metallic spatula, heat a half of solid Z in a Bunsen burner flame.
| Observation | Inferences |
|
( 1 ½ marks ) |
( 1 marks |
(b) Dissolve the remaining portion of solid Z into 10cm3 of distilled water in a boiling tube.
Divide the resulting solution into four portions.
| Observation | Inferences |
|
( 1 mark ) |
( ½ mark ) |
(c ) To 1st portion, add 3 drops of acidified potassium manganate (VII)
| Observation | Inferences |
|
( ½ mark ) |
( 1 ½ marks ) |
(d) To the 2nd portion, add 3 drops of acidified potassium dichromate (VI) and warm.
| Observation | Inferences |
|
( ½ mark ) |
( 1 ½ marks ) |
(e) To the 3rd portion, add all the NaHCO3 provided.
| Observation | Inferences |
|
( 1 mark ) |
( ½ mark ) |
(f) To the 4th portion, add 3 drops of universal indicator and determine the pH value.
| Observation | Inferences |
|
( 1mark ) |
( ½ mark ) |
inferences in the spaces provided.
(a) Place half of solid Y in a boiling tube and heat. Test any gas produced with litmus paper.
| Observation | Inferences |
|
( 1 ½ marks ) |
( 1 mark ) |
(b) Place the remaining solid Y into a boiling tube. Add about 10cm3 distilled water and
shake. Divide the resulting solution into 5 portions.
| Observation | Inferences |
|
( ½ mark ) |
( ½ mark ) |
(c ) To the 1st portion, add NaOH(aq)dropwise till in excess.
| Observation | Inferences |
|
( 1mark ) |
( ½ mark ) |
(d) To the 2nd portion, add NH3 (aq) dropwise till in excess.
| Observation | Inferences |
|
( 1 mark ) |
( ½ mark ) |
(e) To the 3rd portion, add 1cm3 of acidified hydrogen peroxide followed by
NaOH(aq) dropwise till in excess.
| Observation | Inferences |
|
( 1 mark ) |
( ½ mark ) |
(f) To the 4th portion, add 3 drops of lead (II) nitrate and then filter.
| Observation | Inferences |
|
( 1 mark ) |
( 1 mark ) |
(g) To the 5th portion, add 3 drops of acidified barium nitrate solution.
| Observation | Inferences |
|
( ½ mark ) |
( ½ mark ) |
MOCKS 1 2023
233/3
CHEMISTRY
PAPER 3
PRACTICAL
021
Kenya Certificate of Secondary Education
CHEMISTRY
PAPER 3
CONFIDENTIAL TO SCHOOLS
The information contained in this paper is to enable the head of Institution and the teacher in charge of Chemistry to make adequate preparations for the Form 4 entrance examination. NO ONE ELSE should have access to this paper or acquire knowledge of its contents. The teacher in charge of Chemistry should NOT perform any of the experiments in the same room as the candidates NOR make the results of the experiments available to the candidates or given any other information related to the experiments to the candidates. Doing so will constitute an examination irregularity which is punishable.
In addition to the apparatus and fittings found in a Chemistry Laboratory, each candidate will require the following:
NOTE:
MOCKS 1 2023
MARKING SCHEME
233/2
CHEMISTRY
PAPER 2
JULY/AUGUST
Question 1
(ii) The atomic radius of element G is greater than that of B. √1mk
2 : 1
Moles of H2SO4 17.5 x0.5 = 0.00875moles √½ mk
1000
Moles of IOH 0.00875 ÷ 2 = 0.004375 moles √½ mk
Molarity of IOH = 1,000 x 0.004375
20
= 0.21875M √1mk
Concentration = 0.21875 moles/litre √½ mk
Question 2
Or
Existence of sulphur in more than one form in the same physical state.√1mk
(ii) Transition temperature √1mk
Y- Heat exchanger √1 ½ mk
Z – Burner √1 ½ mk
(ii) Vandalism (v) catalyst √1 ½ mk
Temperature – 5000C √1 ½ mk
Pressure – 200atm √1 ½ mk
II- Lose heat and pre-heat incoming gases √1mk
Step 3: SO3(g) + H2SO4(l) H2S2O7 (l) √1mk
Step 4: H2S2O7(l) + H2O(l) 2 H2SO4(l) √1mk
1 : 1 : 1
1 mole of oleum = 178,000 = 1,000moles
178
1 mole at s.t.p = 22.4L
1,000moles = ? √½ mk
= 1000 x 22.4 = 22,400 litres √1mk
Question 3
Conditions: – Nickel catalyst √1mk
– I50-2500C (temperature) √1mk
H C C C H
Br H H
I – Bromopropane √1mk
Step III – substitution √1mk
(ii) H H H
H C C C H
Br Br H √1mk
Question 4
Change in temp. ΔT = 38.5 – 23.5
= 150C
Heat produced. = MCDT
= 100 x 4.2 x 15 √ ½ mk
1000
= 63.1KJ.
Molar mass of ethanol (CH3OH) = 32 √ ½ mk
Molar heat of combustion = 63.1 x 32 √ ½ mk
0.38
= 5313.68 KJ mol-1
Hence the value is lower √ 1 mk
3C(s) + 4H2(g) C3H8(g) √ 1 mk
Heat of formation = 3 (-406) + 4 (-286) + 2209) √ 1 mk
= -1218 – 4576
= – 3585 KJmol-1 √ 1 mk
Question 5
A white solid is formed √ 1 mk
A ‘pop’ sound is produced.
This confirms that the gas is hydrogen √ ½ mk
| Indicator | Colour in | |
| Acidic solution | Alkaline solution | |
| Phenolphthalein | Colourless √ ½ mk | |
| Methyl Orange | Yellow √ ½ mk | |
3Mg(s) + N2(g) Mg3N2(s) √ 1 mk
Question 6
1mole 2moles 1mole
0.12g 0.12 x 24dm3√ 1 mk
100
= 0.0288dm3 √ ½ mk
Question 7
7(a) Na2CO3(s) + 2HNO3(aq) 2NaNO3(aq) + CO2(g) +H2O(l)
(b) 1 mole of CO2(g) 22400cm3
56cm3
=0.0025 moles
(c) Mole ratio Na2CO3 : CO2
1 : 1
? 0.0025moles
0.0025moles of Na2CO3
1Mole of Na2CO3= 106g
0.0025moles = ?
= = 0.265g
(d) Mass of water= 0.715-0.265=0.45g
(e)
| Na2CO3 | H2O | |
| Mass | 0.265g | 0.45g |
| RFM | 106 | 18 |
| Moles | 0.265/106=0.0025 | 0.45/18=0.025 |
| Mole ratio | 0.0025/0.0025=1 | 0.025/0.0025=10 |
X=10
233/3
CHEMISTRY
PAPER 3
PRACTICAL
021
Kenya Certificate of Secondary Education
CHEMISTRY
PAPER 3
MOCKS 1 2023
MARKING SCHEME
Table 1
| 1 | 2 | 3 | |
| Final burette reading | 25.1 | 35.0 | 39.9 |
| Initial burette reading | 0.0 | 10.0 | 15.0 |
| Volume of solution B used | 25.1 | 25.0 | 24.9 |
( Compare / use the teacher’s value )
Award marks as follows:
A: Complete table ( 1 mark )
Conditions
Complete table with three titration 1 mark
Incomplete table with two titrations ½ ark
Incomplete table with one titration 0 mark
B: Decimal place ( 1 mark)
Conditions :
Accept only one or two decimal places used consistently.
If two decimal place the 2nd decimal place MUST be either o or 0.5
C : Accuracy 1 mark
Compare the student’s titre value with teacherstitre values.
Conditions
At least within ± 0.1 1 mark
At least within ± 0.2 ½ mark
Above ± 0.2 0 mark
D: Principles of averaging 1 mark
Values averaged must be shown and within ± 0.1 of each other
3
= 25.0cm3Ö ½
(b) Moles of acid that reacted
If 1000cm3 0.2 moles
Then 25cm3 25.0 x 0.2 Ö1
1000
= 0.005 moles Ö ½
Procedure II
Table II
Award according to procedure I table I
| 1 | 2 | 3 | |
| Final burette reading | 12.6 | 25.1 | 37.6 |
| Initial burette reading | 0.0 | 12.6 | 25.1 |
| Volume of solution B used | 12.6 | 12.5 | 12.5 |
( Compare / use the teacher’s value )
(c ) 12.6 + 12.5 + 12.5 Ö ½
3
= 12.5333cm3Ö ½
(d) (i) Moles of the acid
If 1000cm3 0.2 moles
12.5333cm3 0.2 x 12.5333 Ö1
1000
= 0.002506 moles Ö ½
(ii) Moles of sodium hydroxide
Mole ratio
H+ :OH–
1 : 1 Ö1
\0.002506 : 0.002506
Ie 0.002506 moles Ö ½
(e) Moles of acid that reacted with sodium carbonate
Ans (b) – ans (dii)
0.005 moles – 0.002506 moles Ö1
= 0.002494 moles Ö ½
(f) Molarity of A in terms of NaOH
If 25.0cm3 0.002506
The 1000cm3 0.002506 x 1000 Ö1
25cm3
= 0.10024M Ö ½
(g) CO32- (aq) + 2H+(aq) H2O (l) + CO2 (g)
[Unbalanced 0 mark
Missing state symbol ½ mark]
(h) (i) Moles of sodium carbonate
½ x 0.002494 Ö ½
= 0.001247 moles Ö ½
(ii) Molarity of A in terms of Na2CO3
If 25cm3 0.001247
Then 1000cm3 0.001247 x 1000 Ö1
25
= 0.04988M Ö ½
| Observations | Inferences | |
| 2. (a) | – Solid melts Ö ½ – Burns in yellowÖ ½ sooty flame Ö ½ Max 1 ½ marks | – Presence of either C = C Ö ½ or C CÖ Max 1 mark |
| (b) | – Dissolves Ö ½ into a colourless Ö ½ solution Max 1 mark | – Solid is polar Ö ½ Max ½ mark |
| (c ) | – Acidified KMnO4 changes from purple to colourless Ö ½ Max ½ mark | – Presence of either C CÖ ½ or C CÖ ½ or R – OH Ö ½ Max 1 ½ marks |
| (d) | – Acidified K2Cr2O7 changes from orange to green Ö ½ Max ½ mark | – Presence of either C CÖ ½ ; C CÖ ½ or R – OH Ö ½ Max 1 ½ marks |
| (e) | – EffervencesÖ ½ production of colourless gas Ö ½ Max 1 mark | H+Ö ½ / O
C OH present Max ½ mark |
| (f) | pH 4 Ö ½ Max ½ mark | Weakly acidic Ö ½ Max ½ mark |
| 3 (a) | – Colourless liquid on cooler parts of test tube Ö ½ – Colourless gas with pungent smell Ö ½ – Gas turns moist red litmus blue Ö ½ / Blue litmus remains blue Ö ½ | – Hydrated salt Ö ½ – Presence of NH4+Ö ½ 1 |
| (b) | – Solid dissolves into pale green solution Ö ½ – 1 | – Polar solid Ö ½ – Presence of Fe2+Ö ½ 1 |
| (c ) | -Green Ö ½ ppt insoluble Ö ½ in excess – PPt turns brown on exposure to air Ö ½ 1 ½ | – Presence of Fe2+Ö ½ – Fe2+ oxidized to Fe3+ by air Ö 1 1 |
| (d) | – Green ppt Ö ½ insoluble in excess – Ppt turns brown on exposure to air Ö ½ 1 ½ | – Presence of Fe2+Ö ½ – Fe2+ oxidized to Fe3+ by air Ö ½ 1 |
| (e) | – Pale green solution turns yellow Ö ½ – Brown Ö ½ ppt insoluble Ö ½ in excess 1 ½ | – Presence of Fe3+Ö ½ ½ |
| (f) | – Formation of white ppt Ö ½ as residue and green solution as filtrate Ö ½ 1 | – Presence of Fe2+Ö ½ – Presence of either Ce–Ö ½ , SO42-Ö ½ or SO32-Ö ½ , CO32-Ö ½ Max 2 marks |
| (g) | – White ppt forms Ö ½ ½ | SO42- present Ö ½ ½ |
MOCKS 1 2023
MARKING SCHEME
233/1
CHEMISTRY
PAPER 1
THEORY
M 2.6 √1mk
(b) K 2+
M 2-
2 (a) Anode: I– (l) I2 (g) +2e–√1mk
Cathode: Pb2++ 2e– Pb (s)√1mk
Filter(1/2) to remove copper (ii) oxide and sodium chloride as the filtrate
Evaporate the filtrate to saturation and cool to obtain sodium crystals (√1mk)
Dry them between filter papers/leave them in the open to dry (√1mk)
Brown fumes are observed (√1mk)
Reaction between oxygen gas and ammonia gas over platinum wire is exothermic. (1/2mrk)
Ammonia is oxidized to nitrogen (ii) oxide which reacts with excess oxygen to form nitrogen (iv) oxide,(1/2 mrk)
b)
Non – luminous flame does not produce soot (1mrk) Any one (√1mk)
XOH = 2 = 15 X M
H2SO4 1 20 X 0.045 (√1mk)
M = 2 X 20 X 0.045
1 X 15 = 0,12 moles/l (1/2)
1mole = 2.88
0.12 = 24 (1/2)
RFM of XOH = X + 16 + 1 (1/2)
X = 7 1/2)
X = 6
2 = 3 (1/2)
Y = 12
3 = 6 (1/2)
MF = C3H6 (1/2)
Structural fprmula = CH3 CH = CH2 (1/2)
(ii) Concetrating the mineral ore by making impurities to sink atb the bottom. √1mk
(b) Exp 2 – Wet cloth turned white due to bleaching as chlorine dissolves in water to form hypochloric(I) acid√1mk
(b) Mass of solid Y = 30.4 – 26.2
= 4.2g (1/2mrk)
Mass of water in the solution = 42.4 – 30.4
= 12g (1/2 mrk)
12 g of water dissolve 4.2g solid Y
100g of ware will dissolve 100 x 4.2 (1/2mrk)
12
= 35g / 100g of water (1/2mrk)
-105 + AH0c(CO) = -393 √1mk
AH0c(CO) = -393 + 105 √1mk
= -188kJMol-1 √1mk
b)The delivery tube should first be removed √1mk to avoid sucking back of liquid M √1mk
a = 4 √1mk
Atomic number 1 + 1 = b + 0
b = 2 √1mk
V1 = 4dm3 ; P1 = 152mmHg ; V2 = 2dm3 ; T1 = 250K ; T2 = 500K P2 = ?
152 x 4 = P2 x 2
250 500 √1mk
P2 = 152 x4 x 500
250 x 2 √1mk
= 608mmHg √1mk
(ii) Gas G is Nitrogen (i) oxide √1mk
Equilibrium shift to the left/ backward reaction is favoured to replace OH– that react with H+ √1mk
Element Ba S O
% composition 58.81 13.72 24.47
RAM 13732 16
No. of moles 58.81 13.72 24.47 (1/2mrk)
137 32 16
0.4293 0.4281 1.768
Mole ratio 1 1 4 (1/2mrk)
E . F. BaSO4(√1mk)
(ii) NH3 (g) + HCl(g) NH4Cl(s) (√1mk)
The forward reaction is accompanied by a decrease in volume(1/2mrk). Equilibrium shifts to the right following increase of the forward reaction (1/2mrk)
(ii) The yield of AB is increased’ (√1mk)
The forward reaction is exothermic.(1/2mrk) Decrease in temperature favours the forward reaction, equilibrium shifts to the right.(1/2mrk)
(b) Efflorescence (√1mk)
(c) Hygroscopy (√1mk)
B – Oxgygen/ O2(√1mk)
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