skema mrsm2014...alloy p p3. alloy p/bronze is harder than pure copper p4. atomic size in pure...

CONFIDENTIAL

2

NO. MARKING SCHEME

PAPER 2 (SECTION A)

MARK

SUB TOTAL

1

(a) (i)

(ii)

(iii)

M : soap

N : detergent

Cleansing agent Observation

M Oily stain remain //Cloth is still dirty

N No oily stain // Cloth is clean

N / Detergent

1

1

1

1

1

2

2

1

(b)(i)

(ii)

(iii)

P : Aspirin/ paracetamol/ codeine

Q : Antibiotic

To relieve/ reduce pain // as pain killer

[r: to cure pain]

To make sure all the bacteria are killed//bacteria become

immune/resistant to the given medicine

[r: virus/ microorganism/ germs/ fungi]

1

1

1

1

2

2

TOTAL 9

SKEMA MRSM2014

https://cikguadura.wordpress.com/

CONFIDENTIAL

3

NO. MARKING SCHEME MARK

SUB TOTAL

2(a) Na / Mg / Al 1 1

(b) 2.8 / 2,8

[r : 2:8]

1 1

(c)(i)

(ii)

(iii)

Forces of attraction between nucleus and electrons is stronger in

chlorine atom // nuclei attraction towards electron is stronger in

chlorine atom

[r: is higher]

1. Correct number of electrons and shell with nucleus

2. Label the charge and symbol of atom (can infer from nucleus)

Ionic compound

[a : ionic]

[r : ionic bond]

1

1

1

1

1

2

1

(d) Semiconductor// to produce microchips// to make concrete /

cement / ceramic/ pottery/ enamels

1 1

(e) (i)

(ii)

Chemically inert // argon atom achieves stable octet electron

arrangement

Volume of argon gas = 0.002 x 24 / / 0.048 dm3 / / 48 cm3

1

1

2

TOTAL 9

CONFIDENTIAL

4


SUB TOTAL

3(a) Ionic compound/substance that can conduct electricity in molten

state or aqueous solution and undergoes chemical changes.

1 1

(b) K + , I- , H+ ,OH- 1 1

(c) (i)

(ii)

Iodide ion and hydroxide ion

2I- → I2 + 2e

1

1

2

(d) (i)

(ii)

(Colourless) gas bubbles releases

1. H+ ion is lower than K+ ion in the Electrochemical Series 2. H+ ion will be selectively discharged to form hydrogen

gas

1

1

1

1

2

(e) (i)

(ii)

Oxygen

1. No of mole = 24 / 24000 = 0.001 mol

2. No of molecule = 0.001 x 6.02 x 1023 = 6.02 x 1020

1

1

1

1

2

TOTAL 10

CONFIDENTIAL

5


SUB TOTAL

4(a)(i)

(ii)

(iii)

(iv)

Copper(II) oxide

[a: copper oxide]

Green

Sodium carbonate / potassium carbonate/ ammonium carbonate

[a : formula]

Na2CO3 + Cu (NO3)2 → CuCO3 + 2 NaNO3

[based on answer in a(iii)]

1. Correct formulae of reactants and products

2. Balanced equation

1

1

1

1

1

3

2

(b)

(c)

1. Blue precipitate 2. Dissolves in excess ammonia solution

1. Add 2 cm3 dilute sulphuric acid followed by 2 cm3 of iron ( II) sulphate solution and shake

2. Concentrated sulphuric acid is added slowly/drop by drop 3. Brown ring is formed

1

1

1

1

1

2

3

TOTAL 10

CONFIDENTIAL

6

No. MARKING SCHEME MARK

SUB TOTAL

5(a)

Volume of gas liberated // Mass of calcium carbonate 1

1

(b) (i)

(ii)

(iii)

(iv)

(v)

No of mole = 25 x 0.1 / 1000 = 0.0025 mol

1. 0.0025 mol HNO3 produce 0.00125 mol CO2

2. Volume of CO2 = 0.00125 x 24

= 0.03 dm3

Note : 1. [ecf from (b)(i)]

2. Point 2 must be correct with unit

1. Correct curves with label 2. Axis with label & units

1. Rate of reaction of Exp II is higher than Exp I

1. Exp II has higher temperature than Exp. I 2. Kinetic energy of particles is higher. 3. Frequency of collision between H + and CaCO3 / CO32-

is higher

4. Frequency of effective collision is higher

1

1

1

1

1

1

1

1

1

1

3

2

1

4

TOTAL 11

30 cm3

Volume of gas/ cm3

Isipadu gas / cm3

Time/ s

Masa /s

II

III

I

CONFIDENTIAL

7

NO MARKING SCHEME MARK

SUB TOTAL

6 (a) To allow the movement of ions 1 1

(b)

Shown in the diagram (from electrode L to electrode M

through external circuit).

[r: answers in words]

1 1

(c) (i)

(ii)

Iron(III) ion / Ferum(III) ion / Iron(III) sulphate

[r: formula]

1

3

Alternative 1

1. Add sodium hydroxide/ammonia solution.

2. Brown precipitate is formed.

Alternative 2

1. Add potassium hexacyanoferrate(II) solution.

2. Dark blue precipitate is formed.

Alternative 3

1. Add potassium thiocyanate solution.

2. Blood red colouration is formed.

1

1

(d) Br2 + 2e 2Br-

1 3

(e) 1. Orange solution turns green

2. Cr2O72- ion undergoes reduction to form Cr3+ ion

1

1

(f)

Functional apparatus

Label apparatus and material

Label positive and negative terminal

1

1

1

3

TOTAL 11


CONFIDENTIAL

8

NO MARKING SCHEME SUB

MARK

TOTAL

MARKS

7(a) (i)

(ii)

P1. Empirical formula is (chemical) formula that shows the

simplest ratio of hydrogen and carbon atoms in benzene //

Formula that shows the simplest ratio of hydrogen to

carbon atoms in benzene is 1:1

P2. Molecular formula is (chemical) formula that shows the

actual number of hydrogen and carbon atoms in benzene//

Formula that shows the actual number of atoms which are

6 hydrogen and 6 carbon in benzene.

1

1 4

ethanol,C2H5OH // water,H2O // magnesium oxide, MgO

[a: any compound where its empirical formula and molecular

formula is the same]

1+1

(b) (i)

(ii)

Empirical formula of the compound is CH3

[CH3]n = 30

[12+3(1)]n = 30

n = 2

Molecular formula of the compound is C2H6

C2H6 + 7/2 O2 2CO2 + 3H2O

Correct formula of reactants and products

Balanced equation

Element C H

Number of mole 80/12 = 6.67 20/1 = 20

Simplest ratio of

moles

6.67/6.67 = 1 20/6.67 = 3

1

1

1

1

1

Max 4

1+1

6

(c) (i)

(ii)

Hydrocarbon Homologous

series

General

formula Name

E alkane CnH2n+2

n=1,2.. butane

J alkene CnH2n

n = 2,3,… But-2-ene

3

3

6

P1. J produces more soot than E.

P2 & P3. Calculation

E J

% of carbon

4(12) x 100 = 82.76

58

4(12) x 100 = 85.71

56

% of carbon by mass in J is higher than E

1

1+1

1

4

TOTAL 20

CONFIDENTIAL

9

NO EXPLANATION SUB

MARK

TOTAL

MARKS

8 (a) (i)

(ii)

Substance P Q

Name Bronze Borosilicate glass

Specific

properties

Shiny surface //

Does not corrode

easily//hard and

strong

Withstand high

temperature//

Resistant to chemicals

//low thermal

expansion

1+1

1+1

4

P1& P2. Diagram

tin

Pure copper

copper

Alloy P

P3. Alloy P/bronze is harder than pure copper

P4. Atomic size in pure copper is the same , atomic size in

alloy P/bronze is different.

P5. Atoms in pure copper are orderly arranged while atoms in

alloy P/bronze are not orderly arranged.

P6. When force is applied, layer of atoms in pure copper is

easier to slide than in alloy P/ bronze.

1+1

1

1

1

1

6

(b) (i)

(ii)

(iii)

P1. Sulphur dioxide burnt/react with excess oxygen to produce

sulphur trioxide.

P2. Passed over vanadium(V) oxide catalyst at 450oC, 1 atm

P3. Chemical equation : 2SO2 + O2 2SO3

1

1

1

1

1

1

1

3

4

P1. Step 1: Sulphur trioxide is dissolved in concentrated

sulphuric acid to form oleum.

P2. SO3 + H2SO4 H2S2O7

P3. Step 2: Oleum is diluted with water to produce sulphuric

acid.

P4. H2S2O7 + H2O 2H2SO4

P1. Moles of S = 48 / 32 =1.5 mol

P2. 1 mol of S produce 1 mol SO2

Thus, 1.5 mol of S produce 1.5 mol SO2

P3.Volume of SO2 = 1.5 24 dm3 = 36 dm3

1

1

1

3

TOTAL 20

CONFIDENTIAL

10

NO EXPLANATION MARK ∑ MARKS

9 (a) (i) P1.Correct suggestion for solution T

Eg. Copper (II) sulphate solution /

Sodium chloride/ zinc sulphate solution /

sulphuric acid

P2. Correct suggestion for solution V

Eg. Glucose solution/benzene/ ethanol/methylbenzene

P3. Solution T has free moving ions.

P4. Solution V is made up of neutral molecules//

does not have free moving ions

1

1

1

1

4

(ii) P1. Zinc is more electropositive than copper.

P2. Zinc atom releases electron to form zinc ion Zn2+

P3. Half equation: Zn Zn2+ + 2e

P4. Electrons flow through external circuit to the copper

electrode

If T is copper(II) sulphate solution

P5. Copper(II) ion receives electron to form

copper atom

P6. Half equation: Cu2+ + 2e Cu

OR

If T is sodium chloride solution / sulphuric acid

P5. Hydrogen ion receives electron to form hydrogen

P6. Half equation: 2H+ + 2e H2

1

1

1

1

1

1

6

(b) P1. Materials : 2.0 mol dm-3 sodium hydroxide,

distilled water

P2. Apparatus : volumetric flask100 cm3, pipette 25 cm3,

[note : P1 & P2 can be inferred from procedure]

Calculation :

M1V1 = M2V2

P3. 2.0 x V1 = 0.5 x 100

P4. V1 = 25 cm3

Procedure :

P5. Pipette 25 cm3 2.0 mol dm-3 of sodium hydroxide

solution.

P6. Transfer the solution into a volumetric flask.

P7. Add distilled water into the volumetric flask.

P8. until reaches the calibration mark.

P9. Close/ Put stopper on the volumetric flask.

P10.Shake the volumetric flask.

1

1

1

1

1

1

1

1

1

1

10

TOTAL 20

CONFIDENTIAL

11

NO EXPLANATION MARK ∑ MARKS

10

(a)(i)

P1. L : Mg / Magnesium

P2. R : Zn / Zinc

[Note : 1. L must be more electropositive than R

2. Both L and R must be more electropositive than Cu]

P3. The heat of displacement for Experiment I is

higher than Experiment II.

P4. The distance between L / Cu is bigger/ further in the

Electrochemical Series // L is more electropositive

than R

1

1

1

1

4

(ii) Mg + CuSO4 MgSO4 + Cu P1. Correct chemical formula of reactant and product

P2. Balance chemical equation.

[adp metal L but should be more electropositive than Cu ]

P3. Number of mole of CuSO4 = 0.2 x 50 // 0.01

1000

P4. 1 mol of Cu displaced release 336000 J heat

P5. .: 0.01 mol of Cu 336000 J x 0.01 mol heat

1 mol

= 3360 J

P6. Change in temperature = 3360

50 X 4.2

= 16 oC

[ecf P5 from P3]

1+1

1

1

1

1

6

(b) P1. Apparatus: Polystyrene cup , thermometer,

measuring cylinder ( suitable volume)

P2. Materials : [suitable salt solutions to produce

precipitate/any insoluble salt]

Eg. Lead(II) nitrate solution, sodium

sulphate solution

Procedure:

P3. Measure [25-200] cm3 of [0.1-2.0] mol dm-3 lead(II)

nitrate solution.

P4. Pour into a polystyrene cup.

P5. Measure [25-200] cm3 of [0.1-2.0] mol dm-3 sodium

sulphate solution.

P6. Pour into a different polystyrene cup.

P7. Measure the initial temperature of both solutions.

P8. Pour sodium sulphate solution quickly into lead(II)

nitrate solution.[a: vice versa]

P9. Stir the mixture.

P10. Record the highest temperature.[r: final temperature]

1

1

1

1

1

1

1

1

1

1

10

TOTAL 20

4541/3 SULIT

3

QUESTION 1 (18 marks)

Question Mark Scheme Mark

1 (a)

[Able to record all the initial and final readings of the titration

accurately in 2 decimal places].

Answer:

Titration

I

Titration

II

Titration

III

Initial 0.10 24.90

1.00

Final 24.90 49.50

25.70

3

[Able to record (any five of the initial and final readings of the

titration accurately) // (six readings in one decimal place)]

2

[Able to record (any four of the initial and final readings of the

titration accurately) // (six readings with no decimal place)] 1

No response or wrong response 0


1 (b)

[Able to construct a table with correct headings and units to

record the initial and final readings of the titration ]

Answer:

Titration I II III

Final reading of burette /cm3 24.90 49.50

25.70

Initial reading of burette /cm3 0.10 24.90

1.00

Volume of hydrochloric acid /cm3 24.80 24.60 24.70

3

[Able to construct a table that contains the following]

1. Title without unit

2. Reading

[Note: 1. If data in 1(a) not accurate/incorrect, maximum score 2]

2

[Able to construct a table that contains the following ]

1. at least one title with one set of reading

1


SKEMA MRSM2014


4541/3 SULIT

4


1 (c)(i)

[Able to show the calculation for the average volume (with

unit) of hydrochloric acid accurately with the answer given in

2 decimal places]

Answer : 24.80 + 24.60 + 24.70 = 24.70 cm3

3

3

[Able to show the calculation for the average volume (without

unit) of hydrochloric acid less accurately with the answer

given in 2 decimal places]

//[Able to show the volume (with unit) of hydrochloric acid

used without showing the calculation steps]

Sample answer :

1. 24.80 + 24.60 + 24.70 = 24.70

3

2. 24.70 cm3

3. 24.7 cm3

2

[Able to show the volume (without unit) of hydrochloric acid

used, without showing the calculation steps]

Sample answer : 24.7

1


Question Marc Scheme Mark

1 (c)(ii)

[Able to write chemical equation and calculate the

concentration of sodium hydroxide with units accurately]

Answer :

P1. HCl + NaOH NaCl + H2O

P2. Number of mole of NaOH = number of mole of HCl

= 0.1 x 24.7 = 0.00247 mol

1000

P3. Concentration of NaOH = 0.00247 mol = 0.0988 mol dm-3

0.025 dm3

[a: P2 = 0.0025 mol , P3 = 0.1 mol dm-3]

3

[Able to state any two steps correctly]

Note : [ecf number of mole from P2]

2

[Able to state any step correctly] 1


4541/3 SULIT

5


1(d)

[Able to predict the volume of sulphuric acid accurately to

2 decimal places with unit and explain correctly]

Answer :

P1. 12.35 cm3

P2. Hydrochloric acid is a monoprotic/monobasic acid while

sulphuric acid is a diprotic/dibasic acid

P3. The concentration of H+ ions in sulphuric acid is twice the

concentration of H+ ions in hydrochloric acid

3

[Able to predict the volume of sulphuric acid without unit and

state any step P2/P3 correctly] 2

[Able to state P2 or /and P3 correctly] 1


Question Mark scheme Mark

1 (e)

[Able to classify all the 6 acids into strong acids and weak

acids correctly]

Answer :

Strong Acid Weak Acid

Nitric acid

Sulphuric acid

Hydrochloric acid

Methanoic acid

Ethanoic acid

Carbonic acid

3

[Able to classify at least four acids into strong acids and weak

acids correctly] 2

[Able to classify at least two of the acids into strong acids and

weak acids correctly// able to classify the two groups of acids

inaccurately]

Sample answer:

Strong Acid Weak Acid

Methanoic acid

Ethanoic acid

Carbonic acid

Nitric acid

Sulphuric acid

Hydrochloric acid

1


4541/3 SULIT

6

QUESTION 2 (15 MARKS)


2(a) [Able to state the inference for both experiments correctly]

Sample answer :

Rate of reaction of experiment II is higher//vice versa

// Reaction in Experiment II is faster // vice versa.

3

[Able to state the inference for both experiments less correctly]

Sample answer :

1. Rate of reaction of experiment II is faster// vice versa

2. Frequency of effective collision is higher in Experiment II//vice

versa

2

[Able to state an idea of inference]

Sample answer :

1. Rate of reaction is affected by the size.

2. Experiment II use smaller/ powdered marble chips // vice versa

3. Time taken for Experiment II is shorter // vice versa

1

No response given / wrong response. 0


2(b) [Able to state all the three variables for both experiment correctly]

Sample answer:

P1. Manipulated variable : Size of marble chips // total surface

area of marble chips/ calcium

carbonate // large marble chips and

marble chips powder

P2. Responding variable : Rate of reaction// Time taken for a

10 cm3 carbon dioxide is collected.

P3. Constant variable : Volume and concentration of

hydrochloric acid //

Mass of marble chips /calcium

carbonate// Temperature

3

[Able to state any two variables correctly] 2

[Able to state any one variable correctly] 1


4541/3 SULIT

7


2(c) [Able to state the hypothesis correctly by stating the two following

information]

Manipulated variable:

Size of calcium carbonate/marble chips//total surface area of calcium

carbonate/marble chips

Direction of responding variable:

Increase / Decrease in the rate of reaction

// Increase / Decrease in time taken for the reaction to be complete

Sample answer:

1. The smaller the size of calcium carbonate/marble chips, the higher

the rate of reaction//vice versa.

2. The larger the total surface area of calcium carbonate, the

higher the rate of reaction//vice versa.

3. When the size of calcium carbonate is smaller, the rate of

reaction increases// vice versa.

3

[Able to state the hypothesis less correctly by stating the relation

between the manipulated variable and the responding variable]

Sample answer:

1. The smaller the size of calcium carbonate/marble chips, the faster / slower the rate reaction.

2. The larger the total surface area of calcium carbonate/marble

chips, the faster / slower the rate reaction.

3. The smaller the size, the higher the rate of reaction

4. The rate of reaction is higher when the size of calcium

carbonate /marble chips is smaller

2

[Able to give an idea of the hypothesis]

Sample answer:

1.The size /total surface area affects / changes the rate of

reaction

2. Different size of calcium carbonate/marble chips , different rate of

reaction

1


4541/3 SULIT

8


2(d) [Able to give the operational definition for rate of reaction with the

following aspects:]

1. what you do

2. what you observe with direction

Sample answer:

1.When marble chips powder is reacted with hydrochloric acid, the

time taken to collect 10 cm3 /certain volume of gas is shorter //

2.When large marble chips is reacted with hydrochloric acid, the

time taken to collect 10 cm3 /certain volume of gas is longer

3

[Able to state the operational definition for rate of reaction

incompletely with any one of the following aspects]

1. what you do

2. what you observe with direction

Sample answer:

1. Marble chips powder/large marble chips is reacted with

hydrochloric acid //

2. Time taken to collect 10 cm3 /certain volume of gas in Experiment

II is shorter //

3. Time taken for a 10 cm3 /certain volume of gas collected when the Marble chips is reacted with hydrochloric acid //

4. The change of volume of gas per unit time

2

[Able to give an idea of operational definition for rate of reaction]

Sample answer:

The change of amount of reactant per time taken // rate of reaction is

influenced by the size of marble chips

// Rate of reaction = volume of gas

Time taken

1


4541/3 SULIT

9


2(e) [Able to state the relationship between temperature of surrounding

and the rate of drying clothes correctly].

Sample answers:

The higher the temperature of the surrounding, the higher the rate of

drying.//

The lower the temperature of the surrounding, the lower the rate of

drying.

3

[Able to state the relationship between temperature of surrounding

and the rate of drying clothes less correctly].

Sample answer:

The higher the temperature, the higher the rate of drying// vice versa

// The temperature is directly proportional to the rate of drying //

Clothes dry faster at higher temperature

2

[Able to give an idea of relationship between temperature of

surrounding and the rate of drying clothes].

Sample answer:

Different temperature, different rate of drying

//Temperature affect the rate of drying // Clothes dry at different

time at different temperature

1

No response given / wrong response 0

4541/3 SULIT

10

QUESTION 3 (17 MARKS)


3 (a) [Able to give the problem statement correctly].

Sample answer:

How to differentiate hexane and hexene/ liquid P and liquid Q using

chemical test/bromine water/ acidified potassium manganate(VII)

solution?

Note: P and Q could be either hexane or hexene [vice versa]

3

[Able to give the problem statement incorrectly or able to give the

aim of experiment].

Sample answers :

1. How to differentiate hexane and hexene /liquid P and liquid Q?//

2. To differentiate between hexane and hexene/liquid P and liquid Q//

3. How to differentiate hexane and hexene/ liquid P and liquid Q

using chemical test/bromine water/ acidified potassium

manganate(VII) solution.//

4. How to identify hexane and hexene/ liquid P and liquid Q

using chemical test/bromine water/ acidified potassium

manganate(VII) solution?

2

[Able to give an idea of the problem statement]

Sample answer:

How to identify/determine/differentiate alkane and alkene /

To identify/determine/differentiate alkane and alkene.

Hexane and hexene can cause changes to bromine water/ acidified

potassium manganate(VII) solution

1

No response given or wrong response 0


3(b) [Able to state the three variables correctly].

Sample answer:

P1. Manipulated variable: Hexane and Hexene/ Liquid P and Q/ Type

of hydrocarbon

P2. Responding variable : Colour change of bromine water/ acidified


P3. Constant variable : bromine water/ acidified


3

Able to state any two variables correctly 2

Able to state any one variables correctly 1


4541/3 SULIT

11


3(c)

[Able to state the relationship between the manipulated variable and

the responding variable correctly].

Sample answer:

1. If hexane is used, purple colour of acidified potassium

manganate(VII) solution remains unchanged.

If hexene is used, purple colour of acidified potassium

manganate(VII) solution becomes colourless. //

2. If hexane is used, brown bromine water remains unchanged.

If hexene is used, brown bromine water becomes colourless.

[Note : P and Q can replace hexane and hexene]

3

[Able to state the relationship between the manipulated variable and

the responding variable less correctly].

Sample answer:

1. If hexane is used, purple colour of acidified potassium

manganate(VII) solution remains unchanged.//

2. If hexene is used, purple colour of acidified potassium

manganate(VII) solution becomes colourless. //

3. If hexane is used, brown bromine water remains unchanged.//

4. If hexene is used, brown bromine water becomes colourless.//

5. Purple colour of acidified potassium manganate(VII) solution

remains unchanged if hexane is used.//

6. Purple colour of acidified potassium manganate(VII) solution

becomes colourless if hexene is used. //

7. Brown bromine water remains unchanged if hexane is used.//

8. Brown bromine water becomes colourless if hexene is used.

2

[Able to state the idea of hypothesis].

Sample answer:

1. Change of colour of acidified potassium manganate(Vll) //

2. Hexane does not change the colour of acidified potassium

manganate(VII) solution //

3. Hexene will decolourised colour of acidified potassium

manganate(VlI) solution

1


4541/3 SULIT

12


3(d) [Able to give list of materials and apparatus completely].

Sample answer:

Materials:

1. Liquid P/ hexane

2. Liquid Q/ hexene

3. Acidified potassium manganate(VII) solution / bromine water

Apparatus:

1. Test tube / boiling tube

2. Dropper


3

[Able to give a list of all materials and suitable container].

Sample answer:

Materials:

1. Liquid P/ hexane

2. Liquid Q/ hexene


Apparatus:

1. Test tube / boiling tube / beaker

2

[Able to give an idea of materials and apparatus].

Sample answer:

Materials:

1. Liquid P/ Q / hexane/hexene


Apparatus:

1. [Any suitable container]

1


4541/3 SULIT

13


3(e) [Able to state all steps of procedure correctly]

Sample answer:

1. Pour [1-5] cm3 of liquid P/ hexane into a test tube/ boiling tube

2. Add a few drops of acidified potassium manganate(VII) solution /

bromine water.

3. Shake/ stir the mixture.

4. Record the observation.

5. Repeat steps 1 to 4 by using liquid Q/ hexene to replace liquid P/

hexane


3

[Able to state steps 1, 2, 4 and 5] 2

[Able to state steps 1 and 2] 1



3(f)

[Able to construct a table that consist of the following information]

1. Heading for manipulated variable

2. Heading for observation

Sample answer:

Liquid Observation

P/ hexane

Q/ hexene


2

[Able to exhibit the tabulation of data that includes the following

three information:]

1. Heading for manipulated variable [only 1 liquid listed]

2. Heading for observation

Sample answer :

Liquid Observation

P

1


END OF MARKING SCHEME


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