snr chemistry 07 as acid equil

8/18/2019 Snr Chemistry 07 as Acid Equil

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Supervised assessment: Acid and bases,

and equilibrium

This sample demonstrates on-balance judgments within criteria. It provides information about

student achievement where the indicative response matches the qualities of the A standards.

Criteria assessed

• Knowledge and conceptual understanding

• Investigative processes

• Evaluating and concluding

Assessment instrument

The response presented in this sample is in response to assessment items.

The task is a supervised assessment on the topic of acids and bases, and equilibrium.

Students have had the learning experiences of conducting titrations

Chemistry 2007 Annotated indicative response and judgments within criteria


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Students have had the learning experiences of conducting titrations

Chemistry 2007

Annotated indicative response and judgments using marks within criteria

Queensland Curriculum & Assessment Authority

July 2014

Page 2 of 22

Instrument-specific criteria and standards

Standard A Standard B Standard C Standard D Standard E

K n o w l e d g e a

n d c o n c e p t u a l u n d e r s t a n d i n g

The student work has thefollowing characteristics:





• reproduction andinterpretation of complexand challengingconcepts, theories andprinciples

• reproduction andinterpretation of complexor challenging concepts,theories and principles

• reproduction of concepts,theories and principles

• reproduction of simpleideas and concepts

• reproduction of isolatedfacts

Q4

• comparison andexplanation of complexconcepts, processes andphenomena

• comparison andexplanation of concepts,processes andphenomena

• explanation of simpleprocesses and phenomena

• description of simpleprocesses andphenomena

• recognition of isolatedsimple phenomena

Q6 Q3

• linking and application ofalgorithms, concepts,principles, theories andschema to find solutionsin complex andchallenging situations

• linking and application ofalgorithms, concepts,principles, theories andschema to find solutionsin complex or challengingsituations

• application of algorithms,principles, theories andschema to find solutions insimple situations

• application of algorithms,principles, theories andschema

• application of simplegiven algorithms

Q9 Q2, 8, 9 Q1, 5, 7, 9

Preliminarygrade

boundaries39–24 24–16 16–13 13–9 8–1

Note: Preliminary grade boundaries are determined based on the school’s experience with similar assessment instruments and the relative number of A, B and C marksavailable (see above).

All grade boundaries must be confirmed once they have been applied to student responses and compared to syllabus standards.


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Chemistry 2007



July 2014

Page 3 of 22

Standard A Standard B Standard C Standard D Standard E

Investigative

processes






• systematic analysis ofsecondary data to

identify relationshipsbetween patterns andtrends

• analysis of secondarydata to identify patterns

and trends

• analysis of secondarydata to identify obvious

patterns and trends

• identification of obviouspatterns

• recording of data

Q13, 14 Q13, 14 Q11, 12, 13, 14

Preliminarygradeboundaries

12–11 10–9 8–7

Evaluatingandconcluding






• exploration of

scenarios and possibleoutcomes with justification ofconclusions

• explanation of

scenarios and possibleoutcomes withdiscussion ofconclusions

• description of

scenarios and possibleoutcomes withstatements ofconclusion

• identification of

scenarios or possibleoutcomes

• statements about

outcomes

Q15 Q15 Q15

Preliminarygrade

boundaries10–9 8–7 6–5

Note: Preliminary grade boundaries are based on the school’s experience with similar assessment instruments and on the relative number of marks available based on thetask-specific descriptors (drawn from the syllabus exit standards). All grade boundaries must be confirmed once they have been applied to student responses and matched

to syllabus standards.


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Indicative response — Using marksRelevant exitstandarddescriptor

Question 1 (KCU C: 2 marks)

Calculate the pH of the following solution [H+] = 1.0 x 10

-6M.

State whether it is acidic, basic or neutral.Expected response

[ ]

[ ]6

101log

log

6

3

=

×=

=

−

+O H pH

The solution is below 7 which is neutral, so it is slightly acidic. Question 2 (KCU B: 3 marks)

A solution has a pH of 12.68. Calculate the [H+] and [OH

-].

Expected Response

[ ][ ]

[ ][ ][ ]

[ ]OH

OH

OH H

H

H

pH

pH

13

14

1413

14

13

68.12

100892

101

10110089.2

101

10089.2

10

10

68.12

−

−

−

−−−

−−+

−

−+

−+

×

×

=

×=××

×=

×=

=

=

=

Markingscheme

application ofalgorithms andtheories to findsolutions insimple situations

reproduction ofa concept

Question 1 KCU:2 marks C

1 mark for thealgorithm andcorrect solution1 mark for thestatementregarding acidity

Question 2 KCU:

2C marks

2B marks

1 mark for thedefinition of pH

1 mark for thecalculation of[H+]


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Relevantexitstandard

descriptor

Question 4 (KCU C: 1 mark)Which of the following will determine whether an acid will be strong orweak?

(a) its concentration

(b) its ability to donate protons(c) its solubility in water(d) its pH

Question 5 (KCU C: 4 marks)

Based on the pH values shown in the figure, which of the followingstatements about the concentration of hydrogen ions is correct?

(a) It is twice as great in milk as it is in lemon juice.

(b) It is 1 000 000 times greater in soap than in wine.

(c) It is three times greater in wine than in bleach solution.

(d) It is 1 000 times greater in distilled water than in soap.

Show reasoning that applies the concept of the pH scale to supportyour answer:Expected response

The pH scale is a logarithmic scale. The hydrogen ion concentration atthe positions indicated is:Milk [H

+] = 10

-6

L j i [H+] 10 3

Markingscheme

reproductionof a concept

Question 4KCU:1mark C1 mark forthe correctresponse

Question 5KCU: 4

marks C


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Relevant exitstandarddescriptor

Question 6 (KCU B 3 marks) Write TWO chemical equations to show that the dihydrogenphosphate ion (H2PO4

-) is amphiprotic. Compare the

equations to explain the term amphiprotic.Expected response

A substance is amphiprotic if it acts like both an acid and abase.In the following equation a hydrogen ion is being donated,showing that the dihydrogen phosphate ion is acting as anacid.

(aq)

_

4)()(42−+−

+→ HPO H PO H aqaq

In the following equation a hydrogen ion is being accepted,showing that the dihydrogen phosphate ion is acting as a base.

(aq)43)()(42 PO H H PO H aqaq →+ +−

Question 7 (KCU C: 3 marks) A titration is carried out in which 0.40 mol L

-1 potassium

hydroxide (KOH) solution in a burette is titrated against10.00mL of hydrochloric acid (HCl) solution. The volume ofbase required to reach the end point was 12.5mL.Write a balanced equation for this reaction then calculate theconcentration of the hydrochloric acid solution. Show allworking.

Expected response

Markingscheme

comparison andexplanation ofconcepts

Question 6 KCU: 3marks B1 mark forexplanation of theterm amphiprotic

1 mark for writingeach equation andexplaining whether

the H+ ion iddonated oraccepted


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Question 8 (KCU A: 4 marks)

Borax or sodium tetraborate decahydrate,

Na2B4O7.10H2O, is a base which can easily be obtainedin very pure form. (This salt dissociates to form anion,cation and 10H2O) It reacts with strong acids such ashydrochloric to form the extremely weak boric acid.

B4O72-

(aq) + 2 H+

(aq) + 5 H2O(l) 4 H3BO3 (aq)

Borax is frequently used as a primary standard in acid-base titrations.

In one particular experiment 7.738 g of borax wasdissolved in water and made accurately to 250mL.25 mL of this solution required 27.65 mL of a solution ofhydrochloric acid for exact neutralization.

Calculate the concentrations of the borax andhydrochloric acid solutions.

Expected response( ) ( )

lfbbfiC

moles

boraxof Moles

boraxof weight Molecular

0498.0

17.155

738.7

17.155

93.11124.43

799.15481.10

=

=

=

+=

×+×=

Marking scheme

linking andapplication ofconcepts, principles,theories and schema

to find solutions in a

Question 8 KCU: 4marks AThis is a multi-step(i.e. complex) andunfamiliar (i.e.challenging)situation.

1 mark for thecalculation of themolecular weight1 mark for thecalculation of thenumber of moles ofborax

1 mark for thecalculation of theconcentration ofborax1 mark for thecalculation of theconcentration of theacid


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Question 9 (KCU A: 10 marks) Most swimming pools use 'chlorine' as the sanitizing agent tokill bacteria and viruses. There are a variety of ways of

generating the 'chlorine' in pool water. The most widely usedmethod is the addition of hypochlorite compounds. The activechemical produced in each method is HOCl or hypochlorousacid, sometimes written as HClO.Once in the water, an equilibrium is established between thestrong oxidant HOCl and the weaker OCl

- ion.

1100.3 8)()(3)(2)( Eqn xK OClO H O H HOCl aaqaqaqaq−−

=+⇔+

This equilibrium system is pH dependent and the followinggraph shows how the concentrations of HOCl and OCl

- change

with the pH.

Markingscheme

Question 9 KCU10 marks AThis is a multi-step(i.e. complex) andunfamiliar (i.e.challenging)situation.4 mark C3 marks B3 marks A


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Relevantexitstandard

descriptor

A solution of known concentration of thiosulphate ions (S2O32-

) isreacted with a sample of pool water containing OCl

- ions and when

the reaction is complete the number of moles of OCl- can be

calculated.Refer to the information on the previous page in answering thefollowing questions: A Year 12 student invites some of her friends over for a pool partyduring schoolies week. The pool is regularly maintained by thestudent’s parents and so the pH level was at 7.5 and the 'freechlorine' at the recommended level of 2ppm (ie; [HOCl] + [OCl

-] =

3.84 x 10-5

M) before the students used the pool. After the party was over the student tested the water and found thatthe pH was 8.5 and the 'free chlorine' level was 0 – 0.5ppm.The student also collected a 100mL sample of the pool water tocarry out a titration to check if the chlorine level was as low asindicated by the test kit. A volume of 38mL of 5.0 x 10

-6M

thiosulphate solution was found to react with the OCl- ions in the

pool water sample.(a) Only OCl

- is shown as reacting with thiosulphate ions in

equations (1) & (2) for the titration. Explain how titrating for the OCl-

ion only, enables the amount of 'free chlorine' to be determined andthen show that the concentration of ‘free chlorine’ in the pool wateris 9.5 x 10-6 mol/L.(b) Show that the ratio of [HOCl]: [OCl

-] agrees with what is

indicated in the graph when the pH is 8.5, and then determine theconcentration of both HOCl and OCl

- that are present.

Expected response

(a) The overall equation that occurs during the titration is

422 )(64)(2)()(32)()( EqnOS O H ClOS H OCl aqlaqaqaqaq−−−−−+−

++⇒++

h OCl t ith 2S O 2 i th ti 1 2

Markingscheme

1 mark for thecombination ofequation 2 and3 into an

o erall


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( )

10681059

][105.9][

106.8

11.1

105.9][

105.9][11.1

105.9][][11.0

][11.0][

11.09

1

][

][

105.9][][

105.95.8

2

109.1

38.0100.5

,

100.5][

38.038

,

66

6

6

6

6

6

6

6

32

6

6

32

6

32

32

OCl HOCl And

M

OCltherefore

OCl

OClOCl

ngSubstituti

OCl HOCltherefore

graphthe fromOCl

HOCl

OCl HOCl Now

M is pH at ionconcentrat chlorine freetheand

OClmolesof number OS molesof Number

titrationthe for equationtheFrom

moles

x

Vol xConcOS molesof number therefore

M OS and

LmLOS of Volume

titrationactualthe In

××=

−×=

×=

×=

×=

×=+

×=

==

×=+

×=

×=

×=

×=

=

×=

==

−−

−−

−

−

−

−−

−−−

−

−

−−

−

−−−

−

−

−−

−−−

−−

Markingscheme

linking andapplication ofalgorithms andconcepts to findsolutions in acomplex andchallengingsituation

1 mark for thecalculation ofthe freechlorine

1 mark forsubstituting inequations

1 mark for thecalculation ofthe [OCl-]

1 mark for thecalculation of[HOCl]


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Question 10 (KCU A: 5 marks)In red wine production the grape skins are left in the crushing. Theskin contains compounds called anthocyanins which have

complex molecular shapes. They react with water as follows:

colourlessis form AOH red is form A

Equation H AOH O H A

aq

aqaqlaq

,

1

)(

)()()(2)(

+

+++⇒+

Sulphur dioxide is added to the process to kill any bacteria andwild yeast and SO2 enters an equilibrium with water as follows:

2)(3)()(2)(2 Equation HSO H O H SO aqaqlg−+

+⇔+

The )(3 aq HSO −

also reacts with anthocyanins to form colourless

compounds.

colourlessis form H ASOred is form A

Equation H ASO HSO A

aq

aqaqaq

3)(

)(3)(3)(

,

3

+

−+⇔+

By analysing and evaluating the information given above, explainhow the addition of SO2(g) can alter the ‘taste’ of a wine, byreferring to pH changes that might occur.Expected response

The red A+

(aq) form when added to water will eventually becomemore acidic with the production of H+ (aq) ions. pH woulddecrease. This may make the wine taste sour.

When the sulphur dioxide is added, the solution becomes evenmore acidic with the production of H

+ (aq) ions. As more H

+ (aq)

ions are produced Equations 1 and 2 will move in the reversedirection. Moving to the left will produce A

+(aq) which is red and

use up the H+ions. Some SO2(g) may stay in solution.

Some A+(aq) will react with )(3 aq HSO

−

in Equation 3 causing

l l ASO H t b d d R d l ill l

Markingscheme

reproductionandinterpretation of

l d

Question 10KCU5 marks A

1 mark forreference to

Equation 1producing H+ions

1 mark forreference toEquation 21 mark forreference tothe interaction

f b th


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The following table shows the colours of some indicators inacidic, alkaline and neutral solutions.

Indicator

Colour in

acidicsolution

Colour in

alkalinesolution

Colour in

neutralsolution

RB red blue purple

VR violet red red

RY red yellow orange

YB yellow blue blue

CR colourless red colourless

CB colourless blue colourless

Question 11 (IP C: 1 mark) The leaves of certain cabbage plants are purple when boiled inpure water, red when “pickled” in vinegar and blue or green

when boiled in water containing baking soda (an alkalinesolution). Which one of the following statements could explainthe observed effects?

The cabbage leaves contain A. RBB. VRC. RYD. YB

Q ti 12 (IP C 1 k)

Markingscheme

analysis ofsecondary data toidentify obviouspatterns and trends

Question 10IP C 1 mark

1 mark forthe correctresponse

Question 11IP C 1 mark

1 mark forthe correctresponse


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Question 13 (IP A: 5 marks)

The following figure shows the curves for weak acid-strong base

titrations for the titration of 50.0mL of 0.1.M HA of varyingconcentrations with 0.100M NaOH. The pKa values for HA are (a) 1,(b) 3, (c) 5, (d) 7, (e) 9, and (f) 11 respectively.

Systematically analyse the data to identify the relationships and/ortrends of the curves and relate this to the pKa values of the acids.Relate acid (f) to another acid in your response.

Markingscheme

Question 13IP A 5 marks

http://chemwiki.ucdavis.edu/@api/deki/files/12528/Figure9.10.jpghttp://chemwiki.ucdavis.edu/@api/deki/files/12528/Figure9.10.jpg


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Item-specific marking scheme

Standard Marks Description of response

A 5 Systematic analysis of secondary data to identify relationship betweenpatterns, e.g.

• The equivalence point occurs at different pHs for the various strengthacids.

• The weaker the acid the higher the pH at the equivalence point.

• The equivalence point for the acid solution with pKa=11 is not detectableand would not be seen by an indicator.

B 4 Analysing data to identify pattern, e.g.

• The length of the lines varies at the equivalence point i.e. at the pointwhen 50mL of NaOH is added.

• This is related to the strength of the acid.

C 3 Analysing data to identify an obvious pattern, e.g.

• Graphs have similar slopes.

• Equivalence point occurs when 50 mL of NaOH has been added..

D 2 Identifying obvious patterns, e.g.

• Graphs have the same general shape.

E 1 R di d t

Note: This shows an expected A standard response. Marks will also be awarded for alternative responsesusing valid analytical approaches. Responses will be marked using the table below. It is assumed that anyresponse will contain information recorded in the table for the standard below it.


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Question 14 (IP A: 5 marks)

The following diagrams and information relate to the titration ofphosphoric acid and sodium hydroxide.

Figure 1: Relative fractions against pH

Markingscheme

Question 14IP: 5 marksThis questioninvolvesseveral itemsof data andrequires asystem by thestudent toanalyse

2n

equivalence point

3r

equivalence point


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Figure 3: Acid base indicator colours

Systematically analyse the data to identify a combination ofindicators that could be used to find the first two equivalencepoints. Discuss the possibility of using an indicator to find thethird equivalence point.

Expected response

The equivalence points occur at pHs equal to 4.67, 9.46 and11.93. Methyl orange shows a distinct colour change from red toyellow-orange at approximately 4.7. At pHs below 4.67 thisindicator would be red and changing to yellow orange atapproximately 4.7. Bromothymol blue would not be suitablebecause it is already into its violet blue tones. Bromocresol greenis just at the start of any change phase. So, methyl orange wouldbe suitable to show the first equivalence point.

Markingscheme

systematicanalysis ofsecondary data to

id tif

This question alsoprovidesopportunities forstudents to linkand applyalgorithms to findsolutions (KCU).The school haschosen not to

assess this, butconsidered it partof the analysisrequired for thedemonstration ofthe IP standards.


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Item-specific marking scheme

Standard Marks Descrip tion of response

A 5 Systematic analysis of secondary data to identify relationship between patterns,e.g.

• Correctly identifying the two indicators and the pH range.• Identifies that only two indicators are required.

• Identifies the relationship of the pH and Pka at the half equivalence point.

• Identifies the relationship of the pH and Pka at the three equivalence points.

B 4 Analysing data to identify pattern, e.g.

• Correctly identifying the two indicators and the pH range.

• Identifies that at the half equivalence point the pH equals the pKa value.

C 3 Analysing data to identify an obvious pattern, e.g.

• Identifies that two indicators are needed but may not be correct.

• Identifies that at any half equivalence point half of the acid has dissociated.

D 2 Identifying obvious patterns, e.g.

• Comments about the shape of the curve.

• Identifies one indicator correctly.

E 1 Recording data, e.g.

R d th t th i l i t t H th h

Note: This shows an expected A standard response. Marks will also be awarded for alternative responsesusing valid analytical approaches. Responses will be marked using the table below. It is assumed that anyresponse will contain information recorded in the table for the standard below it.


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Question 15 (EC A: 5 marks)

The legal limit of tin contamination in canned fruit is 250 ppm(parts per million). A pawpaw canning plant in northern

Queensland encountered a sudden, unexpected tincontamination problem. The reaction involving thecontamination has been suggested to be:

)(2)()(4)()()(3 34104 laqaqaqsaq O H Sn NH H Sn NO ++⇒++ ++++−

A comprehensive investigation was conducted on differentbatches of pawpaw of varying acidity. Pawpaws vary inacidity due to the conditions when picked. Nitrateconcentrations may vary due to the amount of nitrate in the

soil as the plant grows.Nitrate ion concentrations in the investigation were 10ppmand 40ppm. The results are summarised in the graphsbelow:

Markingscheme

Question 15 EC5 marks A


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Expected response

Nitrates are present in fruit due to the natural uptake by plants of nitrates from the soil. All nitrate compounds are soluble. Nitrates in fertilisers are absorbed by plants and soif a plant has been subjected to excessive fertilisers the concentration in the fruit willbe higher than normal. The nitrate ion is a strong oxidising agent and when present insufficient quantities will cause the corrosion of the tin can.

The equation suggests that any nitrate (NO3 -) present combines with the tin Sn(s) to

form tin ions Sn2+

(aq) in acidic solution i.e. when H+ ions are present. Even after one

week of storage the cans at 40ppm nitrate are starting to show an increase in tincontamination.

At the lower level of nitrates at 10 ppm, the limit of 250ppm of Sn2+

(aq) is not reachedfor any of the pH readings. At the lower nitrate concentration of 10ppm, the graphshows there is an initial increase in Sn

++ ions for pHs of 3.5, 3.8 and 4.2. The pH level

remains the same for pH of 4.8.

If the nitrate concentration is at 40ppm then the contamination occurs i.e. Sn2+

(aq)ions are formed. This means that there are sufficient nitrate ions and hydrogen ions tocause the reaction to move to the right. At the higher level of nitrate at 40 ppm thelimit is reached only for pH levels at 3.8 or lower.

The equation shows that as the nitrate concentration is raised the reaction is movedto the right producing more Sn

2+(aq) ions. As the reaction moves to the right, more

H+(aq) ions are used up raising the pH. The pH during the time of testing was not

recorded.

Once the can is sealed, it is imposible to adjust the pH inside the can. The pH and thenitrate level of the actual fruit needs to be taken into account before canning. Fruit of avery low initial pH e.g.3.8 or lower should not be used or could have sugar added tolessen the acidity. It is unclear whether or not the curves will continue to increase forlonger storage periods.It would seem unlikely as the curves seem to be levelling offas the H

+ ions are used up.

Further investigations could be carried out to see if the pH of the fruit changes overthe same or longer storage periods. The pH is that of the fruit when picked.

exploration ofscenarios andpossiblerecommendations

justification ofrecommendations


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Chemistry 2007



July 2014

Page 21 of 22

Marking scheme

Q u e s t i o n

Knowledge and conceptual understanding

A B C Total

reproductionandinterpretationof complex andchallengingconcepts,theories andprinciples

comparisonandexplanation ofcomplexconcepts,processes andphenomena

linking andapplication ofalgorithms,concepts,principles,theories andschema to findsolutions incomplex andchallengingsituations

reproductionandinterpretationof complex orchallengingconcepts,theories andprinciples

comparisonandexplanation ofconcepts,processes andphenomena

linking andapplication ofalgorithms,concepts,principles,theories andschema to findsolutions incomplex orchallengingsituations

reproduction ofconcepts,theories andprinciples

explanation ofsimpleprocesses andphenomena

application ofalgorithms,principles,theories andschema to findsolutions insimplesituations

1 2

2 2

3 1

4 1

5 4

6 3

7 3

8 4 4

9 10

10 5

Totals 19 9 11 39

The Marking sc heme on pages 21 and 22 shows the match of marks to standard descriptors .


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Chemistry 2007



July 2014

Page 22 of 22

Investigative processes

A B C

Question systematic analysis of secondary data toidentify relationships between patterns and

trends

analysis of secondary data to identify patterns and trends analysis of secondary data to identify obviouspatterns and trends

11 1

12 1

13 5

14 5

IP totals 10 2 12

Evaluating and concluding A B C

exploration of scenarios and possibleoutcomes with justification of conclusions

explanation of scenarios and possible outcomes withdiscussion of conclusions

description of scenarios and possible outcomes withstatements of conclusion

15 5

EC total s 5 5

snr chemistry 07 as acid equil

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