Chemistry 9701/34 — October/November 2013

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

FB 1 is $0.125\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FB 2 is an aqueous solution containing sodium hydroxide, $\text{NaOH}$ , and sodium carbonate, $\text{Na}_2\text{CO}_3$ .
bromophenol blue acid-base indicator

By carrying out titrations, you are to determine the percentage by mass of sodium carbonate in the mixture of sodium hydroxide and sodium carbonate in solution FB 2.

(a)

Titration

Fill a burette with FB 1.
Pipette $25.0\text{ cm}^3$ of FB 2 into a conical flask.
Add a few drops of bromophenol blue indicator.
Titrate the mixture in the flask with FB 1 until the blue-violet colour of the solution changes to yellow.
Perform a rough titration and record your burette readings in the space below.

The rough titre is .................... $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make certain any recorded results show the precision of your practical work.
Record in a suitable form below all of your burette readings and the volume of FB 1 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Carry out a rough titration and record the rough titre, e.g. $29.20\text{ cm}^3$ .

Then carry out at least two accurate titrations, adding FB 1 dropwise near the end-point until the blue-violet colour changes to yellow.

Record all burette readings in a table with headings and units, e.g.:

	Rough	Accurate 1	Accurate 2
final burette reading / $\text{cm}^3$	$29.20$	$28.90$	$28.95$
initial burette reading / $\text{cm}^3$	$0.00$	$0.05$	$0.00$
titre / $\text{cm}^3$	$29.20$	$28.85$	$28.95$

(These values are examples only; candidate's own readings must be recorded.) Record every burette reading to $0.05\text{ cm}^3$ . The accurate titres should be concordant, ideally within $0.10\text{ cm}^3$ .

Final answer

Candidate-dependent: record rough and at least two accurate titres, with all burette readings to 0.05 cm3 in a headed table.

Detailed explanation

Background Concept

In a titration, a solution of known concentration is added from a burette to a measured volume of the solution being analysed until an indicator signals that the reaction is complete. Here FB 1 is $0.125\text{ mol dm}^{-3}$ HCl and FB 2 contains a mixture of NaOH and Na2CO3. Bromophenol blue is blue-violet in alkaline/neutral solution and yellow in acidic solution, so its end-point is reached only after the acid has neutralised all the NaOH and converted all the carbonate to CO2. This title is therefore the total acid needed.

Understanding the Question

This part asks you to perform and record the titration data. No calculation is asked for yet. The marks are for technique and recording: a rough titre, at least two accurate titres, readings to 0.05 cm3, and a clear table with headings and units.

Approach

Fill the burette with FB 1, making sure the jet is filled. Pipette exactly $25.0\text{ cm}^3$ of FB 2 into a conical flask and add a few drops of bromophenol blue. Do a quick rough titration to find the approximate end-point, then repeat accurately, adding the acid dropwise near the end-point. Record the initial and final burette readings for each accurate titration.

Step-by-Step Reasoning

Rinse the burette with FB 1, fill it contrasting with the solution, and read the initial volume. It is better not to start at $50.00\text{ cm}^3$ ; use a more convenient reading and always read to the nearest $0.05\text{ cm}^3$ .
Use a pipette and pipette filler to transfer exactly $25.0\text{ cm}^3$ of FB 2 into the conical flask. Add a few drops of bromophenol blue.
Titrate quickly for the rough titre until the blue-violet colour just changes to yellow. Record the rough titre.
Refill the burette and repeat accurately. Near the end-point, add acid dropwise, swirling, until one drop causes a permanent yellow colour.
Perform at least two accurate titrations. For each, titre = final reading - initial reading. The accurate titres should agree within $0.10\text{ cm}^3$ where possible.
Record all results in a table with headings such as 'initial burette reading / cm3', 'final burette reading / cm3' and 'titre / cm3'. Note: the mark scheme requires the title to 0.05 cm3 but the titre itself may be in 0.05 or 0.1 increments.

Key Takeaways

Precision in burette readings and the use of concordant titres are essential in titration practicals. A clear table with headings and units also earns display marks.

Common Mistakes

Using the rough titre as one of the accurate titres.
Recording burette readings only to 1 decimal place instead of 0.05 cm3.
Starting the burette at 50.00 cm3, or having any burette reading greater than 50.00 cm3.
Recording readings without any headings or units.
Carrying out only one accurate titration, or using titres that differ by more than 0.20 cm3.

Things to Be Careful About

All burette readings must be recorded to $0.05\text{ cm}^3$ .
The rough titration is not used for calculating the mean.
Make sure the burette jet is filled before taking the initial reading, and remove any air bubbles.
The indicator colour change is blue-violet to yellow; use the first permanent colour change.

Techniques used

fill and use a burette correctlycarry out a rough and then accurate titrationsrecord burette readings to 0.05 cm3tabulate initial/final readings and titres with units

(b)

From your accurate titration results, obtain a suitable value to be used in your calculations. Show clearly how you have obtained this value.

$25.0\text{ cm}^3$ of FB 2 required .................... $\text{cm}^3$ of FB 1.

DifficultyMedium-Easy

Worked solution

Working

Select the two accurate titres that are closest together; ignore the rough titre.

For example:

\text{mean titre} = \frac{28.85 + 28.95}{2} = 28.90\text{ cm}^3

Show your selection by ticking or circling the two titres used.

Answer

$28.90\text{ cm}^3$ (example; candidate-dependent using your own concordant titres).

Final answer

28.90 cm3 (example; candidate-dependent)

Detailed explanation

Background Concept

A mean titre is the average of two or more accurate titrations that agree closely. In CIE practical work, the two chosen accurate titres should normally be within $0.10\text{ cm}^3$ of each other, and the mean is calculated to 2 decimal places.

Understanding the Question

You must choose a single suitable titre value from your accurate results to use in all later calculations. The mark is not for the calculation itself but for showing clearly which values you averaged.

Approach

Look at your accurate titres, ignore any labelled 'rough', and choose two that are very close. Calculate their mean. If their total spread is more than $0.20\text{ cm}^3$ , you would not be able to award the mean-titre mark; ideally the chosen titres differ by no more than $0.10\text{ cm}^3$ .

Step-by-Step Reasoning

List your accurate titres, e.g. $28.85$ , $28.90$ , $28.95\text{ cm}^3$ .
Choose the two closest, e.g. $28.85$ and $28.95\text{ cm}^3$ ; their spread is $0.10\text{ cm}^3$ .
Average them: $(28.85 + 28.95)/2 = 28.90\text{ cm}^3$ .
Check significant figures: the mean should be quoted to 2 decimal places unless a special case applies (e.g. a value ending in 0.025 or 0.075 may be quoted to 3 decimal places).

Key Takeaways

A correct mean titre requires concordant titres, clear selection, and proper rounding. It is the foundation for every later mole calculation.

Common Mistakes

Including the rough titre in the average.
Averaging titres whose spread exceeds $0.20\text{ cm}^3$ .
Giving the mean to only 1 decimal place when the readings support 2 decimal places.
Not showing which titres were used, so the marker cannot see the selection.

Things to Be Careful About

If all the burette readings are integers, the mean-titre mark may not be awarded.
Do not use $0.05\text{ cm}^3$ precision reading rule for the mean; the mean is rounded to 2 decimal places.
The chosen titres must be from accurate titrations, not the rough one.

Techniques used

select concordant accurate titrescalculate a mean titreround the mean to the appropriate number of decimal places

(c)

Calculations

When the titrations were repeated using phenolphthalein as the indicator, $25.0\text{ cm}^3$ of FB 2 required $23.25\text{ cm}^3$ of FB 1.

The following explains why different results are obtained using two different indicators.

When phenolphthalein is used as the indicator, the following reactions have taken place at the end-point of the titration.
1. $\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$
2. $\text{Na}_2\text{CO}_3\text{(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{NaHCO}_3\text{(aq)}$
When bromophenol blue is used as the indicator in (a), the following reactions have taken place at the end-point of the titration.
1. $\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$
2. $\text{Na}_2\text{CO}_3\text{(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{NaHCO}_3\text{(aq)}$
3. $\text{NaHCO}_3\text{(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$

Show your working and use appropriate significant figures in the final answer to all steps of your calculations.

(i)

Calculate the number of moles of hydrochloric acid in the volume of FB 1 calculated in (b).

moles of $\text{HCl}$ in volume in (b) = .................... mol

DifficultyEasy

Worked solution

Working

Using the mean titre from (b), $V = 28.90\text{ cm}^3$ (representative value).

n(\text{HCl}) = \frac{0.125 \times 28.90}{1000} = 0.0036125\text{ mol}

= 0.00361\text{ mol} \text{ (3 sf)}

Answer

$0.00361$ mol (using representative $V = 28.90\text{ cm}^3$ ).

Final answer

0.00361 mol (using V = 28.90 cm3)

Detailed explanation

Background Concept

The number of moles of a solute is given by:

n = c \times V

where $c$ is concentration in $\text{mol dm}^{-3}$ and $V$ is volume in $\text{dm}^3$ . Burette readings are usually in $\text{cm}^3$ , so they must be divided by 1000 before substitution.

Understanding the Question

Part (i) asks for the moles of HCl in the volume of FB 1 used in the mean titre from part (b). Since FB 1 is $0.125\text{ mol dm}^{-3}$ HCl, the calculation is direct.

Approach

Substitute the mean titre into $n = cV$ , dividing the volume by 1000 to convert $\text{cm}^3$ to $\text{dm}^3$ .

Step-by-Step Reasoning

Write the concentration: $c = 0.125\text{ mol dm}^{-3}$ .
Use the mean titre, e.g. $V = 28.90\text{ cm}^3 = 0.02890\text{ dm}^3$ .
Calculate: $n = 0.125 \times 0.02890 = 0.0036125\text{ mol}$ .
Give the answer to 3 or 4 significant figures: $0.00361\text{ mol}$ .

Key Takeaways

Always convert $\text{cm}^3$ to $\text{dm}^3$ in mole calculations. Use unrounded intermediate values when carrying results into later parts.

Common Mistakes

Forgetting to divide by 1000.
Using the rough titre rather than the mean accurate titre.
Rounding too early and producing a slightly different later answer.

Things to Be Careful About

The answer must have the unit mol.
Keep the unrounded value, e.g. $0.0036125$ , for later subtraction and multiplication.

Techniques used

convert cm3 to dm3calculate moles from concentration and volume

(ii)

Calculate the number of moles of hydrochloric acid in $23.25\text{ cm}^3$ of FB 1.

moles of $\text{HCl}$ in $23.25\text{ cm}^3$ = .................... mol

DifficultyEasy

Worked solution

Working

n(\text{HCl}) = \frac{0.125 \times 23.25}{1000} = 0.00290625\text{ mol}

= 0.002906\text{ mol} \text{ (4 sf)}

Answer

$0.002906$ mol

Final answer

0.002906 mol

Detailed explanation

Background Concept

This is the same mole calculation as in part (i), but using the fixed phenolphthalein titre, $23.25\text{ cm}^3$ , given in the question.

Understanding the Question

The question wants the moles of HCl in exactly $23.25\text{ cm}^3$ of FB 1. This titre corresponds to the acid consumed when phenolphthalein is used, i.e. neutralisation of the NaOH and conversion of carbonate only to hydrogencarbonate.

Approach

Use $n = cV$ with $V = 23.25\text{ cm}^3 / 1000$ .

Step-by-Step Reasoning

$V = 23.25\text{ cm}^3 = 0.02325\text{ dm}^3$ .
$n = 0.125 \times 0.02325 = 0.00290625\text{ mol}$ .
Quote to 3 or 4 significant figures: $0.002906\text{ mol}$ or $0.00291\text{ mol}$ .

Key Takeaways

The same formula is reused throughout the calculation; keeping unrounded values is important for accuracy.

Common Mistakes

Dividing by 100 incorrectly.
Using the mean titre from part (b) instead of the fixed $23.25\text{ cm}^3$ .

Things to Be Careful About

The mark scheme accepts $0.002906$ or $0.00291$ . Use the unrounded value in the next subtraction.

Techniques used

calculate moles from concentration and volumeconvert cm3 to dm3

(iii)

Use the following formula to calculate the number of moles of hydrochloric acid that react with the $\text{Na}_2\text{CO}_3$ in the titration using phenolphthalein indicator.

moles $\text{HCl}$ = answer (i) $-$ answer (ii) = .................... mol

DifficultyMedium-Easy

Worked solution

Working

Using the unrounded values from (i) and (ii):

n(\text{HCl}) = 0.0036125 - 0.00290625 = 0.00070625\text{ mol}

= 0.000706\text{ mol} \text{ (3 sf)}

Answer

$0.000706$ mol

Final answer

0.000706 mol

Detailed explanation

Background Concept

With bromophenol blue the acidic end-point means that NaOH is neutralised and Na2CO3 is converted all the way to CO2. With phenolphthalein the end-point occurs after Na2CO3 has been converted only to NaHCO3. Therefore the difference between the two acid volumes corresponds to the HCl needed for the second proton of carbonate, which is one mole of HCl per mole of Na2CO3.

Understanding the Question

The question gives the direct formula: moles HCl = answer (i) - answer (ii). This subtraction isolates the acid that reacted with the hydrogencarbonate stage.

Approach

Use the unrounded values from (i) and (ii) to avoid rounding errors, subtract, and record the result.

Step-by-Step Reasoning

Total acid used with bromophenol blue: $0.0036125\text{ mol}$ .
Acid used with phenolphthalein: $0.00290625\text{ mol}$ .
Difference: $0.0036125 - 0.00290625 = 0.00070625\text{ mol}$ .
This value is also the number of moles of Na2CO3 in the $25.0\text{ cm}^3$ sample.

Key Takeaways

The difference in titre between the two indicators is a stoichiometric measure of carbonate. Understanding why the indicators give different end-points is the key idea.

Common Mistakes

Subtracting rounded values, e.g. $0.00361 - 0.00291$ , giving a less accurate result.
Not realising that the answer also represents moles of Na2CO3.
If answer (i) were smaller than answer (ii), the difference would be negative; this would indicate an error.

Things to Be Careful About

Keep all decimal places until the final subtraction.
The mark scheme allows this answer to be given to any sensible number of significant figures.

Techniques used

subtract moles of acidinterpret the two-indicator titration difference

(iv)

Use your answer to (iii) to calculate the mass of sodium carbonate present in $25.0\text{ cm}^3$ of FB 2.
$[A_r: \text{C}, 12.0; \text{O}, 16.0; \text{Na}, 23.0]$

mass of $\text{Na}_2\text{CO}_3$ in $25.0\text{ cm}^3$ FB 2 = .................... g

DifficultyMedium-Easy

Worked solution

Working

From (iii), $n(\text{Na}_2\text{CO}_3) = 0.00070625\text{ mol}$ .

M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0

m(\text{Na}_2\text{CO}_3) = 0.00070625 \times 106.0 = 0.0748625\text{ g}

= 0.0749\text{ g} \text{ (3 sf)}

Answer

$0.0749$ g

Final answer

0.0749 g

Detailed explanation

Background Concept

The mass of a substance is related to its amount by:

m = n \times M_r

For sodium carbonate, $M_r = 2(23.0) + 12.0 + 3(16.0) = 106.0$ .

Understanding the Question

Part (iii) gave the moles of HCl corresponding to the second proton of carbonate, which is equal to the moles of Na2CO3 in the $25.0\text{ cm}^3$ sample. This part asks for the mass of that carbonate.

Approach

Use $n(\text{Na}_2\text{CO}_3)$ directly from (iii) and multiply by its molar mass.

Step-by-Step Reasoning

$n(\text{Na}_2\text{CO}_3) = 0.00070625\text{ mol}$ .
$M_r = 106.0$ .
$m = 0.00070625 \times 106.0 = 0.0748625\text{ g}$ .
Quote to 3 significant figures: $0.0749\text{ g}$ .

Key Takeaways

If a stoichiometric relation is 1:1, moles of one species can be used directly for another. Always use the unrounded mole value.

Common Mistakes

Using $M_r = 53$ instead of $106$ , forgetting that the formula has two sodium atoms.
Using the rounded $0.000706\text{ mol}$ , giving $0.0748\text{ g}$ ; this is still usually acceptable, but unrounded is better.

Things to Be Careful About

The unit is grams, not moles.
The mark scheme requires answers to 3 or 4 significant figures for this step.

Techniques used

calculate relative molecular massconvert moles to mass

(v)

The overall equation for the reaction of $\text{Na}_2\text{CO}_3$ with $\text{HCl}$ when bromophenol blue is used as indicator is given below.

\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

Calculate the number of moles of $\text{HCl}$ that reacted with the $\text{Na}_2\text{CO}_3$ in the above equation in $25.0\text{ cm}^3$ of FB 2.

moles of $\text{HCl}$ = .................... mol

DifficultyEasy

Worked solution

Working

From (iii), $n(\text{Na}_2\text{CO}_3) = 0.00070625\text{ mol}$ .

The overall equation shows $\text{Na}_2\text{CO}_3 : \text{HCl} = 1 : 2$ .

n(\text{HCl}) = 2 \times 0.00070625 = 0.0014125\text{ mol}

= 0.00141\text{ mol} \text{ (3 sf)}

Answer

$0.00141$ mol

Final answer

0.00141 mol

Detailed explanation

Background Concept

The overall reaction when carbonate is fully neutralised is:

\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

So each mole of Na2CO3 uses 2 moles of HCl.

Understanding the Question

Part (iii) gave the moles of Na2CO3 in the sample. Part (v) asks for the moles of HCl that react with this carbonate in the overall equation, i.e. after both protons of carbonate have been neutralised.

Approach

Multiply the moles of Na2CO3 by 2, using the stoichiometric ratio from the balanced equation.

Step-by-Step Reasoning

$n(\text{Na}_2\text{CO}_3) = 0.00070625\text{ mol}$ .
$n(\text{HCl})$ for carbonate $= 2 \times 0.00070625 = 0.0014125\text{ mol}$ .
Quote to 3 significant figures: $0.00141\text{ mol}$ .

Key Takeaways

The balanced equation is the source of the 1:2 ratio. The moles of HCl used by carbonate are two times the moles of carbonate.

Common Mistakes

Forgetting to multiply by 2.
Multiplying the mass of Na2CO3 by 2 instead of using the moles.

Things to Be Careful About

Use the mole value from (iii) for this step.
Keep unrounded values for use in part (vi).

Techniques used

apply stoichiometric ratio from the overall equationmultiply moles by 2

(vi)

Use your answers to (i) and (v) to calculate the mass of sodium hydroxide in $25.0\text{ cm}^3$ of FB 2.
$[A_r: \text{H}, 1.0; \text{O}, 16.0; \text{Na}, 23.0]$

mass of $\text{NaOH}$ = .................... g

DifficultyMedium-Easy

Worked solution

Working

The total HCl used with bromophenol blue, (i), reacts with all the NaOH and all the Na2CO3. The HCl used by carbonate is (v), so:

n(\text{NaOH}) = 0.0036125 - 0.0014125 = 0.0022000\text{ mol}

M_r(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0

m(\text{NaOH}) = 0.0022000 \times 40.0 = 0.08800\text{ g}

= 0.0880\text{ g} \text{ (3 sf)}

Answer

$0.0880$ g

Final answer

0.0880 g

Detailed explanation

Background Concept

The bromophenol blue titre measures total acid needed to neutralise NaOH and to convert all carbonate to CO2. Thus the total moles of HCl in (i) equals moles of NaOH plus 2 times moles of Na2CO3. Part (v) gave the second term, so subtraction leaves the moles of NaOH.

Understanding the Question

Use answer (i) and answer (v) to calculate the mass of NaOH in $25.0\text{ cm}^3$ of FB 2.

Approach

Subtract the HCl used by carbonate from the total HCl to find the HCl used by NaOH. Because NaOH and HCl react 1:1, this is also the moles of NaOH. Then multiply by $M_r = 40.0$ .

Step-by-Step Reasoning

Total HCl, (i): $0.0036125\text{ mol}$ .
HCl used by carbonate, (v): $0.0014125\text{ mol}$ .
$n(\text{NaOH}) = 0.0036125 - 0.0014125 = 0.0022000\text{ mol}$ .
$M_r(\text{NaOH}) = 23 + 16 + 1 = 40.0$ .
$m(\text{NaOH}) = 0.0022000 \times 40.0 = 0.08800 = 0.0880\text{ g}$ .

Key Takeaways

A difference method can isolate the amount of one component from the total. The 1:1 NaOH : HCl stoichiometry is essential.

Common Mistakes

Subtracting (v) from the wrong value, e.g. from (ii).
Using $M_r(\text{NaOH}) = 40$ but then confusing units.
Forgetting that (i) is total acid, not acid for NaOH only.

Things to Be Careful About

The final answer should be in grams.
Quote to 3 or 4 significant figures; $0.0880\text{ g}$ is correct for 3 significant figures.

Techniques used

deduce moles of NaOH by differencecalculate mass of NaOH from moles and Mr

(vii)

Calculate the percentage by mass of sodium carbonate in the mixture of sodium hydroxide and sodium carbonate in FB 2.

FB 2 contains .................... % by mass $\text{Na}_2\text{CO}_3$

DifficultyMedium-Easy

Worked solution

Working

\% \text{Na}_2\text{CO}_3 = \frac{m(\text{Na}_2\text{CO}_3)}{m(\text{Na}_2\text{CO}_3) + m(\text{NaOH})} \times 100

Using the unrounded masses:

= \frac{0.0748625}{0.0748625 + 0.08800} \times 100 = 45.97\%

= 46.0\% \text{ (3 sf)}

Answer

$46.0\%$

Final answer

46.0%

Detailed explanation

Background Concept

Percentage by mass is the mass of one component divided by the total mass of the mixture, multiplied by 100. Here the mixture is only Na2CO3 and NaOH, so the denominator is the sum of the two masses.

Understanding the Question

Use the mass of Na2CO3 from (iv) and the mass of NaOH from (vi) to find what percentage of the total solute mass is sodium carbonate.

Approach

Add the two masses to get the total mass, then calculate the fraction of Na2CO3 and convert to a percentage.

Step-by-Step Reasoning

$m(\text{Na}_2\text{CO}_3) = 0.0748625\text{ g}$ .
$m(\text{NaOH}) = 0.08800\text{ g}$ .
Total mass $= 0.0748625 + 0.08800 = 0.1628625\text{ g}$ .
$\% = (0.0748625 / 0.1628625) \times 100 = 45.97\%$ .
Quote to 3 significant figures: $46.0\%$ .

Key Takeaways

Percentage by mass uses masses, not moles. It is a direct ratio of the component mass to the total mass.

Common Mistakes

Using mole ratio instead of mass ratio.
Forgetting to multiply by 100.
Using only the mass of Na2CO3 in the denominator.

Things to Be Careful About

Ensure both masses are in the same unit (grams).
If rounded masses are used, the percentage should still be about 46.0%; use unrounded values for best precision.
The mark scheme expects 3 or 4 significant figures in the final answer.

Techniques used

calculate percentage by masscombine masses of components

The rest of this paper

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Q2Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation11M
Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation15M

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