Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
Fig. 1 shows a student measuring a sulphuric acid solution with a pipette before adding it to a flask containing sodium hydroxide. The student uses a suitable indicator and then records three titres from the burette. The sodium hydroxide volume is 20.0 cm3 and its concentration is 0.150 mol dm-3.
| titre number | volume of H2SO4 / cm3 |
|---|---|
| 1 | 14.9 |
| 2 | 15.1 |
| 3 | 15.0 |
H2SO4 + 2NaOH → Na2SO4 + 2H2O
(a) Record the mean titre. [1]
(b) Use the sodium hydroxide data to calculate its amount in the flask. [2]
(c) Use the equation and mean titre to calculate the concentration of sulphuric acid. [3]
(a) The titres are 14.9 cm3, 15.1 cm3 and 15.0 cm3, so their mean is:
\[\frac{14.9+15.1+15.0}{3}=15.0\text{ cm}^3\]
The mean titre is 15.0 cm3. [1 mark]
(b) Convert the sodium hydroxide volume:
\[20.0\text{ cm}^3=0.0200\text{ dm}^3\]
Use \(n=cV\):
\[n=0.150\times0.0200=0.00300\text{ mol}\]
The amount of sodium hydroxide is 0.00300 mol. [2 marks]
(c) The equation shows that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide:
\[n(\mathrm{H_2SO_4})=\frac{0.00300}{2}=0.00150\text{ mol}\]
\[15.0\text{ cm}^3=0.0150\text{ dm}^3\]
\[c=\frac{n}{V}=\frac{0.00150}{0.0150}=0.100\text{ mol dm}^{-3}\]
The sulfuric acid concentration is 0.100 mol dm-3. [3 marks]
(a) The titres are 14.9 cm3, 15.1 cm3 and 15.0 cm3, so their mean is:
\[\frac{14.9+15.1+15.0}{3}=15.0\text{ cm}^3\]
The mean titre is 15.0 cm3. [1 mark]
(b) Convert the sodium hydroxide volume:
\[20.0\text{ cm}^3=0.0200\text{ dm}^3\]
Use \(n=cV\):
\[n=0.150\times0.0200=0.00300\text{ mol}\]
The amount of sodium hydroxide is 0.00300 mol. [2 marks]
(c) The equation shows that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide:
\[n(\mathrm{H_2SO_4})=\frac{0.00300}{2}=0.00150\text{ mol}\]
\[15.0\text{ cm}^3=0.0150\text{ dm}^3\]
\[c=\frac{n}{V}=\frac{0.00150}{0.0150}=0.100\text{ mol dm}^{-3}\]
The sulfuric acid concentration is 0.100 mol dm-3. [3 marks]
Question 2 Report
This experiment is used to make sodium chloride from sodium hydroxide solution and hydrochloric acid. Fig. 1 shows a burette above a conical flask containing sodium hydroxide and an indicator. A student first carries out a rough titration. In later runs, the student adds acid slowly near the end-point, using the rough result as a guide. No excess acid or alkali can be removed by filtration because both reactants and the salt are soluble in water. The student repeats the accurate titration without indicator before evaporating the neutral solution.
(a) Give the name of the apparatus that delivers measured hydrochloric acid in Fig. 1. [1]
(b) Describe the colour change used to identify the end-point when phenolphthalein is used. [1]
(c) Use the reason why a second titration is done without indicator. [1]
(d) Draw a balanced symbol equation for sodium hydroxide reacting with hydrochloric acid. [2]
The diagram shows a gardener making calcium nitrate solution for a small hydroponic system. Dilute nitric acid is placed in a large beaker, and powdered calcium carbonate is tipped in slowly. The mixture fizzes at first. When excess solid remains, the gardener filters the mixture. This avoids putting carbonate particles into the irrigation pipes. The filtrate is a salt solution that can be concentrated carefully. The gardener notices that the reaction becomes less vigorous when the acid has all reacted.
(a) Give the name of the gas seen during the fizzing. [1]
(b) Describe a test for this gas. [2]
(c) Use Fig. 1 to give the salt made by this reaction. [1]
(d) Give one reason for adding calcium carbonate slowly. [1]
Sodium chloride preparation
(a) The apparatus delivering measured hydrochloric acid is a burette. [1]
(b) With phenolphthalein, the end-point colour change is pink to colourless. Alkali is pink with this indicator; neutralisation removes the alkaline colour. [1]
(c) The second titration is performed without indicator because indicator would contaminate the sodium chloride crystals. A pure salt solution is required. [1]
(d) NaOH + HCl → NaCl + H2O [2]
Calcium nitrate preparation
(a) The fizzing gas is carbon dioxide. [1]
(b) Bubble the gas through limewater. It turns milky or cloudy if carbon dioxide is present. [2]
(c) The salt made is calcium nitrate. [1]
(d) Add calcium carbonate slowly to prevent rapid frothing or overflow and to control the reaction safely. [1]
Sodium chloride preparation
(a) The apparatus delivering measured hydrochloric acid is a burette. [1]
(b) With phenolphthalein, the end-point colour change is pink to colourless. Alkali is pink with this indicator; neutralisation removes the alkaline colour. [1]
(c) The second titration is performed without indicator because indicator would contaminate the sodium chloride crystals. A pure salt solution is required. [1]
(d) NaOH + HCl → NaCl + H2O [2]
Calcium nitrate preparation
(a) The fizzing gas is carbon dioxide. [1]
(b) Bubble the gas through limewater. It turns milky or cloudy if carbon dioxide is present. [2]
(c) The salt made is calcium nitrate. [1]
(d) Add calcium carbonate slowly to prevent rapid frothing or overflow and to control the reaction safely. [1]
Question 3 Report
This diagram is of a water-treatment tank where sodium carbonate solution is added to acidic water. The operator has 2.50 dm3 of sodium carbonate solution at a concentration of 0.0800 mol dm-3. The tank display records the volume pumped. The carbonate reacts with acid and can form carbon dioxide gas.
(a) Use the concentration and volume to calculate the amount of sodium carbonate pumped into the tank. [2]
(b) Give the volume in cm3 shown by 2.50 dm3. [1]
(c) Describe one observation if excess acid reacts with carbonate in the tank. [2]
Fig. 1 shows a student warming a solid sample of hydrated sodium carbonate before preparing a solution. The student should not heat the compound in this method because the measured mass is needed for an accurate molar concentration. The student dissolves 14.3 g of Na2CO3.10H2O and makes 500 cm3 of solution. Its relative formula mass is 286.
(a) Use the mass and relative formula mass to calculate the amount of hydrated sodium carbonate. [2]
(b) Give the concentration of the solution in mol dm-3. [2]
(c) Describe why heating the hydrated solid before weighing would give an unreliable concentration. [1]
Sodium carbonate treatment solution
(a) Use \(n=cV\):
\[n=0.0800\times2.50=0.200\text{ mol}\]
The amount pumped into the tank is 0.200 mol of sodium carbonate. [2 marks]
(b)
\[2.50\text{ dm}^3=2500\text{ cm}^3\]
The displayed volume is 2500 cm3. [1 mark]
(c) Bubbles or effervescence would be observed because carbon dioxide gas is produced when excess acid reacts with carbonate. [2 marks]
Hydrated sodium carbonate solution
(a)
\[n=\frac{m}{M_r}=\frac{14.3}{286}=0.0500\text{ mol}\]
The amount of hydrated sodium carbonate is 0.0500 mol. [2 marks]
(b)
\[500\text{ cm}^3=0.500\text{ dm}^3\]
\[c=\frac{0.0500}{0.500}=0.100\text{ mol dm}^{-3}\]
The concentration is 0.100 mol dm-3. [2 marks]
(c) Heating can remove water of crystallisation, changing the mass and composition of the hydrated solid. The amount calculated from the original formula mass would then be unreliable. [1 mark]
Sodium carbonate treatment solution
(a) Use \(n=cV\):
\[n=0.0800\times2.50=0.200\text{ mol}\]
The amount pumped into the tank is 0.200 mol of sodium carbonate. [2 marks]
(b)
\[2.50\text{ dm}^3=2500\text{ cm}^3\]
The displayed volume is 2500 cm3. [1 mark]
(c) Bubbles or effervescence would be observed because carbon dioxide gas is produced when excess acid reacts with carbonate. [2 marks]
Hydrated sodium carbonate solution
(a)
\[n=\frac{m}{M_r}=\frac{14.3}{286}=0.0500\text{ mol}\]
The amount of hydrated sodium carbonate is 0.0500 mol. [2 marks]
(b)
\[500\text{ cm}^3=0.500\text{ dm}^3\]
\[c=\frac{0.0500}{0.500}=0.100\text{ mol dm}^{-3}\]
The concentration is 0.100 mol dm-3. [2 marks]
(c) Heating can remove water of crystallisation, changing the mass and composition of the hydrated solid. The amount calculated from the original formula mass would then be unreliable. [1 mark]
Question 4 Report
The table below shows results from a factory trial using cobalt(II) chloride solution as a catalyst in the reaction between sodium hydrogen sulfite solution and hydrogen peroxide. Fig. 1 shows the gas syringe used to collect the gas. Each trial used the same volumes and concentrations of both solutions at 25 °C.
| mass of cobalt(II) chloride / g | time to collect 30.0 cm3 gas / s |
|---|---|
| 0.00 | 180 |
| 0.10 | 92 |
| 0.20 | 55 |
| 0.30 | 53 |
(a) Describe the effect of increasing the mass of cobalt(II) chloride from 0.00 g to 0.20 g. [2]
(b) Use the results for 0.20 g of cobalt(II) chloride to calculate the mean rate of gas collection in cm3/s. [3]
(c) Give two variables that must be controlled to make this comparison valid. [2]
(d) Describe what should be done to the cobalt(II) chloride after a reaction if the factory wishes to reuse it. [2]
(a) Increasing the mass of cobalt(II) chloride from 0.00 g to 0.20 g decreases the time from 180 s to 55 s. Therefore the reaction rate increases: the catalyst speeds up the reaction. [2]
(b) mean rate = volume ÷ time
= 30.0 cm3 ÷ 55 s
= 0.545... cm3 s-1
= 0.55 cm3 s-1. [3]
(c) Keep the temperature constant and keep the volume or concentration of hydrogen peroxide constant. The volume or concentration of sodium hydrogen sulfite, and total solution volume, are also valid control variables. Any two. [2]
(d) Filter to separate the solid catalyst from the solution, then wash and dry it before reuse. [2]
(a) Increasing the mass of cobalt(II) chloride from 0.00 g to 0.20 g decreases the time from 180 s to 55 s. Therefore the reaction rate increases: the catalyst speeds up the reaction. [2]
(b) mean rate = volume ÷ time
= 30.0 cm3 ÷ 55 s
= 0.545... cm3 s-1
= 0.55 cm3 s-1. [3]
(c) Keep the temperature constant and keep the volume or concentration of hydrogen peroxide constant. The volume or concentration of sodium hydrogen sulfite, and total solution volume, are also valid control variables. Any two. [2]
(d) Filter to separate the solid catalyst from the solution, then wash and dry it before reuse. [2]
Question 5 Report
This experiment is a school demonstration of a carbonate reacting with acid. Fig. 1 shows sodium carbonate solution in a flask connected to a test tube of limewater. Dilute hydrochloric acid is injected through the bung. The limewater turns cloudy as gas reaches it. The student must keep the syringe tip below the bung to prevent acid leaking out.
(a) Give the name of the gas that turns limewater cloudy. [1]
(b) Use the substances to give the word equation for the reaction in the flask. [2]
(c) Describe the visible change in the flask while the reaction is taking place. [1]
(a) The gas that turns limewater cloudy is carbon dioxide. [1]
(b) The word equation is:
sodium carbonate + hydrochloric acid → sodium chloride + water + carbon dioxide [2]
(c) The reaction in the flask causes fizzing, or bubbles to be produced, as carbon dioxide gas forms. [1]
(a) The gas that turns limewater cloudy is carbon dioxide. [1]
(b) The word equation is:
sodium carbonate + hydrochloric acid → sodium chloride + water + carbon dioxide [2]
(c) The reaction in the flask causes fizzing, or bubbles to be produced, as carbon dioxide gas forms. [1]
Question 6 Report
Fig. 1 shows a gas-monitoring flask used by an environmental group beside a lake. A water sample is acidified so that dissolved carbonate releases carbon dioxide into a calibrated gas bag. The group wants to compare the amount of carbonate in water from two different streams.
Stream A produces 96.0 cm3 carbon dioxide and stream B produces 144 cm3, using equal volumes of water. At room conditions, molar gas volume is 24 dm3 mol-1. Assume each mole of carbonate produces one mole of carbon dioxide when acid is added.
(a) Calculate the amount, in mol, of carbon dioxide from stream A. [2]
(b) Use the information to calculate the amount, in mol, of carbonate in stream B. [2]
(c) Describe which stream contains more dissolved carbonate and use data to support your answer. [2]
The calculation uses \(n=\frac{V}{24}\) when gas volume is in dm3 and molar gas volume is \(24\text{ dm}^3\text{ mol}^{-1}\).
The calculation uses \(n=\frac{V}{24}\) when gas volume is in dm3 and molar gas volume is \(24\text{ dm}^3\text{ mol}^{-1}\).
Question 7 Report
A boat-repair company is selecting a metal for a valve body in a seawater cooling pump. Fig. 1 compares a regular row of copper atoms with atoms in brass. Brass is an alloy containing mostly copper with zinc added. The lower drawing is not to scale, but it shows that zinc atoms are a different size from copper atoms. The company wants a valve that resists wear when sand passes through the pump.
Table 1 gives data from samples made by the company.
| sample | mass of copper / g | mass of zinc / g | mean scratch width / mm |
|---|---|---|---|
| A | 90 | 10 | 0.42 |
| B | 70 | 30 | 0.31 |
| pure copper | 100 | 0 | 0.58 |
(a) Describe two differences between the particle arrangements in pure copper and brass shown in Fig. 1. [3]
(b) Use Fig. 1 to explain why brass is harder than pure copper. [3]
(c) Use Table 1 to calculate the percentage by mass of zinc in sample B. [3]
(d) What conclusion about wear resistance can be made from the scratch-width results? [2]
(e) How could the company improve this scratch test so that the comparison is more reliable? [3]
(f) Give two properties of metals that allow brass to be made into a shaped valve body. [2]
(a) Pure copper contains one type of atom [1]. Brass contains both copper and zinc atoms [1], and the atoms are different sizes, making the arrangement less regular [1].
(b) Different-sized atoms distort the metal layers [1]. The layers cannot slide past one another as easily [1], so brass is harder and more resistant to scratching [1].
(c) \[\text{total mass}=70+30=100\text{ g}\] [1]
\[\text{percentage zinc}=\frac{30}{100}\times100=30\%\] [2]
(d) Sample B has the narrowest scratch, so it is the most wear resistant [1]. Pure copper has the widest scratch, so it is the least wear resistant [1].
(e) Repeat each scratch test and calculate a mean [1]. Use the same load on the scratching point [1] and the same type and size of point [1]. Keeping scratch distance constant and using the same calibrated measuring instrument are also valid controls.
(f) Brass is malleable, so it can be shaped [1], and it has a high melting temperature [1]. Strength or conductivity are also acceptable metal properties.
(a) Pure copper contains one type of atom [1]. Brass contains both copper and zinc atoms [1], and the atoms are different sizes, making the arrangement less regular [1].
(b) Different-sized atoms distort the metal layers [1]. The layers cannot slide past one another as easily [1], so brass is harder and more resistant to scratching [1].
(c) \[\text{total mass}=70+30=100\text{ g}\] [1]
\[\text{percentage zinc}=\frac{30}{100}\times100=30\%\] [2]
(d) Sample B has the narrowest scratch, so it is the most wear resistant [1]. Pure copper has the widest scratch, so it is the least wear resistant [1].
(e) Repeat each scratch test and calculate a mean [1]. Use the same load on the scratching point [1] and the same type and size of point [1]. Keeping scratch distance constant and using the same calibrated measuring instrument are also valid controls.
(f) Brass is malleable, so it can be shaped [1], and it has a high melting temperature [1]. Strength or conductivity are also acceptable metal properties.
Question 8 Report
Fig. 1 shows limestone chips being added to hydrochloric acid in a school investigation. Carbon dioxide passes through a delivery tube into a gas syringe. The student repeats the experiment using equal masses of calcium carbonate with different chip sizes. Table 1 shows the time needed to collect the same volume of gas. The acid is in excess, so all the calcium carbonate can react. The student should wear eye protection because hydrochloric acid is corrosive.
| calcium carbonate form | time to collect 30 cm3 carbon dioxide / s |
|---|---|
| large chips | 84 |
| small chips | 35 |
(a) Give the formula of carbon dioxide. [1]
(b) Use particle collision ideas to account for the result for small chips. [2]
The diagram shows a particle model for a reaction between aqueous sodium carbonate and hydrochloric acid. In beaker A the acid is more concentrated than in beaker B, but both have the same volume. Each small circle represents an acid particle. A student measures carbon dioxide produced in the first 30 s. Table 1 gives the results. Sodium carbonate is a compound that reacts with acid to make carbon dioxide, water and a salt.
| beaker | carbon dioxide in first 30 s / cm3 |
|---|---|
| A | 42 |
| B | 18 |
(a) Describe one difference between the particle models. [1]
(b) Use Table 1 to calculate the difference in gas volume. [1]
(c) How does a more concentrated acid change the number of collisions? [1]
(d) Give one observation that shows a reaction is taking place. [1]
(e) Draw a conclusion linking concentration and rate. [1]
Calcium carbonate chips
Acid concentration
Calcium carbonate chips
Acid concentration
Question 9 Report
Fig. 1 shows an electron-shell model used by a laboratory that checks the purity of bottled argon for welding. Argon is a noble gas in Group 0. The laboratory compares it with chlorine, which is used to disinfect water, and sodium, which is a reactive metal. All three elements are in Period 3. The student is asked to use their electron arrangements to predict their behaviour.
(a) Give the number of outer-shell electrons in chlorine and argon. [2]
(b) Describe why argon is much less reactive than chlorine. [3]
(c) Use the electron arrangements to describe the ions formed when sodium reacts with chlorine. [3]
(d) Draw a dot-and-cross diagram to show the bonding in sodium chloride. Include the charges on the ions. [3]
(a) Chlorine has 7 outer-shell electrons [1] and argon has 8. [1]
(b) Argon has a full outer electron shell. [1] This arrangement is stable [1], so argon does not need to gain, lose or share electrons. [1] Chlorine has one fewer outer electron and can gain one to obtain a full shell.
(c) Sodium loses one electron [1], forming \(\mathrm{Na^+}\). [1] Chlorine gains that electron, forming \(\mathrm{Cl^-}\). [1]
(d) Sodium chloride contains oppositely charged \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions. The transferred electron is shown as a cross in chlorine's full outer shell. [3]
(a) Chlorine has 7 outer-shell electrons [1] and argon has 8. [1]
(b) Argon has a full outer electron shell. [1] This arrangement is stable [1], so argon does not need to gain, lose or share electrons. [1] Chlorine has one fewer outer electron and can gain one to obtain a full shell.
(c) Sodium loses one electron [1], forming \(\mathrm{Na^+}\). [1] Chlorine gains that electron, forming \(\mathrm{Cl^-}\). [1]
(d) Sodium chloride contains oppositely charged \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions. The transferred electron is shown as a cross in chlorine's full outer shell. [3]
Would you like to proceed with this action?