49 questions. Try each one, then open the answer to check yourself and read the explanation.
Question 1
Separation of Mixtures
A brand of ink containing cobalt (II), copper (II) and iron (II) ions can best be separated into its various components by
A.fractional crystallization
B.fractional distillation
C.sublimation
D.chromatography
Show the answer and explanation
Answer: D (chromatography)
Explanation
The coloured ions in ink move at different speeds through paper or another adsorbent, so chromatography separates them. The other methods cannot separate several dissolved coloured substances present in tiny amounts.
Sea-water is water with many different salts dissolved in it, in no fixed proportion, so it is a mixture. Sugar and sodium chloride are compounds and iron is an element.
The number of molecules of carbon (IV) oxide produced when 10.0 g of CaCO3 is treated with 0.2 dm3 of 1 M HCl in the equation CaCO3 + 2HCl → CaCl2 + H2O + CO2 is [Ca = 40, O = 16, C = 12, NA = 6.02 × 1023, H = 1, Cl = 35.5]
A.1.00 × 1023
B.6.02 × 1023
C.6.02 × 1022
D.6.02 × 1024
Show the answer and explanation
Answer: C (6.02 × 1022)
Explanation
10.0 g of CaCO3 (M = 100) is 0.1 mol, and 0.2 dm3 of 1 M HCl is 0.2 mol, exactly enough. So 0.1 mol of CO2 forms: 0.1 × 6.02 × 1023 = 6.02 × 1022 molecules.
In the reaction CaC2(s) + 2H2O(l) → Ca(OH)2(s) + C2H2(g) what is the mass of solid calcium carbide needed to produce 1000 cm3 of acetylene gas at S.T.P.? [C = 12, Ca = 40, G.M.V = 22400 cm3]
A.3.8 g
B.2.9 g
C.2.0 g
D.1.0 g
Show the answer and explanation
Answer: B (2.9 g)
Explanation
1 mol CaC2 (64 g) gives 22400 cm3 of C2H2. For 1000 cm3: 64 × 1000/22400 = 2.86 g, about 2.9 g.
If the quantity of oxygen occupying a 2.76 litre container at a pressure of 0.825 atmospheres and 300 K is reduced by one-half, what is the pressure exerted by the remaining gas?
A.1.650 atm
B.0.825 atm
C.0.413 atm
D.0.275 atm
Show the answer and explanation
Answer: C (0.413 atm)
Explanation
At fixed volume and temperature, pressure is proportional to the amount of gas. Halving the gas halves the pressure: 0.825/2 = 0.413 atm.
An isotope has an atomic number of 17 and a mass number of 36. Which of the following gives the correct number of neutrons and protons in an atom of the isotope?
A.Neutrons 53, Protons 17
B.Neutrons 17, Protons 36
C.Neutrons 19, Protons 17
D.Neutrons 36, Protons 17
Show the answer and explanation
Answer: C (Neutrons 19, Protons 17)
Explanation
Protons = atomic number = 17. Neutrons = mass number − atomic number = 36 − 17 = 19.
A stream of air was successively passed through three tubes X, Y and Z containing a concentrated aqueous solution of KOH, red hot copper powder and fused calcium chloride respectively. What was the composition of the gas emanating from tube Z?
A.CO2 and the inert gases
B.N2, CO2 and the inert gases
C.N2 and the inert gases
D.water vapour, N2 and the inert gases.
Show the answer and explanation
Answer: C (N2 and the inert gases)
Explanation
KOH absorbs carbon (IV) oxide, hot copper removes oxygen and fused calcium chloride removes water vapour. Only nitrogen and the inert gases are left.
Which of the following water samples will have the highest titre value when titrated for the Ca2+ ions using soap solution?
A.Permanently hard water after boiling
B.Temporarily hard water after boiling
C.Rain water stored in a glass jar for two years
D.Permanently hard water passed through permutit.
Show the answer and explanation
Answer: A (Permanently hard water after boiling)
Explanation
Boiling does not remove permanent hardness, so that sample still has its Ca2+ ions and needs the most soap. Boiling removes temporary hardness, permutit removes permanent hardness and rain water is soft.
The solubility of Na3AsO4(H2O)12 is 38.9 g per 100 g H2O. What is the mass percentage of Na3AsO4 in the saturated solution? [As = 75, Na = 23, O = 16, H = 1]
A.87.2%
B.38.9%
C.19.1%
D.13.7%
Show the answer and explanation
Answer: D (13.7%)
Explanation
The hydrate has M = 424, of which Na3AsO4 is 208. So 38.9 g of hydrate holds 38.9 × 208/424 = 19.1 g of Na3AsO4 in 138.9 g of solution: 19.1/138.9 × 100 = 13.7%.
Which is the correct set of results for tests conducted respectively on fresh lime juice and ethanol?
A.Add crystals of NaHCO3: fresh lime juice, gas evolved; ethanol, no gas evolved
B.Test with methyl orange: fresh lime juice, turns colourless; ethanol, no change
C.Taste: fresh lime juice, bitter; ethanol, sour
D.Add a piece of sodium: fresh lime juice, no gas evolved; ethanol, hydrogen evolved
Show the answer and explanation
Answer: A (Add crystals of NaHCO3: fresh lime juice, gas evolved; ethanol, no gas evolved)
Explanation
Lime juice contains citric acid, which reacts with hydrogencarbonate to give carbon (IV) oxide; ethanol is not acidic enough to do this. Methyl orange turns red (not colourless) in acid, lime juice is sour, and sodium gives hydrogen with both liquids.
A quantity of electricity liberates 3.6 g of silver from its salt. What mass of aluminium will be liberated from its salt by the same quantity of electricity? [Al = 27, Ag = 108]
A.2.7 g
B.1.2 g
C.0.9 g
D.0.3 g
Show the answer and explanation
Answer: D (0.3 g)
Explanation
3.6 g Ag is 3.6/108 = 1/30 mol, needing 1/30 F. Al3+ needs 3 F per mole, so 1/90 mol Al forms: 27/90 = 0.3 g.
Which of the following statements is CORRECT if 1 Faraday of electricity is passed through 1 M CuSO4 solution for 1 minute?
A.The pH of the solution at the cathode decreases
B.The pH of the solution at the anode decreases
C.1 mole of Cu will be liberated at the cathode
D.60 moles of Cu will be liberated at the anode.
Show the answer and explanation
Answer: B (The pH of the solution at the anode decreases)
Explanation
At the anode OH− ions are discharged as oxygen, leaving H+ ions behind, so the solution there becomes more acidic and its pH falls. 1 F deposits only 0.5 mol Cu, at the cathode.
What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hrs. 30 mins through molten magnesium chloride? [1 Faraday = 96500 coulombs, Mg = 24]
A.1.12 g
B.2.00 g
C.2.24 g
D.4.48 g
Show the answer and explanation
Answer: C (2.24 g)
Explanation
Q = 2 × 9000 s = 18000 C. Mg2+ needs 2 × 96500 C per 24 g, so mass = 24 × 18000/193000 = 2.24 g.
Z is a solid substance which liberates carbon (IV) oxide on treatment with concentrated H2SO4. A warm solution of Z also decolorizes acidified KMnO4. The solid substance, Z, is
A.sodium hydrogen trioxocarbonate (IV)
B.methanoic acid
C.iron (II) trioxocarbonate (IV)
D.ethanedioic acid (oxalic acid).
Show the answer and explanation
Answer: D (ethanedioic acid (oxalic acid).)
Explanation
Concentrated H2SO4 dehydrates solid oxalic acid to CO and CO2, and warm oxalic acid is oxidised by KMnO4, decolourising it. Methanoic acid is a liquid and NaHCO3 is not a reducing agent.
5 g of ammonium trioxonitrate (V) on dissolution in water cooled its surrounding water and container by 1.6 kJ. What is the heat of solution of NH4NO3? [N = 14, O = 16, H = 1]
A.+51.4 kJ mol−1
B.+25.6 kJ mol−1
C.+12.9 kJ mol−1
D.−6.4 kJ mol−1
Show the answer and explanation
Answer: B (+25.6 kJ mol−1)
Explanation
M(NH4NO3) = 80, so 5 g is 0.0625 mol. Heat absorbed per mole = 1.6/0.0625 = 25.6 kJ; it cools the water, so it is endothermic (+25.6 kJ mol−1).
Tetraoxosulphate (VI) acid is prepared using the chemical reaction SO3(g) + H2O(l) → H2SO4(l). Given the heats of formation for SO3(g), H2O(l) and H2SO4(l) as −395 kJ mol−1, −286 kJ mol−1 and −811 kJ mol−1 respectively, the enthalpy change accompanying this reaction is
The times taken for iodine to be liberated in the reaction between sodium thiosulphate and hydrochloric acid at various temperature are as follows: Temp °C: 25, 35, 45 Time (seconds): 72, 36, 18 These results suggest that
A.for a 10° rise in temperature, rate of reaction is doubled
B.for a 10° rise in temperature, rate of reaction is halved
C.time taken for iodine to appear does not depend on temperature
D.for a 10° rise in temperature, rate of reaction is tripled.
Show the answer and explanation
Answer: A (for a 10° rise in temperature, rate of reaction is doubled)
Explanation
Each 10 °C rise halves the time (72 → 36 → 18 s). Rate is inversely proportional to time, so the rate doubles.
The reaction between sulphur (IV) oxide and oxygen is represented by the equilibrium reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −196 kJ. What factor would influence increased production of SO3(g)?
A.Addition of a suitable catalyst
B.Increase in the temperature of the reaction
C.Decrease in the temperature of the reaction
D.Decrease in the concentration of SO2(g).
Show the answer and explanation
Answer: C (Decrease in the temperature of the reaction)
Explanation
The forward reaction is exothermic, so lowering the temperature shifts the equilibrium to the right, giving more SO3. A catalyst only speeds up reaching equilibrium; it does not change the yield.
Hot concentrated alkali turns chlorine into chlorate(V) and chloride in the ratio 1 : 5. The chlorate ion stays dissolved, so it is (aq), not (s). Option A is the reaction with cold dilute alkali.
Magnesium ribbon was allowed to burn inside a given gas P leaving a white solid residue Q. Addition of water to Q liberated a gas which produced dense white fumes with a drop of hydrochloric acid. The gas P was
A.nitrogen
B.chlorine
C.oxygen
D.sulphur (IV) oxide
Show the answer and explanation
Answer: A (nitrogen)
Explanation
Magnesium burns in nitrogen to form magnesium nitride, Mg3N2. Water turns this into ammonia, which forms white fumes of NH4Cl with HCl.
The best treatment for a student who accidentally poured concentrated tetraoxosulphate (VI) acid on his skin in the laboratory is to wash the skin with
A.cold water
B.sodium hydroxide solution
C.iodine solution
D.sodium trioxocarbonate (IV) solution.
Show the answer and explanation
Answer: A (cold water)
Explanation
Washing at once with plenty of cold water dilutes and removes the acid quickly. Sodium hydroxide is itself corrosive and would cause more harm.
Which of the following gases can best be used for demonstrating the fountain experiment? (i) Nitrogen (ii) Ammonia (iii) Nitrogen (I) oxide (iv) Hydrogen chloride
A.(ii) and (iii)
B.(i) and (iii)
C.(ii) and (iv)
D.(ii) only.
Show the answer and explanation
Answer: C ((ii) and (iv))
Explanation
The fountain experiment needs a gas that is extremely soluble in water. Ammonia and hydrogen chloride both are; nitrogen and nitrogen (I) oxide are only slightly soluble.
C.metals below hydrogen in the electrochemical series
D.all metals.
Show the answer and explanation
Answer: B (electropositive metals)
Explanation
Corrosion is the oxidation of a metal by air and water. Electropositive (reactive) metals lose electrons easily, so they corrode; unreactive metals such as gold and platinum do not.
Inspite of the electronic configuration, 1s22s22p2, carbon is tetravalent because
A.the electrons in both 2s and 2p orbitals have equal energy
B.the electrons in both 2s and 2p orbitals are equivalent
C.both the 2s and 2p orbitals hydrodize
D.the six orbitals hybridize to four.
Show the answer and explanation
Answer: C (both the 2s and 2p orbitals hydrodize)
Explanation
One 2s electron is promoted to 2p and the 2s and three 2p orbitals mix (hybridise) into four equal sp3 orbitals, each with one unpaired electron, so carbon forms four bonds.
Which of the following compounds will give a precipitate with an aqueous ammoniacal solution of copper (I) chloride?
A.CH3CH = CHCH3
B.CH3C ≡ CCH3
C.HC ≡ C–CH2CH3
D.CH2 = CH–CH = CH2
Show the answer and explanation
Answer: C (HC ≡ C–CH2CH3)
Explanation
Only alkynes with a terminal ≡C–H (an acidic hydrogen) form a red precipitate of copper(I) acetylide. But-1-yne has one; but-2-yne and the alkenes do not.
A palm wine seller stoppered a bottle of his palm wine in his stall and after a few hours the bottle exploded. Which of the following equations represents the reaction that occurred?
A.C6H12O6 → 2C2H5OH + 2CO2(g) (enzymes)
B.C2H5OH → CH2 = CH2(g) + H2O
C.C2H5OH + dil H2SO4 → C2H5OSO2OH
D.2C6H12O6 → C12H22O13 + H2O.
Show the answer and explanation
Answer: A (C6H12O6 → 2C2H5OH + 2CO2(g) (enzymes))
Explanation
Yeast enzymes keep fermenting the sugar in palm wine to ethanol and carbon (IV) oxide. The gas builds up pressure in the closed bottle until it bursts.
When water drops are added to calcium carbide in a container and the gas produced is passed through a jet and lighted, the resultant flame is called an
A.oxyethylene flame
B.oxyhydrocarbon flame
C.oxyacetylene flame
D.oxymethane flame
Show the answer and explanation
Answer: C (oxyacetylene flame)
Explanation
Water and calcium carbide produce ethyne (acetylene). Burnt with oxygen it gives the very hot oxyacetylene flame.