48 questions. Try each one, then open the answer to check yourself and read the explanation.
Question 1
Separation of Mixtures
Which of the following can be obtained by fractional distillation?
A.Nitrogen from liquid air
B.Sodium chloride from sea water
C.Iodine from a solution of iodine in carbon tetrachloride
D.Sulphur from a solution of sulphur in carbon disulphide.
Show the answer and explanation
Answer: A (Nitrogen from liquid air)
Explanation
Liquid air is a mixture of liquids with close boiling points (nitrogen −196°C, oxygen −183°C), which is exactly what fractional distillation separates.
Which of the following are mixtures? i. Petroleum ii. Rubber latex. iii. Vulcanizer's solution iv. Carbon (II) sulphide
A.i, ii and iii
B.i, ii and iv
C.i and ii only
D.i and iv.
Show the answer and explanation
Answer: A (i, ii and iii)
Explanation
Petroleum, rubber latex and vulcanizer's solution (rubber dissolved in a solvent) are all mixtures. A carbon sulphide is a compound with a fixed formula.
In two separate experiments, 0.36 g and 0.71 g of chlorine combined with a metal X to give Y and Z respectively. An analysis showed that Y and Z contain 0.20 g and 0.40 g of X respectively. The data above represents the law of
A.multiple proportion
B.conservation of mass
C.constant composition
D.reciprocal proportion
Show the answer and explanation
Answer: C (constant composition)
Explanation
1 g of chlorine combines with about 0.56 g of X in both samples. The same ratio by mass each time shows the law of constant composition.
Calculate the volume of carbon (IV) oxide measured at s.t.p, produced when 1 kg of potassium hydrogen trioxocarbonate (IV) is totally decomposed by heat. [G.M.V. at s.t.p = 22.4 dm3, K = 39, O = 16, C = 12, H = 1].
A.28 dm3
B.56 dm3
C.112 dm3
D.196 dm3
Show the answer and explanation
Answer: C (112 dm3)
Explanation
2KHCO3 → K2CO3 + H2O + CO2: 200 g gives 22.4 dm3 of CO2, so 1000 g gives 5 × 22.4 = 112 dm3.
A sample of a gas exerts a pressure of 8.2 atm when confined in a 2.93 dm3 container at 20°C. The number of moles of gas in the sample is [R = 0.082 litre atm / deg mole].
Atoms of element X (with 2 electrons in the outer shell) combine with atoms of Y (with 7 electrons in the outer shell). Which of the following statements is FALSE? The compound formed
A.has formula XY
B.is likely to be ionic
C.contains X2+ ions
D.contains Y− ions.
Show the answer and explanation
Answer: A (has formula XY)
Explanation
X loses 2 electrons to form X2+ and each Y gains 1 to form Y−, so the formula is XY2, not XY.
The ions X− and Y+ are isoelectronic, each containing a total of 10 electrons. How many protons are in the nuclei of the neutral atoms of X and Y respectively?
A.10 and 10
B.9 and 9
C.11 and 9
D.9 and 11
Show the answer and explanation
Answer: D (9 and 11)
Explanation
X− has one more electron than protons, so X has 9 protons. Y+ has one fewer electron than protons, so Y has 11.
A quantity of air was passed through a weighed amount of alkaline pyrogallol. An increase in the weight of the pyrogallol would result from the absorption of
A.nitrogen
B.neon
C.argon
D.oxygen
Show the answer and explanation
Answer: D (oxygen)
Explanation
Alkaline pyrogallol absorbs oxygen, which is why it is used to find the percentage of oxygen in air.
The solubility of copper (II) tetraoxosulphate (VI) is 75 g in 100 g of water at 100°C and 25 g in 100 g of water at 30°C. What mass of the salt would crystallize, if 50 g of copper (II) tetraoxosulphate (VI) solution saturated at 100°C were cooled to 30°C?
A.57.5 g
B.42.9 g
C.28.6 g
D.14.3 g
Show the answer and explanation
Answer: D (14.3 g)
Explanation
175 g of saturated solution at 100°C holds 75 g of salt, and 50 g of solution is 2/7 of that. Cooling 175 g of solution drops 50 g of salt, so 50 g drops 50 × 50/175 = 14.3 g.
A sample of temporary hard water can be prepared in the laboratory by
A.dissolving calcium chloride in distilled water
B.saturating lime water with carbon (IV) oxide
C.saturating distilled water with calcium hydroxide
D.dissolving sodium hydrogen trioxocarbonate (IV) in some distilled water.
Show the answer and explanation
Answer: B (saturating lime water with carbon (IV) oxide)
Explanation
Excess CO2 in lime water forms calcium hydrogen trioxocarbonate(IV), Ca(HCO3)2, which causes temporary hardness. Calcium chloride gives permanent hardness.
50 cm3 of sulphur (IV) oxide, 800 cm3 of ammonia, 450 cm3 of hydrogen chloride, 1.0cm3 of carbon (IV) oxide will respectively saturate 1.0 cm3 of water at 15°C. Which of the following is suitable for demonstrating the fountain experiment?
A.Sulphur (IV) oxide and hydrogen chloride.
B.Carbon (IV) oxide and ammonia
C.Ammonia and hydrogen chloride
D.Carbon (IV) oxide and sulphur (IV) oxide.
Show the answer and explanation
Answer: C (Ammonia and hydrogen chloride)
Explanation
The fountain experiment needs a gas that dissolves very quickly and in large amounts. Ammonia (800 volumes) and hydrogen chloride (450 volumes) are the most soluble.
A simple cell (figure above): a zinc plate and a copper plate dip into substance X, connected to a bulb. Which of the following substances could be satisfactorily used as X in the above figure?
A.Ammonia and potassium hydroxide
B.Potassium hydroxide and sodium chloride.
C.Ammonia and ethanoic acid
D.Ethanoic acid and sodium chloride.
Show the answer and explanation
Answer: B (Potassium hydroxide and sodium chloride.)
Explanation
X must be an electrolyte that conducts well. Potassium hydroxide and sodium chloride are both strong electrolytes; ammonia and ethanoic acid are only weak electrolytes.
What volume of CO2 at s.t.p would be obtained by reacting 10 cm3 of 0.1 M solution of anhydrous sodium trioxocarbonate (IV) with excess acid? [G.M.V at s.t.p = 22.4 dm3]
If a current of 1.5 A is passed for 4.00 hours through a molten tin salt and 13.3 g of tin is deposited, what is the oxidation state of the metal in the salt? [Sn = 118.7, F = 96500 C mol−1].
A.1
B.2
C.3
D.4
Show the answer and explanation
Answer: B (2)
Explanation
Charge = 1.5 × 4 × 3600 = 21 600 C = 0.224 F. Moles of tin = 13.3 ÷ 118.7 = 0.112. Faradays per mole = 0.224 ÷ 0.112 = 2, so the tin is +2.
1.1 g of CaCl2 dissolved in 50 cm3 of water caused a rise in temperature of 3.4°C. The heat of reaction, ΔH for CaCl2 in kJ per mole is [Ca = 40, Cl = 35.5, specific heat of water is 4.18 JK−1].
A.−71.1
B.−4.18
C.+71.1
D.+111.0
Show the answer and explanation
Answer: A (−71.1)
Explanation
Heat given out = 50 × 4.18 × 3.4 ≈ 711 J. Moles of CaCl2 = 1.1 ÷ 111 ≈ 0.01, so ΔH ≈ −0.711 ÷ 0.01 ≈ −71 kJ/mol. It is negative because the temperature rose.
NO(g) + CO(g) ⇌ ½N2(g) + CO2(g) ΔH = −89.3 kJ. What conditions would favour maximum conversion of nitrogen (II) oxide and carbon (II) oxide in the reaction above?
A.Low temperature and high pressure
B.High temperature and low pressure
C.High temperature and high pressure
D.Low temperature and low pressure.
Show the answer and explanation
Answer: A (Low temperature and high pressure)
Explanation
The forward reaction is exothermic, so low temperature favours it. It goes from 2 moles of gas to 1.5 moles, so high pressure also pushes it forward.
Initial concentration of NO in moles: 0.001, initial rate (moles / sec): 3.0 × 10−5 Initial concentration of NO in moles: 0.002, initial rate (moles / sec): 1.2 × 10−4 The data in the table above shows the rate of reaction of nitrogen (II) oxide with chlorine at 25°C. It can be concluded that doubling the initial concentration of NO increases the rate of reaction by a factor of
A.two
B.three
C.four
D.five
Show the answer and explanation
Answer: C (four)
Explanation
1.2 × 10−4 ÷ 3.0 × 10−5 = 4, so the rate becomes four times faster.
Which of the following salts can be melted without decomposition?
A.Na2CO3
B.CaCO3
C.MgCO3
D.ZnCO3.
Show the answer and explanation
Answer: A (Na2CO3)
Explanation
Sodium trioxocarbonate(IV) is thermally stable and melts without breaking down. The carbonates of calcium, magnesium and zinc decompose to the oxide and CO2 on heating.
Which of the following metals can be prepared in samples by the thermal decomposition of their trioxonitrate (V) salts?
A.Copper and mercury
B.Silver and copper
C.Mercury and silver
D.Magnesium and mercury
Show the answer and explanation
Answer: C (Mercury and silver)
Explanation
Nitrates of metals low in the activity series, such as mercury and silver, decompose all the way to the metal. Copper and magnesium nitrates leave the oxide.
Which of the following compounds can exist as geometric isomers?
A.2-methylbut-2-ene.
B.But-2-ene
C.But-1-ene
D.CI—C—Br with I above and I below the carbon
Show the answer and explanation
Answer: B (But-2-ene)
Explanation
Geometric (cis–trans) isomers need two different groups on each carbon of the C=C. But-2-ene has CH3 and H on each carbon; the others have two identical groups on one carbon or no C=C.
A certain liquid has a high boiling point. It is viscous, non-toxic, miscible with water being very hygroscopic. This liquid is most likely to be
A.CH3CH2CH2OH
B.CH3CHOHCH3
C.CH3CH2CHOHCH3
D.CH2OHCHOHCH2OH
Show the answer and explanation
Answer: D (CH2OHCHOHCH2OH)
Explanation
Glycerol (propane-1,2,3-triol) has three –OH groups, giving strong hydrogen bonding. That makes it viscous, high-boiling, hygroscopic and fully miscible with water.