paper 12 chemistry answers section -i · 1 paper 12 chemistry answers section -i answer 1 a)...
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Paper 12 Chemistry
Answers Section -I
Answer 1 a) Select the correct answer: 1. Iodine. 2. Oxidation. 3. Magnesium. 4. Sulphur (S). 5. Sunlight. b) Name the following: 1. Hydrogen. 2. Fluorine, chlorine. 3. Copper & Aluminium. 4. Element A (having electronic configuration 2,8,1,). 5. Sodium is more reactive than hydrogen. Copper is less reactive than hydrogen. c) Balancing equations: 1. C + H2O o1000 C⎯⎯⎯→ CO + H2
_ Δ . water gas
2KH (potassium hydride). 2. 2K + H2 Δ⎯⎯→
3. 3Ca + N2 → Ca3 N2 (calcium nitride). 4. 2S + C Δ⎯⎯→ CS2 (carbon disulphide). 5. 2 Na + Cl2→ 2NaCl (sodium chloride). d) Change the following statements by replacing the underlined words: 1. Oxidation reaction is the removal of Hydrogen. 2. Promoters. 3. Neutrons. 4. Law of triads. 5. Sulphurous acid. e) Mention the terms used for each of the following : 1. Boiling point. 2. Boyle’s law. 3. Radicals. 4. Modern Periodic law. 5. Inner transition elements. f) Pick the ODD one out : 1. Cu3+. 2. Ionisation. 3. Electrons. 4. Combustible.
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g) 1. What do you observe when: a) Heated iron is introduced in a jar of Cl2 it combines to form reddish brown Iron III
chloride 2Fe + 3Cl 2 → 2FeCl 3
b)On addition of water the quick lime absorbs, swells, crumbles, the water is converted to steam since reaction is exothermic.
2. Give reasons for the following: a) Carbon burns in a limited supply of air forming carbon monoxide. Qb) Due to suffocation ie., want of oxygen.
3. Complete and balance the following equations:
a) PbO2 + 4HCl → PbCl2 + 2H2O + Cl2. b) 3CuO + 2 NH 3→ 3Cu + 3H2O +N2.
h) 1. Write chemical formula of the following: a) CaCN2 + C. b) CaSO4. 2. Write chemical names of the following compounds:
a) Potassium hypo-chlorite. b) Potassium per-chlorate.
3. Molecular weight of Ca(OH)2 = 40×1+16×2+1×2 = 74. i) Hydrogen. ii) 2H2O + O2 →2H2O.
Section II
Answer 2 a) 2, 8, 7, 18, 3, 5.
b) Property
Rhombic Sulphur
Plastic Sulphur
Colour
Pale yellow
Dark brown.
Nature
Crystalline
Amorphous.
Crystal Shape
Octahedral
No definite shape.
Solubility
Soluble in CS2
Insoluble in CS2.
Answer 3 a) Because it gives a fizz to the effervescent drink & on.
1. Opening the bottle, CO2 escapes out due to fall in pressure. 2. Because it destroy germs due to its strong oxidizing action.
3. Q it is combustible, burns without smoke and produces a large amount of heat on combustion.
4. Q a layer of inactive nitrogen prevents the oxidation of food and retards bacterial growth.
b) 1. KClO3 & MnO2 are strongly heated KClO3 melts and later decomposes evolving O2 – makes easy flow of O2.
2. MnO2. 3. Other wise water may enter the test tube.
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4. 2KClO3 2MnO⎯⎯⎯→ 2KCl + 3O2 (g). Δ c) CuO – Oxidation Cu – Reduced.
FeCl2 – Reduced FeCl3 – Oxidised. Br2 – Oxidised HBr – Reduced. KI – Reduction I2 – Oxidation.
Answer 4
a) 1. – B) 2. – E) 3. – D) 4. – C) 5. – A). b) Initial Condition Final Condition P1 = 760 mm of Hg P2 = 400mm of Hg. V1 = 700 ml V2 = x ml T1 – 0oC = 273 K T2 = 15 + 273 = 288 K.
Equation , 1 1
1
P VT
= 2 2
2
P VT
substituting 1 1 22
2 1
P V .TVP . T
= = 760 700 288400 273× ×
× = 1403.07 ml.
Answer 5 a) 1. “At constant, Temperature, the volume of a given mass of dry gas is inversely proportional
to its pressure”.
V 1P
∝ [T Constant]
2. “ The physical and chemical properties of elements are periodic functions of their atomic weights”.
b) + -Na + Cl [Na] + [ Cl ] NaCl .. ... . : : :.. ..→ →
[2,8,1] [2,8,7] [2,8] [2,8,8] Sodium chloride. c) i) – E). ii) – A). iii) – C). iv) – D). v) – B). Answer 6 a) 1. CuSO4.5H2O. 2. Blue.
3. The process by which crystals are separated or deposited from a hot saturated solution of a substances on cooling gently is called crystallization.
4. CuSO4.5H2O Δ⎯⎯→ CuSO4 + 5H2O. 5. Amorphous. b) 1. transference, covalent 2. electronegativities. 3. low , high.
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Answer 7 a) 1. 11. 2. 17. 3. 7 . 4. 9 . 5. 19. 6. 21. b) Urea – Ammonia & carbon di oxide. C.A.N - Ammonium nitrate, lime stone.
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Paper 13 Chemistry Answers Section -I
Answer 1 a) Select the correct answers: 1. Aluminium (Al). 2. CS2 carbon di sulphide. 3. Reduction . 4. Carbon [C]. 5. Dilute HNO3. b) Name the following: 1. Fluorine. 2. III A (Aluminium). 3. Sodium. 4. Metal B. 5. Potassium. c) Balancing chemical equations: 1. CO + H2 + H2O
o450 CFe O2 3⎯⎯⎯→ CO2 + 2H2 + Δ .
2. 2Na + H2 Δ⎯⎯→ 2NaH ( Sodium hydroxide).
3. 3Mg + N2 → Mg3N2 (Magnesium nitride) 4. H2 + S → H2 S (Hydrogen sulphide). 5. 2Fe + 3Cl2 → 2FeCl3 ( Iron (III) chloride).
d) Change the following statements by replacing the underlined words: 1. Oxidation reaction is addition of electronegative elements. 2. Inhibitors. 3. Negative charge. 4. Every eighth element was a repetition of the first element. 5. Sulphuric acid. e) Mention the terms used for each of the following: 1. Vapourisation. 2. Boyle’s Law. 3. Chemical formula. 4. Mendeleeff’s periodic Law. 5. Actinide series. f) Pick the ODD one out : 1. Fe+1. 2. Moseley. 3. Neutrons. 4. Combustible. g) 1. What do you observe when:
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a) A piece of white phosphorus when placed in a jar of Cl2 first melts and later burns spontaneously forming thick white fumes of phosphorus trichloride (PCl3) & a small amount of PCl5.
b) It melts around 1650oC and emits a bright white light called ‘lime light’.
2. Give reasons for the following: a) Q carbon mono-oxide is present in the exhaust fumes.
b) Q it is almost as heavy as air. 3. Complete and balance the following equations: a) 2NaOH + CO2 → Na2CO3 + H2O. b) 2H2O + 2Cl2 → 4HCl + O2. h) 1. Write chemical formula of the following : a) Bleaching powder – CaOCl2. b) Marble chips – CaCO3. 2. Write chemical names : a) Urea. b) Glucose. 3. Molecular weight of NaCl = 1 23 + 35.5 1 = 58.1 x x
a) Water vapour (H2O). b) CuSO4 + 5H2O → CuSO4 5H2O.
Section- II
Answer 2 a) 1. Metals. 2. 14Si 28. 3. 13 Al27. 4. 18Ar40. 5. 17Cl35.5. 6. 13Al27. b)
Sl.No. Property CO CO2
1. Litmus test Neutral. Acidic.
2. Solubility 130 vols of H2O dissolves 3 vol.
1 vol of H2O dissolves 1 vol.
3. Combustibility Combustible . non combustible.
4. Lime water Not absorbed in lime H2O.
Absorbed in lime H2O.
Answer 3 a) 1. Q gaseous NH3 & CO2 at 200oC & 100 to 200 atmospheric pressure it reacts easily to form urea & is used as nitrogenous fertilizer. 2. Q it acts as a good reducing agent. 3. Qbleaching action of Cl2 is vigorous to liberate [O]. 4. Q it combines with metals to form sulphide which is safety.
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b) 1. Concentrated H2SO4 is acting as a dehydrating agent & removes H2O molecules. 2. CO is collected by downward displacement of H2O because it is almost insoluble in H2O & heavier than air. 3. On passing the gas though lime water no effect is observed. 4. HCOOH H SO2 4⎯⎯⎯→ CO + H2O. c) 1. FeCl2 is reduced FeCl3 is oxidised. 2. CuO is oxidised Cu is reduced. 3. Zn reduced ZnSO4 is oxidised. 4. H2S reduced S is oxidised. Answer 4 a) 1. Bluish white to white. 2. Brown to Grey. 3. Yellow to Silvery grey. 4. Black to reddish brown. b) Higher unsaturated oils on heating with hydrogen in the presence of catalyst like nickel or platinum gets converted into saturated fats. This is known as hydrogenation of oils. c) 1. Sodium sulphate. 2. Iron III chloride. 3. Concentrated H2SO4 . 4. Chlorine. Answer 5 a) Initial conditions Final conditions P1 = Pmm of Hg P2= 2Pmm of Hg V1 = 1000ml V2 = 3000ml T1 = 273 + 15 = 288k T2 = ?
Gas equation 1 1
1
P VT
= 2 2
2
P VT
Substituting, P x 1000288
= 2
2P x 1000T
∴ T2 = 2P x 3000 x 288P x 1000
= 1728 K or 1455oC
b) Separation, identification, adsorption, adsorbent, cellulose Answer 6
a) 1. C2H2 + 52
O2 → 2CO2 + H2O.
2. Ca(OH)2 + 2HCl → CaCl2 + 2H2O. 3. 3CaOCl2 + 2NH3 → 3CaCl2 + 3H2O + N2 (g). 4. (NH4)2 Cr2O7 → Cr2O3 + 4H2O+N2(g). 5. 3PbO + 2NH3 → 3Pb + 3H2O + N2(g). b) 1. O
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2. 2 , II 3.
Sl.No. Element E C Valence M / N G/ N M 1. 15PP
31 2, 8, 5 5 N. M 2. 18 Ar 40 2, 8, 8 0 N.G 3. 13 Al 27 2, 8, 3 3 M 4. 6 C 12 2, 4 4 N.M
Answer 7 1. a) KCl. b) KClO. c) KClO2. d) KClO3. e) KClO4. 2. a) Oxidation – Loss of electron. Reduction – Gain of electron. b) Cation – Positively charged ions. Anion – Negatively charged ions. 3. a) Magnesium + Nitrogen → Magnesium nitride. b) Zinc Sulphide + Oxygen → Zinc oxide + Sulphur di oxide. c) Sodium carbonate + hydrochloric acid → Sodium chloride + Water + Carbon dioxide.
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Paper 14 Chemistry Answers
Section - I
Answer 1 a) Select the correct answers: 1. Slaked lime. 2. Sodium. 3. H2. 4. SO2, H2SO3. 5. Sulphur. b) Name the following : 1. Sodium. 2.Group VII A. 3. Sodium. 4. Metal K. 5. Gold. c) Balancing the chemical equations: 1.CH4 + H2O o
Ni800 C
⎯⎯⎯→ CO + 3H2.
2. Ca + H2 Δ⎯⎯→ CaH2 (Calcium hydride). 3. 2Al + N2 2AlN (Aluminium nitride) (ALN). → 4. 2S + Cl2 S→ 2Cl2 (Sulphur monochloride). 5. CO2 + C 2CO (Carbon monoxide). → d) Chance the following statements by replacing the underlined words : 1.Oxidation reaction is removal of electro positive elements. 2. Negative catalyst. 3. Positive charges which are emitted from the anode. 4. Moseley stated that “ Properties of elements are periodic functions of their atomic numbers”. 5. Carbonic acid. e) Mention the terms used for each of the following : 1. Freezing or solidification. 2. Charle’s Law. 3. Symbol. 4. Isotopes. 5. Lanthanide series. f) Pick the ODD one out : 1. Pb3+. 2. Moseley. 3. Electrons. 4. Non combustible.
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g) 1. What do you observe when : a) A burning candle burns in a jar of chlorine with a reddish yellow flame forming black particles of soot & thick white fumes of hydrogen chloride.
CxHy + ½yCl2 xC + yHCl. → Carbon Hydrogen chloride
b) On adding of water slaked lime dissolves very sparingly about 0.2 % of slaked lime dissolves in water & the clear solution of lime water is formed.
2. Give reasons for the following:
a) Because the fumes of burning tobacco contains traces of carbon monoxide. b) Because their outer most shell is saturated. 3. Complete and balance the following equations : a) 2NH4OH + H2SO4 → (NH4)2SO4 + 3H2O. b) Na2SO3 + 2HCl → 2NaCl + H2O + SO2. h) 1. Write chemical formula of the following a) Ca(OH)2. b) SiO2. 2. Write chemical name of the following compounds: a) Nitrous acid. b) Sodium Aluminate CaSO4 = 1 40 + 1 x 32 + 4 x 16 = 136. xi) Oxygen - It absorbs colourless alkaline pyrogallol solution and turns brown . Section II Answer 2 a)
Sl.No. Column Elements
Atomic Number
Electronic Configuration
Group Underline the element present in that group
1. Hydrogen 1 1 I A Be / Na / Ca 2. Carbon 6 2, 4 IV A N / P / Si 3. Nitrogen 7 2, 5 V A C / P / S 4. Oxygen 8 2, 6 VI A N / S / Cl 5. Sulphur 16 2, 8, 6 V A O / N / F 6. Chlorine 17 2, 8, 7 VII A I / O / S
b)
Property Diamond Graphite Colour Colour less Blackish Grey Nature Hardest Softest Conduction non conductor good conductor Density High Low
Answers 3 a) 1. As a fuel.
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2. As a reducing agent. 3. Manufacture of electrodes. 4. Product of black shoe polish. b) 1. A → H2O2 B MnO→ 2. 2. MnO2 acts as catalyst. 3. MnO2 decomposes giving manganous oxide [MnO] and nascent oxygen [O]. 4. 2H2O2
MnO2⎯⎯→ 2H2O + O2 (g). c) 1. Q it is slightly soluble in water & it is almost as heavy as air. 2. H2, N2. Answer 4 a) 1. – B) 2. – A) 3. – B) 4. – C) 5. – B) b) Rusting is slow oxidation of iron into its hydrated oxide by O2 of the air in the presence of moisture. c)
Sl.No. Observations Product Litmus action (i) Glowing
Charcoal burns C with a whitish glow giving bright Spark forming CO2
CO2 Blue litmus solution turns red.
(ii) Burning Sulphur
Burns with a bluish flame forming SO2
SO2 Blue litmus solution turns red.
Answer 5 a) SO2. b) Rhombic S & Monoclinic S.
Rhombic S is most stable. Valcanization – Heating a natural rubber with S to a definite temperature for a known period of time. c) Donor. One. Accepts. Reducing. Answer 6 a) Mg (HCO3)2 Δ⎯⎯→ MgCO3 + CO2 + H2O. b) 1. Dehydrating agent. 2. 2CO + O2 → 2CO2 + . Δ 3. NaHCO3 + tartaric acid + starch baking powder. →
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c) a) Reddish Brown. b) Violet vapours. c) Water vapour. d) Rekindles a glowing splint e) yellow residue. Answer 7 a) V1 = 45 litres V2 = 42 litres T1 = ? T1 = 7oC = 7 + 273 = 280 By Charles Law
1
1
VT
= 2
2
VT
Substituting, 1
45T
= 42280
T1 = 45× 28042
15 201×= = 300 K.
b) 1. Chemical equation. 2. Drying agent. 3. Volume. 4. Electronic configuration. 5. Oxidation.
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Paper 15 Chemistry Answers
Section -A Answer 1 1. a) Carbonate or bicarbonate CO3
2-or HCO3-
b) B → Sulphate – SO42-
c) C → Bromide – Br-
d) D → Chloride – Cl-
e) E → Chloride – Cl-
2. a) Ionic , NaCl, conductor , table salt , basic. b) Polar covalent , unequally, chlorine, +δ , -δ . 3. a) When concentrated sulphuric acid is heated strongly it decomposes giving sulphur di oxide, oxygen and water. b) When CO2 & NH3 in the ratio 1:3 by volume are passed into reactor with a pressure of 378 atmosphere & temperature of 473 K urea is formed. c) The outermost shell in potassium atom is shell . d) CAN is a Nitrogeneous fertilizers. e) Chlorine gas is a powerful oxidising agent. 4. a)
Atomic numbers
Family of elements
9 Halogens. 19 Alkali metals. 4 Alkaline earth metals. 18 Inert gases. 24 Transition metals.
b) Cl- = 18 = 2, 8, 8. Ca2+ = 18 = 2, 8, 8. S4- = 20 = 2, 8,10. O-1 = 9 = 2, 7. N-3 = 10 = 2, 8. 5. a) 1) An orbit is a circular path around the nucleus in which the e-s revolves. An orbital is a region around the nucleus where the probability of finding the e-is maximum. 2) Atomic number is the number of protons or number of e-s present in an atom. Atomic mass is the number of protons and neutrons is the nucleus of the atom. 3) Ground state e- in the same orbit. Excited state e- in the orbit accepts some energy in the form of heat or light and goes to the next orbit.
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b) 1. Cl2 + H2O → 2HCl +[O]. 2. N2 + 3H2 → 2NH3 3. 2S + Cl2→ S2Cl2 4. K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2(SO4) 3 + H2O.
Section B Answer 2 a) 2H2O2
MnO2⎯⎯→ 2H2O +O2. S + O2 → SO2. 2SO2 + O2→ 2SO3. SO3 + H2O → H2SO4. SO2 + H2O → H2SO3.
A : MnO2 B: S C: O2 D : H2O E : H2O.
b) V1 = 7cm3 V2 = ?
T1 = 25oC 25 +273 = 298K T2 = 125oC = 125 + 273 = 398
by Charles law, 1
1
VT
= 2
2
VT
by substituting, V2 = 1 2
1
V TT
= 7 398 199298 149× = 9.3 cm3.
c) N2 +3H2 → 2NH3→ catalytic tower. 2SO2 + O2 2SO3 + contact tower. Δ d) Total molecular mass = KClO3 = 39.1 + 35.45 + 16 x 3 = 122.55 % of K = 39 .1 x 100 = 31.905 122.55
Answer 3 a) 1) Methyl alcohol and water are two miscible liquids & have different boiling points. 2) QSilvery salts under go photochemical reaction under lights & dark coloured bottles do not allow light to penetrate through them. 3) Frasch process is not waste fuel and S of 99.5 % purity is obtained.
4) Because here is a difference in atomic arrangement in the crystal structure of the element. 5) Because table salt is a hydroscopic substance it absorbs water through moisture that is from the atmosphere.
b) 1. Separating funnel. 2. Petrol + water. 3. Water. 4. No, because they are miscible liquids. 5. No, because they are miscible solutions having different boiling points. Answers 4 a) C : H O Empirical formula = C6HO4 6 1 4 b) 1. – D
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2. – F 3. – B 4. – E 5. – C
c) i) S + O2 → SO2. ii) Cl2 + H2O → 2HCl + O. iii) H2SO4 + H2C2O4 → CO2 + CO+ H2SO4.H2O. Answer 5 a) 1. F belongs to VII group. 2. Zinc nitrate do not contain any water molecules in its water of crystallization. 3. A is not a shell of an atom. 4. Sulphuric acid di basic acid. 5. Yellow vitriol none of the salt appears yellow when it absorbs water molecules. b) 1. F, Cl, Br, I. 2. Kr , Ar, Ne, He. 3. Oxidation , catalytic , absorption, dilution. 4. N , M, L , K. 5. CrO2Cl2 , MnCl2, K2Cr2O7 , KMnO4. Answers 6 a) 1. Oxidising reactions, Sulphur is formed.
2. Addition or combination reaction, NaCl is formed. 3. Combustion, CO2 & H2O is formed. b) 1. Jar A filled with H2 & a jar B which is empty because it contains air. 2. Test the jar B by bringing a flame near it, upper jar burns with a ‘pop sound’.
3. It extinguishes a burning candle because H2 is combustible but does not support combustion.
4. Water does not rise up the trough because H2 is slightly soluble in water . 5. Pure H2 burns quietly in air forming water as the only product, H2 air mixture burns with a characteristics ‘pop sound’.
c) i) Na2CO3 . H2O. ii) CaSO4. 2H2O. Answer 7 a) 1. Iodine. 2. Sodium aluminate. 3. SO2 or SO3 4. Ammonia. 5. Nitrogen. 6. Hydrogenation of oils. 7. Amphoteric oxide. 8. Oxygen (49.9%). b) V1 = 450ml V2 = 150ml T1= 300k T2 = ? By Charles law,
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1
1
VT
= 2
2
VT
T2 = 2 1
1
V TV× = 150 300
450× = 100K.
T2 = 100K = [100 -273]oC = -173oC
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Paper 16 Chemistry
Answers Section -I
Answer 1 a) A – 2 B – 3 C – 5 D – 4 E – 1. b) 1) I , Br, F, Cl 2) F, Be, Mg, Na 3) Al, Si, B, O 4) K , Na, Cl, Ar 5) I- , Br- , Cl - , F-. c) 1) Radical. 2) Redox reaction. 3) Drying or desiccating agents. 4) Hydrogenation. 5) Valcanization. d)
Fire extinguisher
Solution in Cylinder
Solution in glass bottle
Solution / Foam ejected from nozzle.
Soda acid type NaHCO3 H2SO4 Na2SO4, H2O, CO2
Foam type NaHCO3 Al2(SO4)3 Na2SO4, Al(OH)3, CO2
e) i) An acid is a substance which donates proton. ii) Organic acid – HCOOH mineral acid sulphuric acid iii) HCOOH iv) HCOOH v) Sulphuric acid because it ionises in to H2SO4 → H + + HSO-
4 f) A : 2 B : 3 C : 4 D : 1 E : 5 g) 1. Golden yellow colour is imparted to the flame . 2. Orange crystals changes to green residue. 3. It fumes because of formation of HCl gas.
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4. Turns moist starch iodide paper blue black. 5. Brisk effervescence is observed because of CO2 liberation. h) 1. C6H6 + 15/2O2 → 6CO2 + 3H2O +∆. 2. Ca (HCO3)2 + 2HNO3 → Ca (NO3)2 + 2CO2 + 2H2O. 3. KMnO4 + 8HCl → KCl + MnCl2 + 4H2O + 3Cl2. 4. MnO2 + 2NaCl + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O +Cl2. 5. 2Fe SO4 + H2SO4 +Cl2 → Fe2(SO4)3 + 2HCl.
Section II Answer 2 a) i) Concentrated HCl. ii) MnO2 acts as oxidising agent. iii) Thistle funnel should dip below the level of the acid in the flask otherwise the Cl2 gas obtained will escape out through the thistle funnel. iv) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2. v) By Passage through 2 wash bottles. 1) - Containing water which absorbs HCl acid vapours. 2) - Containing concentration H2SO4 which absorbs moisture. vi) Because it is heavier than air & moderately soluble in H2O. vii) A moist blue litmus paper is brought near the mouth of jar it turns red & Then gets bleached. viii) Because MnO2 decomposes to give MnO + {O}. [O] Oxidises HCl toCl2. MnO reacts with more acid to give MnCl2 & water. Answer 3. 1.
Dalton’s atomic theory Modern atomic theory a) Atoms are indivisible and in destructible.
Atoms are divisible and destructible.
b) An atom is the smallest indivisible particle of an element.
Atoms are divisible, consisting of sub atomic particles like Protons , electrons & neutrons.
2. Health salts is a mixture of citric or tartaric acid with sodium bicarbonate . 3. 3COO +NH3 3CO + 3H→ 2O + N2 (g). 4. Because heat energy is saved in heating the non reactive air component that is N2.
5. Elements having same chemical properties but different physical properties are called allotropes.
1. Crystalline 2. Amorphous. 6. 2NaOH + CO2 → Na2CO3 + H2O. Na2CO3 + H2O + CO2 → 2NaHCO3. Answer 4 1. i) a) Calcium carbonate [CaCO3]. b) Carbon dioxide CO2. c) Ca(OH)2 calcium hydroxide. ii) Heat is absorbed because it is a exothermic . iii) a) Lime water, CaCO3. b) If gas A that is excess of CO2 is passed through the milky suspension the milkiness
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disappears due to formation of soluble calcium bicarbonate but on heating it reappears because it decomposes to form CaCO3. Answer 5 a) i) 13 Al 27 2, 8, 3 ii) 15 P 31 2, 8, 5 b)
Element Atomic number
Mass number
P N F
Be 4 9 4 5 4 F 9 19 9 10 9 Na 11 23 11 12 11 Al 13 27 13 14 13 P 15 31 15 16 15
Answer 6
1. -273oC. a. Decreases b. Increases c. 1 litre. d. Positive. e. OCC.
2. P1 = 760mm of Hg P2 = 2× 760 V1 = V1 litres V2= ? T1 = 273 K T2 = 3 ×273 K
Gas equation 1 1
1
P VV
= 2 2
2
P V T
Substituting = V2 = 1 1 2
1 2
P V T T × P
V2 = 1760 × V × 3 × 273 2 × 760 × 273
V2 = 1.5 V1 Answer – 7 a) i) Water and sand . ii) Petrol and water. b) i) 2 ii) 1 iii) 4 c) a) (NH4)2 SO4. b) K2Cr2O7. c) Na2CO3. d) Ca3 (PO4)2. d) i) Sublimation. ii) Evaporation iii) Diffusion.
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Paper 17 Chemistry
Answers Section - A
Answer 1 I) a) Ammonium dichromate. b) a) Graphite b) Gas carbon. c) Oxygen gas. d) Phosphorous. e) Natural gas and petroleum gas. II) a) A : ZnCO3 b) B : ZnO c) C : CO2. ZnCO3 ZnO + CO→ 2. b) A : Na2CO3 .10H2O B : Na2CO3 . H2O C : H2O. Na2CO3 .10H2O Na→ 2CO3 . H2O + 9H2O. III) a) 2Pb3O4 → 6PbO +O2. b) CuSO4 5H2O CuSO→ 4 + 5H2O. CuSO4 Δ⎯⎯→ 2CuO + 2SO2 + O2. c) (NH4)2Cr2O7 → Cr2O3 + 4H2O + N2. d) 2HgO 2Hg + O→ 2. e) 2Zn(NO3)2 2ZnO + 4NO→ 2 +O2 IV) a) 2+ b) 2n2,18. c) Atomic number , Neutrons. d) Iron. V) 1. Changes to red and slowly bleaches the colour. 2. Rekindles the glowing splinter. 3. More dense white fumes is liberated due to NH4Cl. 4. It becomes anhydrous white CuSO4. 5. H2 gas is liberated. VI) 1. Because diamond is a hardest substance but graphite is a softest substances with a greasy feel. 2. Because CO2 gas dissolves in lime water to form CaCO3 which is milky in appearance. Even SO2 gas dissolves in lime water to form CaCO4 which is milky in appearance. 3. Because loss of electrons and gain of electrons goes hand in hand. 4. Because galvanized steel is not used for food containers since food acids dissolves zinc forming poisonous compounds. 5. Because Cl2 is more reactive than Br2 so it displaces Br2 from KBr. Section B Answer 2
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a) Used in the manufacture of Urea, Baked food items. i) Solid CO2. ii) NaHCO3, tartaric acid , Starch iii) Passage of CO2 through Water under pressure, addition sugar flavours & colours.
b) Because CO inhaled the hemoglobin combines with it forming a stable compound carboxy hemoglobin which prevents the hemoglobin from taking up O2. A person dies due to lack of O2.
c) 1. a) Respiration in living organisms. b) By burning of C compounds such as wood, coal, petroleum alcohols etc. 2. Passing a mixture through concentrated potassium hydroxide solution.
CO2 dissolves forming potassium bicarbonate. CO remain undissolved and is collected.
Answer 3 1. i) The element with at number 26 is Fe , Its symbol is 55
26 Fe ii) The element with at number 4 is Be , Its Symbol is 9
4 Be2. 27 3. Same atomic number 4. 12-2 = 10 5. 1H1
6. Protons, neutrons. 7. No, the e–s are not stationary in the stationary state of an atom. But the energy associated with the e – s remains the same in the stationary state.
Answer 4 1. 7, 10, 9, 10, 10 2. B3+ ion has smallest radius there ions correspond to first 3 elements present in second period . Since the atomic radius decreases along a period the ionic radius is also expected to follow the same order . 3. Na+, Mg2+, Al3+
4. a) O atom has higher electron affinity because for N the value is nearly zero due to symmetrical configuration.
b) Cl has higher electron affinity because greater atom size and less electron crowding. c) B has highest E -A Because the value of Be is almost zero due to symmetrical configuration. Answer 5 1. i) Because water contains soluble gases which contributes to the taste of water. ii) Common salt is not deliquescent but contains a small amount of Ca & MgCl2 impurities which are deliquescent . Hence the impurities absorbs moisture from air, there by turning common salt sticky & wet.
iii) HNO3 is a powerful oxidizing agent & [O] formed on its decomposition oxidises the H2 to water & oxidising action of the acid is much reduced due to dilution.
iv) Because it is a hardest substance & that is due to strength and uniformity of C – C bonds in diamond.
v) Because wood charcoal and sorbs & accumulates gases, liquids on its outer surface and hence adsorption takes plane on its outer surface. 2.From the elements of group IV A of the periodic table ……….. A : Pb B : with HCl numbers with dilute NaOH – Si, Ge, Sn, Pb.
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C : acidic oxide – Si Neutral oxide – nil. D : Ge , Sn, Pb. E : Si, Ge, Sn, Pb. Answer 6 1. 1. E 2. C 3. F 4. A 5. B 2. i) 20Ca40 Ca 2, 8 ,8 ,2 8O16 O 2, 6
C a [Ca .. O
..:.. → 2+] [O2-] CaO
2,8,8,2 2,6 2,8,8 2,8 ii) NH3
7 N 14 2.5 1 H 2 1 3. Cu shows variable valency 1 and 2 that is Cu+, Cu2+
Answer 7 a) 1. A → cooling chamber B Purified air C → un liquefied air D → liquefied air → 2. Joule – Thompson effect. 3. [liquid N2 ,liquid O2] . 4. Temperature 200oC pressure 200 atmosphere 5. The dried gas is dried, compressed and stored in steel cylinder under pressure because it almost as heavy as air. b) 1. Nitric acid - Ostwald process. 2. H2SO4 - contact process.
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Paper 18 Chemistry Answers
Section -1 Answer 1 a) Ions 1 2 3 4 5 Precipitates E A B C D b) i) Cl2 gas turns moist blue litmus paper red and bleaches the colour of litmus paper. ii) Displacement reaction. iii) When Cl2 gas reacts with KBr , reducation of KCl takes place. iv) Vapourisation. v) Atomic numbers. c) i) - C ii) - A iii) - E iv) - D v) – B d) i) Cu (NO3)2 decomposes forming black CuO, reddish brown fumes of NO2 and water vapour. ii) (NH4)2Cr2O7 decomposes forming green Cr2O3, water & N2 gas. iii) NH3 combines with HCl giving dense white fumes of NH4Cl. iv) Mg wire burns in N2 forming Mg3N2. v) It bleaches the coloured flowers due to oxidation. e) i) Because Cl2 also belongs to the same group in the periodic table & has seven electrons in its outer most shells they all belongs to halogen series. ii) Because different allotropes of carbons in the pure state are chemically identical. iii) Because Cl2 water contains HCl and HOCl. HOCl acid in Cl2 water is unstable and decomposes in the presence of sunlight.
iv) Because they add calculated amount of a sulphate powder in it hence it turns into a hard mass.
v) Because the last shell does not contain either duplet configuration or outer configuration. f) i) e-s in valance shells of atoms. ii) Metals and nonmetals . iii) NH3, PCl5. iv) Nessler’s reagent test. g) i) CaCO3 CaO + CO2 ii) 2NH3 +CO2 (NH→ 2)2 CO+ H2O. iii) 2S + Cl2 S→ 2Cl2
iv) HCOOH 2 4H SO⎯⎯⎯→ CO + H2O v) CO + NaOH → HCOONa h) i) KNO3.
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ii) NH4NO3. iii) Cu(NO3)2 . iv) Ag ( NO3)2. v) Pb (NO3)2.
Section -2 Answer 2 1. i) 2, 4. ii) Carbon, lead. iii) Metalloid. iv) Graphite. v) NO. vi) Fullerenes.
2.i) NH4+ Ion.
ii) H3O+ Ion .
Answers 3 1. i) KMnO4 K(x): Mn(+7): 0(-2) x + 7+ (-2) 4 = 0 x + 7 – 8 = 0 x -1 = 0 x = 1. ii) K2Cr2O7 K(x): Cr ( +6 ) : O(-2) 2x + 6x 2+7(-2) =0 2x +12 -14 = 0 2x = 2 x = 2
2 =1.
iii) KClO4 K(x): Cl (+7): O(-2) x + 7 + 4 (-2) x + 7 -8 = 0 x =1. 2. H2. 3. It increases. 4. Calculation of empirical formula calculation of the ratio by atoms
C : H : O = 6 1 4 : : 12 1 16
= 0.5 : 1 : 0.25
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Dividing by the least number we get 2 : 4 : 1 Therefore Empirical formula = C2H4O Molecular formula = (C2H4O)n Calculation of molecular formula = Molecular formula = 2 x Vapour density
n = molecular weightemprical formula weight
n = 4444
= 1
Therefore empirical formula = Molecular formula = C2H4O. Answer 4 1. i) S + O2 SO→ 2↑ . ii) 2NaOH + SO2 →Na2SO3 + H2O. (limited supply) 2. i) It reduces pink coloured acidified KMNO4 to colourless Manganous sulphate. 2KMnO4 + 3H2SO4 K→ 2SO4 + 2MnSO4 + 3H2O + 5 (O). 5[H2O + SO2 + [O] H→ 2SO4]. Adding, 2KMnO4 + 2H2O + 5SO2 → K2SO4 + 2MnSO4 + 2H2SO4. ii) It reacts with iodides to give hydrogen iodide KI + H2SO4 KHSO→ 4 + HI ↑ 3. i) d. ii) e, a iii) b. Answer 5 1 i) → Barium chloride (b). ii) Concentrated H→ 2SO4 (cold) (c). iii)→ lime water (a). iv)→ Concentrated H2SO4 (hot) (f). v) → Silver nitrate (d) or c. 2. i) Cl + < Cl < Cl-. ii) Na. iii) Mg2+ < Na + < F- < O2- < N3-. iv) Li , Na, K.
v) Cl. Answer 6 1. a) A -NH4Cl B - NH3 C - HCl NH4Cl NH3 + HCl White colourless colourless b) A – (NH4)2 Cr2O7. B – Cr2O3. C – H2O.
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D – N2. (NH4)2 Cr2O7 →Cr2O3 + 4H2O + N2. N2 is unreactive or inert in nature. 2. a) K2SO4. b) H2 gas burns with a pale blue flame producing ‘pop’ sound. c) Golden yellow flame. d) HCl.
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Paper 19 Chemistry Answers
Section - I Answer 1 1. a) Enameling. b) Greasing. c) Galvanising. d) Tin – planting. e) Electroplating. 2. a) Gun-powder. b) Calcium bisulphite. c) Sodium thiosulphate. d) Sulphur & Potassium chlorate. e) Carbon di sulphide. 3. a) N2 + O2 →2NO - ∆ Temperature required 3000oC. b) N2 + 3H2 2NH→ 3 + ∆ Temperature 450 – 500oC. c) CaC2 + N2 →CaCN2 + C Temperature 1000oC. d) 3Mg + N2 →Mg3N2 Dilute red heat. e) 2Al + N2 2AlN Dilute red heat. → 4. a) Oxidation reaction. H2S is oxidized to H2O. b) Reduction reaction CuO is reduced to Cu. c) Reduction reaction H2S is reduced to 2HCl. d) Oxidation reaction H2S is oxidized to S. e) Oxidation reaction Fe2+ is oxidized to Fe3+.
5. Sl.No Atomic
Number Name of element Family of element
a) 9 Fluorine Halogens. b) 19 Potassium Alkali metals. c) 4 Beryllium Alkali earth metals. d) 18 Argon Noble gases e) 24 Chromium Transition metals.
6. a) The pressure of the air is directly proportional to the temperature since the temperature is higher is summer than in winter, the pressure of the air in the tube of the type is likely to be quite high as compared to winter. It is quite likely that the tube
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may burst under high pressure is summer. b) The volume of the gas is inversely propostional to the pressure at a given temperature according to Boyle’s law . As the weather balloon ascends, the pressure tends to decreases. c) Dry air mainly contains gases like N2 & O2 along with other gases wet air mainly contains small amount of H2O vapours now vapour density of H2O vapour is less than that of N2 and O2 because molar mass of H2O (18) is less than that of N2 (28) and O2(32). This means that heavier molecules of N2 & O2 have been replaced by lighter molecules of H2O vapour when the dry air changes to wet air. 7. a) Filtration b) Evaporation. c) Fractional distillation d) Magnetic separation. 8. Fill in the blanks with the correct word form the words in brackets. a) A symbol represent a short form of an ____________ [atom / element / molecules]. b) Aurum is the latin name for the metal ________ [gold / lead / iron / silver]. c) A chemical equation is a short hand form for a ________ [Physical / chemical] change. d) All chemical equations must be balanced to completly with ______. [Avagadro’s law / law of conservation of matter / lussac’s law]. e) Variable valency is exhibited, since electrons are lost from an elements from the _____ [Valence / penultimate shell].
Section II Answer 2 1. a) A – reactants, B – Products. b) To yield, represents direction c) Reactants - are the substance which takes part in a chemical reaction. Products are the substances formed as a result of the chemical change. d) L.H.S - NH4OH + HCl
R.H.S - NH4Cl + H2O 2. a) Heating initiates the reacting particles to move faster and collide more frequently. b) Molecule of the reaction absorbs light energy, become activated and react rapidly. c) Electricity directs the changed particles to opposite electrodes. d) Initiates a change in the rate of the reaction without under going any changes in its chemical composition . e) Sound energy speeds up the reacting molecules atoms or ions causing a reaction to occur. Answer 3 a) a) A – Burning Candle.
B – Anhydrous CaCl2. C – Cotton wool. D – Soda lime.
b) NaOH + CaO. c) When the candle is lit the weight of the apparatus progressively increases as CaCl2
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absorbs H2O and soda lime absorbs CO2. d)There is a gain in weight on burning candle. e) It is connected to beam of a balance. b) A student forgot to add the reaction mixture to the removal bottomed flask at 27oC but put it on the flame. After a lapse of time, he realized his mistake, By using a pyrometer, he found that the temperature of the flask was 477oC what fraction of air would have been expelled out? Since the student was working in the laboratory, there was no change in pressure. Therefore By Charles law, V1 = V dm3 V2 = ? T1 = 27 + 273 = 300K T2 = 477 + 273 = 750K
1
1
VT
= 2
2
VT
or V2 = 1 2
1
V TT
V2 = 3Vdm 750K
300K× = 2.5 vdm3.
Thus volume of air expelled = 2.5 v – v = 1.5v dm3
Fraction of air expelled = 1.5 v2.5 v
= 35
.
Answer 4 1) 1 → D 2 → E 3 → C 4 → B 5 → A. 2) i) Ionic bond is formed between x and y ii) Na – Cl 17 Cl 35 electronic configuration
2, 8, 7 Na → NaCl.....:
+ Cl-
11Na 23 2, 8,1 Na – Cl iii) Yes it is soluble in water iv) Yes is molten state.
Answer 5 1. a)
Carbonate Bicarbonate Salt + water heat the gas liberated is passed through lime water.
Lime water does not turns milky.
Lime water turns milky.
b) Chlorine Bromine Salt + concentrated
H2SO4. Colourless gas on exposing a glass rod dipped in NH4OH more dense white fumes liberated.
Brown gas is liberated.
2. KCl - lilac colour or light violet colour is imparted to the flame
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BaCl2 – apple green colour is imparted to the flame. CaCl2 – Brick red colour is imparted to the flame.
Salt is treated with concentrated HCl made a paste. The paste is held to the flame by means of a platinum wire & viewed through a cobalt blue glass to make the above observations.
Answer 6 1. i) Pure nitrogen gas is obtained from fractional distillations. ii) H2 is obtained by Bosch process. iii) Catalyst is finely divided iron. Promoter is molybdenum. iv) Reversible, endothermic.
v) N2 + 3H2 2NH3 + ∆
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Paper 20 Chemistry Answers
Section -1
Answers 1 I. a) i) Is the property due to which a substance accumulates gases liquids on its outer surface. ii) Certain metals like palladium, platinum etc adsorb large volumes of gas iii) Natural process by which green plants manufacture food by absorbing CO2 in the presence of sunlight and chlorophyll & convert it into glucose. iv) Heating natural rubber with S to a definite temperature for a known period of time. v) It is a series of metals arranged according to their decreasing reactivity, so that the most reactive and the most easily oxidised metal is placed at the top of the series . vi) Addition of H2 to organic compounds in the presence of a catalyst eg. Platinum or Ni is called hydrogenation. b) Chlorine evolved from bleaching powder on addition of dilute acid . CaOCl2 + 2HCl(dil) CaCl→ 2 + H2O+ Cl2 (g). c) BaSO4 + BaCl2 → BaSO4 (white precipitate) this white precipitate do not disappears with dilute HCl because BaSO4 is insoluble. II. a) i) (NH4)2Cr2O7 → Cr2O3 + 4H2O + N2(g).
(NH4)2 Cr2O7 orange crystals are heated it decomposes to form green Cr2O3 and N2 gas is evolved. ii) HOCl → HCl + [O] HOCl decomposes to Hcl and nascent oxygen. iii) The bottle breaks, Al2(SO4)3 mixes with NaHCO3 liberating CO2 gas & Al(OH)3 which together are ejected out under pressure through the nozzle in the form of foam. 6NaHCO3 + Al2(SO4)3 → 3Na2SO4 + 2Al(OH)3 + 6CO2 b) i) Valency – combining capacity of an element so as to form a compound. Valency electron – electrons present in the outer most shell. ii) Chemical formula – A molecule of a substance that is elements or compound could be represented by symbols. Chemical change – A change in which the substance retails its identity and changes in form or state without altering in composition. III. i) Hydrogen – Bosch process. Step (i) C + H2O
o1000 C⎯⎯⎯→ CO + H2 - ∆ (coke)(steam) (water gas)
Step (ii) CO + H2 + H2O 450
Fe O2 3Cr O2 3
oC⎯⎯⎯⎯→ CO2 + 2H2 + ∆
Step (iii) Recovery of H2 from the above mix Removal of CO2 → dissolves above mix in H2O under pressure or caustic
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potash solution. Removal of CO → dissolving in ammonical cuprous chloride. ii) N2 & O2 – liquefaction or fractional evaporation. Step (i) Air freed from CO2 & dust particles is cooled and compressed at about 200 atmospheric pressure is passed through small orifice results in temperature of the air being lowered until the air becomes a liquid at around -2000c Step (ii) Liquid air is a mixture of [liquid N2 B.P.(-196oC) liquid O2 B.P. ( -183oC)]
Liquid air FractionalDistillation⎯⎯⎯⎯→ (low B.P) evolved first leaving behind. 2
2
Liquid NLiquid O
Step ( iii) Collection of N2 liquid N→ 2 on warming turns into a gaseous N2 which is collected & stored. Step (iv) Collection of O2.
Fractional distillation after N2, Liquid O2 is left behind it is dried, compressed and stored in steel cylinders under pressure.
iii) Sulphur – frasch process or sicilian process. The S deposits are situated about 300mts below the earths surface.
Broughtby⎯⎯⎯→ 3 Concentric
pipes o2H O at 170 C
under pressure
1st pipe⎯⎯⎯⎯→ passed melts Swhich⎯⎯⎯→ 2nd pipe
compressedair is passed
⎯⎯⎯⎯→ which makesS forth
frothmiddlepipes
⎯⎯⎯→ collected.
IV i) b ii) d
iii) c iv) b v) b vi) c vii) d viii)b
ix) a x) d
Section -2 Answer 2 a) i) MgO 12 Mg 24 electronic configuration 2, 8, 2 8 O 16 electronic configuration 2, 6
Mg → Mgxx
O..:..
2+ O2- MgO →
Type of bond electrovalence bond or ionic bond ii) CH4 6 C 17 electronic configuration 2, 4 1H2 electronic configuration 1
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iii) NH3 7 N 14 electronic configuration 2, 5 1H 2 electronic configuration 1.
Type – covalent b) Because ionic bond is strong to break the bond high energy is required therefore they have high M.P.& B.P. Answer 3 a) 1) Group. 2) Atomic mass. 3) Repetition of a property at regular intervals . 4) 18. 5) 18. 6) Na. 7) Fe. 8) Cerium. 9) Same number of electrons; 10. 10) N3-. Answer 4 a)
Sl.No. Formula Conventional name
Oxidation number
1. N2O Nitrous oxide +1 2. NO Nitric oxide +2 3. N2O3 Nitrogen sesquioxide +3 4. NO2 Nitrogen dioxide +4 5. N2O5 Nitrogen pent oxide +5
b) a- iv b- iii c- ii d – i c) Lead reacts with dilute HCl and dilute H2SO4 forming an in soluble coating of lead coating of lead chloride [PbCl2] and [PbSO4] lead sulphate and hence further reaction comes to a stop. Answer 5 a) 1) Phenolpthalein 2) To indicate the completion of the reaction. 3) Yellow 4) Phenolpthalein 5) HCl
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b) Both P and V change ; n is constant.
Equation 1 1
1
P VT
= 2 2
2
P VT
T1 = 24 + 273 = 297k T2 = 35 + 273 = 308 P1 = 101.5 Kpa P2 = 102.8 Kpa V1 = 5.41dm3 V2
= ?
V2 = V1 1 2
2 1
P T P T
× ×
= 5.41 dm3 101.5 Kpa 308K 102.8 Kpa 297 K
× × = 5.54dm3.
Answer 6 a) a). 1) A funnel. 2) Anhydrous CuSO4. 3) Limewater. 4) Aspirator.
b). Cx Hy 2burns in O
→ CO2 + water vapour. c). Acts as dehydrating agent CuSO4 + 5 H2O CuSO→ 4 5H2O White blue hydrous. d). Acts as decarboxylation agent Ca(OH)2 + CO2 → CaCO3 ↓ H2O.
turns milky e). Products of burning candle in air contains H2O & CO2. b) A = NH3 gas Nesseler’s reagent test.
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Paper 21 Chemistry Answers
Section -1
Answer 1 I. 1. Nitrogen tri chloride. 2. Galvanising. 3. Na2O2. 4. 3. 5. Wood tar. II. 1. Because due to the absorbed gases which are trapped in its pores on its surface. 2. Because H2-O2 mixture on burning produces a flame having temperature around 2500oC.
3. Because dilute HNO3 is a powerful oxidising agent and the [O] formed on its decomposition oxidises the hydrogen to H2O & reduced to dilution.
4. Because it is almost as heavy as air. 5. Because process is not wasteful & S of 99.5 % Purity is obtained. III. 1. d. 2. a. 3. b. 4. e. 5. c. IV. 1. Fullerene. 2. Destructive distillation. 3. Nitrogen. 4. Argon. 5. 21 %. V. 1. a. 2. c. 3. c. 4. a. 5. d. VI. 1. CaC2 + N2 CaCN→ 2 +C. 2. 2NH3 + CO2
o110 - 150 C200 atoms⎯⎯⎯⎯→ (NH2)2 CO +H2O.
3. 2Na2O2 + 2H2O 4NaOH +O→ 2(g). 4. 2KMnO4 + 16HCl 2KCl + 2MnCl→ 2 + 8H2O + 5Cl2. 5. 2KBr + Cl2 2KCl + Br→ 2. VII. 1. False. 2. False. 3. True. 4. True. 5. True.
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VIII. 1. Valcanization is a process in which calculated amount of sulphur is added to sticky rubber to harden it . 2. Diamond and graphite.
3. The various process by which the percentage of CO2 in the atmosphere remains constant are collectively represent as carbon cycle.
4. Tartaric acid with sodium bicarbonate 5. Symbiotic bacteria directly absorbs N2 from the soil and converts it to soluble nitrates.
Section - 2 Answer 2 1.
Group 1 2 13 14 15 16 17 18 2nd Period Li Be B C N O F NeElectronic configuration 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
V. e- 1 2 3 4 5 6 7 8Valenacy 1 2 3 4 3 2 1 0
Answer 3 1. i) Granulated zinc dilute HCl. ii) Zn + 2HCl → ZnCl2 + H2(g) iii) Impurities are removed by passing through 3 Washer bottles and a U -tube. Washer bottle (1) - Contains AgNO3 Solution absorbs, Arsine , PH3 Washer bottle (2) - Contains Pb(NO3)2 solution absorbs, H2S Washer bottle (3) - Contains KOH solution absorbs, NO2, CO2, SO2 U tube (4) - Contains anhydrous CaCl2 absorbs Moisture.
iv) No leakage of gas should take place & no flame must be near the apparatus. v) Because it is slightly soluble in water. vi) Impure zinc obtained from molten zinc.
Answer 4 1. a) Because they are sublimable substance.
b) Because ammonium nitrite is unstable it decomposes at room temperature & cannot be stored in the laboratory.
c) N2 obtained in the lab from air is slightly heavier than the N2 obtained from chemical compounds due to the presence of inert gases namely argon (1%).
d) NH4Cl + NaNO2 → NaCl + NH4NO2 NH4NO2 2H→ 2O + H2. e) It is slightly soluble in water.
2. (NH4)2Cr2O7 Cr→ 2O3 + 4H2O + N2(g). Answer 5 1. A: Pb3O4 B: PbO C:PbO D:O2 2Pb3O4 → 6PbO + O2. Yellow 2. A: PbO2 B: PbO C: O2 2PbO2 → 2PbO + O2(g). Original colour: chocolate brown.
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Answer 6 1. a→ (iii) b (iv) c (v) d (ii). → → → 2. 1) 7 horizontal rows 2) 18 vertical columns. 3) The physical and chemical properties of all the elements are the periodic function of their atomic number. 4) Francium. 5) Because the atomic size of fluorine is small and the added electronegative faces strong repulsion from other electronegatives.
6) Cl.
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Paper 22 Chemistry Answers
Section -1 Answer 1 I) i) → C ii) C iii) →A iv) →B v) C. → → II) i) 2KHCO3 + H2SO4 → K2SO4 + 2H2O +2CO2. ii) 2CuSO4 Δ⎯⎯→ 2CuO + 2SO2 + O2 iii)CaCO3 + 2HCl CaCl→ 2 + H2O + CO2 iv) NH4Cl + NaNO2 →NH4NO2 + NaCl v) Fe2O3 + 3H2 → 2Fe + 3H2O. III) i) On heating white sublimate is formed. ii) Lilac colour is imparted to the flame. iii) Hydrogen gas is liberated. iv) Forms a thin coating of calcium carbonate. v) Form CO2 & H2O. IV) i) Hydrogen ii) Ernest Rutherford iii) Protons + neutrons iv) Fluorine v) p-block. V) i) Electronic configuration. ii) Penultimate Shell. iii) Re-distribution of electrons. iv) 17Cl37. v) 2. VI) i) Parallel layers of C atoms in graphite are held by weak forces of attraction or layers slide over one another. ii) Sparkling brilliance. iii) Process of carbonization. iv) Incomplete combustion of carbonaccous fuels. v) Acts as a strong reducing agent. VII) i) → a), ii) → b), iii)→d), iv) c), v) → a). → VIII) i) Diffusion ii) Fractional distillation iii) Solvent extraction iv) Fractional Crystallisation v) Filtration.
Section -2 Answer 2 a) Number of protons i) A 40 – 20 = 20 B 19 – 10 = 9 C 20 – 10 = 10 D 23 – 12 = 11 ii) Metal. Non – metal. Noble gas.
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Metal. iii) D Number of electronegatives = 11 : 2, 8, 1. iv) Ionic bonding. Answer 3 a) i) 4 AgOH → 4Ag + 2H2O + O2 ii) 2Zns + 2O2 → 2ZnO + SO2 iii) 2Pb(NO3)2 → 2PbO + 4NO2 + O2. iv) 3 Fe + 4H2O Fe3O4 + 4H2. v) Mg3N2 + 6H2O 3Mg(OH)→ 2 + 2NH3. b) i) 1) Boyle’s law: Temperature remaining constant the volume of a given mass of any gas is
inversely proportional to its pressure. 2) Charle’s law: Pressure remaining constant the volume of a given mass of any gas is directly proportional to its temperature.
ii) Absolute Zero : The temperature -273oC is called absolute zero. iii) V1 = 512 Cm3. V2 = ? T1 = 273K T2 = 27 + 273 = 300K. By Charles law,
1
1
VT
= 2
2
VT
1 22
1
V × TV = T
= 512 × 300 100273 91
= 562 cm3.
Answer 4 a i) Oxidation – Addition of oxygen to a substance is called oxidation. ii) Reduction – Addition of Hydrogen to a substance is called Reduction.
iii) Redox Reaction – A chemical reaction which involves both oxidation and reduction is called redox reaction.
b) i) Starch iodide paper turns blue – black. Cl2 gas displaces the iodine which reacts with the starch giving a blue – black colour. ii) Moist blue litmus turns red and it bleaches the colour. iii) Forms a greenish yellow solution called Cl2H2O. c) i) S + 2H2SO4 3SO→ 2 + 2H2O. ii) C + 2S → CS2 iii) H2 + S → H2S. iv) Fe + S FeS. →Answer – 5
a) i) 7N14 Electronic configuration 2, 5. ii) CH4 6C12 Electronic configuration 2,4. 1H2 Electronic configuration 1.
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b) i) Pass pure H2 gas over heated Cu (II) oxide and light a flame at the end of the jet tube ‘A’. Heating is stopped slowly. But the H2 is continued to pass slowly till the tube cools down. A reddish brown metallic residue is seen in the hard glass test tube. A colourless liquid collects in the evaporating dish.
ii) CuO + H2 Cu + H→ 2O. iii) The reddish brown residue is metallic copper and the colourless liquid on testing is found to be water. Thus H2 reduces black Cu(II) oxide to metallic Cu and is itself oxidized to water.
Answer 6
a) i) Passage of hydrogen under pressure over powdered coal in presence of a catalyst at a suitable temperature. H2 is used for conversion of coal which contain a lower % of H2 by weight to a product similar to petroleum containing a higher % of H2. ii) During the preparation of bone charcoal from animal bone by destructive distillation the residue obtained when animal bones are strongly heated is porous and black. Hence it is called bone black or ivory black.
iii) Nitrolim is a nitrogeneous fertilizer. Preparation CaC2 + N2
o1000 C⎯⎯⎯→ CaCN2 + C. b) a) Because inactive N2 atmosphere filled in the space above the mercury prevents its vapourisation at high temperature (5000oC) b) Because inactive N2 prevents the oxidation of food and retards bacterial growth. c) Because liquid N2 (-196oC) provides very low temperature. d) For protein synthesis in green plants. Answer 7
i) Procedure : The jar of O2 with its open and downwards is inverted in a trough of H2O and kept for a period of time. Water rises V. slightly up in the trough.
Conclusion O2 is slightly soluble in H2O.
ii) Fishes survive in a bowl of water at room temperature on warming the water the fishes die, because O2 is soluble in water,
The solubility decreases with increases in temperature. iii) a) B.P = -183oC b) F.P = -218.4oC. iv) 4P + 5O2 2P→ 2O5 2+6H O⎯⎯⎯→ 4H3PO4 brilliant phosphoric acid. white fumes.
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