Khan Sir Chemistry PDF Notes

Document Details

LeanCouplet

Uploaded by LeanCouplet

null

Khan G. S.

Tags

chemistry atomic structure electronic configuration science

Summary

These notes provide an overview of atomic structure, including different models, electron configurations, and related concepts. Topics covered include atomic models, subatomic particles, and the electronic structure of atoms. The notes may be used for chemistry classes.

Full Transcript

KHAN G. S. RESEARCH CENTRE inkFkks± dk oxhZdj.k (Solid) :– (Liquid) :– (Gas) :– Remark:– = > > = > > = > > (Exp...

KHAN G. S. RESEARCH CENTRE inkFkks± dk oxhZdj.k (Solid) :– (Liquid) :– (Gas) :– Remark:– = > > = > > = > > (Expnsion) = > > (Diffusion) = > > = > > 22 KHAN G. S. RESEARCH CENTRE (iv) (v) (iv) (v) not same 33 KHAN G. S. RESEARCH CENTRE (Atomic Structure)    ATOM  etc.   (i) (ii)  etc     (Nuculius)  (Å) = 10 –10 m (1Å)  (f) = 10 –5 m (1 )  1 (105) Å J. J. Thomson (Watermelon Theory) 44 KHAN G. S. RESEARCH CENTRE (Atomic Model of Rutherford) :– – – + (i) – (ii) – (Positive) (iii) 20,000 –Ray (Nucleous)  (Extraction) Maxwel     55 KHAN G. S. RESEARCH CENTRE Electron J. J. (1897) 9.1  10–31 kg –Ve Proton (1919) 1.6725  10–27 kg +Ve Neutron (1932) 1.6748  10–27 kg No Charge =N>P>E =N>P>E (eº) :– J. J. –31 = 9.1 10 kg (Absolute mass) = (Relative mass) = 0.00054 amu –19 = –1.6 10 1 1 Remark:– 1837 1840 amu (Atomic Mass Unit) – (U):– –12 (C–12) 12 1 amu U 1amu = 1.66×1027 kg (P / H) :– = 1.0072 amu –27 = 1.6725 10 kg –19 +1.6 10 Remarks:– 1 0 n :– 66 KHAN G. S. RESEARCH CENTRE –27 = 1.6748 10 kg = 1.0086 amu =0 Remark:– 0 0 n :–   (Possitron) (e+) :– (Anti-Particle) –31 = 9.1 10 kg –19 = + 1.6 10 Remark:– Anti Particle Atomic Number:– 'Z'  (Z) = (P) Note:– Z=P=e Eg:– 23 39 196 11Na 19K 79Au Z = 11 Z = 19 Z = 79 P = 11 P = 19 P = 79 e = 11 e = 19 e = 79 (Ion):– Electrons 77 KHAN G. S. RESEARCH CENTRE =Z+ = –Ve (Subtract) = +Ve (Add) 24 Eg:– 12 mg ++ Z = P = 12 e = 12 – 2 = 10 26 13 Al+++ Z = P = 13 e = 13 – 3 = 10 16 8 O– – Z=P=8 e = 8 + 2 = 10 35 17 Cl – Z = P = 17 e=Z+ = 17 + 1 = 18 (Atomic Mass) :– = A = N + P/Z A= N + Z = A–Z=N N=A–Z n Radio Active = 1.5 p 26 (A) Al 13 (Z) N = A–Z Z = P = 13 = 26 – 13 = 13 e = 13 (ISOELECTRONIC):– Eg:– 11Na + e = 11 – 1 = 10 12Mg ++ e = 12 – 2 = 10 88 KHAN G. S. RESEARCH CENTRE 13Al +++ e = 13 – 3 = 10 10Ne e = 10 CH4 e = 6 + 4 = 10 (ISO-TOPS):– Eg:– 1 2 3 1H 1H 1H Polonium (Po) 27 Remark:– Eg:– (i) 1 1H – (n = 0) 2 1H – (n = 1) 3 1H – (n = 2) (ii) 235 238 92U 92U n = 143 n = 146 (iii) 12 14 6C 6C n=6 n=8 (i) –14 (C14) (ii) U235 (iii) I131 (iv) Fe59 (v) As74 (vi) Co60 (vii) Na24 (ISO-BAR):– Eg:– 14 14 6C 7N 40 40 18Ar 20Ca (ISO-TONES):– (Iso-tones) Eg:– (i) 14 16 3 4 6C 8O (ii) 1H 2He n=8 n=8 n=2 n=2 (iii) 31 32 15P 16S n = 16 n = 16 (Orbit) (Shell):– (Orbit) K, L, M, N..... 99 KHAN G. S. RESEARCH CENTRE K– Remark:–  2n2 n e = 2n 2 K n=1 e = 2n2 2  12 = 2 L n=2 e = 2n2 2  22 = 8 M n=3 e = 2n2 2  32 = 18 N n=4 e = 2n2 2  42 = 32 (Sub-orbit / Sub-shell) :– s, p, d, f (electron) s 2 p 6 d 10 f 14 s K=2 2 s p L=8 , 2 6 s p d M = 18 , , 2 6 10 s p d f N = 32 , , , 2 6 10 14 10 10 KHAN G. S. RESEARCH CENTRE (ORBITAL):– electron  electrons  s=2 1 p=6 3 d = 10 5 f = 14 7   (Electronic Configuration)  d 4 9 s– 5 10  f 6 13 s– 7 14 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5d 5p 5d 5f 6s 6p 6d 7s 2H = 1s1 2He = 1s2 3Li = 1s2, 2s1 4Be = 1s2, 2s2 5B = 1s2, 2s2 2p1 6C = 1s2, 2s2, 2p2 11 11 KHAN G. S. RESEARCH CENTRE 9F 1s2 2s2 2p5 8 electron (n) = 2 (l) = 1 –1 0 1 (m) = –1, 0, +1 (s) = = –½ Anticlockwise direction Q.N () (P) h h P mv (Valence Electron) (Core Electron) Core Electron Naecleon Note:– 1 8 Remark:– 8 eº – e =3 – e = 10 12 12 KHAN G. S. RESEARCH CENTRE  e– s p– Group–A Eg:– 8O 1s2, 2s2, 2p4 (Group–A) 11Na 1s2, 2s2, 2p6 3s1 (Group–A) e– d f– Group–B Eg:– 26Fe [Ar] 4s2 3d6 (Group-B)  Valence shell or Altimate shell  Penultimate Shell  Anti-Penultimate shell Valence shell Allimate shell Penultimate shell Antipenultimate shell First Shell (Valence):–  Valence Electron Case I :– 1, 2, 3, 4 Eg:– 13Al 2, 8, 3 (Valency = 3) 11Na 2, 8, 1 (Valency = 1) Case II :– Valence electron 5, 6, 7 8 Eg:– 17Cl 2, 8, 7 8–7 = 1 8O 2, 6 13 13 KHAN G. S. RESEARCH CENTRE KHAN G. S. RESEARCH CENTRE Q. CH4 Valency of Carbon = 4 C H4 * Q. CaCO3 Ca CaCo3 = 40 + 12 + 16  3 = 100 40 % of Ca = –— 100 = 40% 100 Q. (NH2 CO NH2) % NH2 CONH2 = 14 + 2 + 12 + 16 + 14 + 2 = 60 14 +14 28 N %= 100 100 46.66 46% 60 60 Q. (Ca3 (PO4)2) Oxizen % Ca3 (PO4)2 A = 40  3 + 2 (31 + 4  16) = 120 + 2 (31 + 64) = 120 + 190 = 310 P = 62 62 % of P = ––– 100 = 20% 310 128 % of O = 100 41.29% 310 Q. 10gm CaCO3 CO2 Soln:– CaCO3 —— CaO + Co2   40 + 12 + 16  3 12 + 16  2 Q. 20 gm CaCO3 gm CaO CaCO3 = 100 Q. 20 gm gm 2H2 + O2 —— 2H2O Q. 60 gm C O2 CO2 C + O2 —— CO2   12 32 —— 12 + 32 = 44 14 14 KHAN G. S. RESEARCH CENTRE =2 Q. 22 = 2 = 2  22 = 44 Q. H2SO4 H2SO4 = 2 + 32 + 16  4 = 34 + 64 = 98 98  = 49 2 2 Q. HCl HCl = 1 + 35.5 = 36.5 36.5  = 18.25 2 2 Q. 164 164 = 82 2 (i) (i) (ii) (ii) Na 2, 8, 1 Na+ 2, 8 (iii) (iii) (iv) (iv) (Redical):– (K+) (F–) (Na+) (Cl–) (Ag+) (ClO3–) (H+) (Br–) (NH4+) (I–) (Co+) (NO3–) (Co++) (NO2–) (Cu++) (N– – –) (Fe++) (OH–) (Fe+++) (S– –) (Mg++) (SO3– –) (Mn++) (SO4– –) (Hg++) (S2O3– –) (Al+++) (O– –) (Cr+++) (CO3– –) (Pb++++) (CrO4– –) (SiO3– –) 15 15 KHAN G. S. RESEARCH CENTRE (1) Aluminium Silicate Al+++ SiO3– – Al2 (SiO3)3 (2) Aluminium Cromate Al+++ CrO4– – Al2(CrO4)3 (3) Aluminium Oxide Al+++ O– – Al2O3 (4) Potassium Chlorate K+ ClO3– KClO3 (5) Potacium Iodide K+ I– KI (6) Sodium Hidroxide Na+ OH– NaOH (7) Sodium Carbonate Na+ CO3– – Na2CO3 CHEMICAL BONDING 8 (i) (Ionic or Electrovalent Bond) (ii) (Co-Valent Bond) (iii) (Coordinate Bond) (i) (Ionic Bond) :– Blood electrons e– Eg :– Na+ Cl– Mg++ O– – = NaCl = MgO (i)   (C6 H6) (C2H5OH) (ii) (Crystaline) (iii) (Condition for Ionic Bond) (i) Bond (ii) Bond (Ionization Potential / Engery) :–  Ionization Potential  16 16 KHAN G. S. RESEARCH CENTRE  (ii) (Electron Affinity) : Electron Affinity (Cl) (iii) (Electron Negativity) : (Pair) Electron Negativity Electro Negativity * (F) (2) Co-valent Bond Bond):– Bond Bond Bond Co-Valent bond (i) Single Bond (ii) Double bond (iii) Triple Bond 17 17 KHAN G. S. RESEARCH CENTRE (i) Single Bond :– Bond Eg:– H2O H — H–O–O H2O CH4 H H H C H— H–C–H H CH4 H NH3 H N H— H–N–H H H (ii) Double Covalent Bond / e– O O O C O O = O, O2 O = C = O, CO2 (iii) Triple Covalent Bond / 3e– N N N N, N 2 15P – 2, 8, 5 P P P P2 Co-valent Bond (i) (Soliable) (ii) (M.P.) (B.P.) (iii) — >=>– — >=>– — –>=> Remarks:– (Compound) Ionic Bond Co-valent Bond Ex:– NaOH Na O H 18 18 KHAN G. S. RESEARCH CENTRE FON) Bond Hydrogen F, O, N Bond Bond dotted line (.....) Eg:– HF, H2O, NH3 Hydrozen bond = HF > H 2 O > NH 3 Note:– (HF) HF (Element) :– (Periodic Table) (Metal) (Non-metal) (Metalloid) Chemistry (Fe) (O) (Hg) (Si) 1. 2. 3. (Quick Silver) 4. 5. 6. 7. 8. 9. 19 19 KHAN G. S. RESEARCH CENTRE Eg :– etc. (Non-Metal) 1. 2. 3. 4. 5. 6. Eg:– etc. Note:– METALLOID (Hole) Memory Card, SIM, PCB Eg:– etc. Note:– Remarks:– O > SI > Al > Fe > Ca > Na  Al > Fe > Ca > Na  O > C > H > N > Ca  Ca (Mn) (COMPOUND):– Eg:– H2O, CH4, CO2, SO2, CFC etc. (ORGANIC) :– Eg:– CO2, CH4, C2H5SH L.P.G. 20 20 KHAN G. S. RESEARCH CENTRE (INORGANIC) : Eg:– H2O, SO2, N2O, NO2 N2O Laughing Gas Co2 = 2 (C, O) = 3 (C = 1, O = 2) = 1 : 2 (C = 1, O = 2) (MIXTURE) Eg:– etc. (HETROGENOUS MIXTURE) Eg:– (HOMOGENOUS MIXTURE) Eg:– (Solution) (SOLVENT) (SOLUTE) Eg:– etc. (Dilute Solution) (Consantrate Solution).............. 21 21 KHAN G. S. RESEARCH CENTRE 1. (Unsaturated Solution):– 2. (Saturated Solution):– 3. (Super Saturated):– Remark:– Example Cold Drink ——————— 100 Q. 5 kg 2 kg = (2 kg) = (5 kg) 2 kg 20 = ×100 = 40 5 kg * (Dispersion):– Eg:– * (Suspension):– 10–5 m (Filter Paper) Eg:– * (Colloid):– 10–7 m 10–5 m Microscope 22 22 KHAN G. S. RESEARCH CENTRE Eg:– Blood, Note:– (Dialysis) Remark:– * (Emulsion):– Eg:– * (Solution) 10–7 m Microscope Eg:– * (zig – zag Random) Brownium movement Brownium.......... MOLE CONCEPT  = ————— Q. 10 l 45 g =? = ————— 45 45 45 1 = = = = 0.25 C6 H13O6 72 +12 + 96 180 4 Q. 6l 116 gm =? NaCl 23 + 35 = 58 116 2 58 23 23 KHAN G. S. RESEARCH CENTRE (1l) Q. 3l 12 kg 12 4 kg 3 Or 3l ——— 12 kg 12 1l ——— 4 kg 3  (i) (Molarity) : Q. 12 mole 3l 12 4 mole/l 3 Q. 4l 116 gm 116 116 2 NaCl 58 2 1 0.5 4 2 Q. 5l 180 gm 180 180 180 1 C6 H12 O6 72 12 96 180 1 0.2 m/l 5 * MOLALITY Mole kg Q. 3 mole 15 kg 3 = 0.2 m/kg 15 5 Q. 20 kg 5 kg 5 kg 5000 58 58 24 24 KHAN G. S. RESEARCH CENTRE mole kg 5000 58 (ACID & BASE) (ACID) :– (1) (2) (3) (4) Eg:– Zn + H2SO4 —— ZnSO4 + H2 (5) pH 7 Ex:– HCl, H2SO4, HNO3, CH3COOH, HCOOH  1. Eg:– HCl, HF 2. Acid:– Eg:– H2SO4, HNO3, CH3COOH Remark:– 50% Eg:– CO2, SO2, NO2  (CO) 50%  H+   (BASE):– 1. 2. Paper 3. 4. 5. PH 7 Eg:– Ca(OH)2, Mg(OH)2, NaOH Note:– (Alkali) 25 25 KHAN G. S. RESEARCH CENTRE Eg:– NaOH, KOH, Mg(OH)2, Ca(OH)2 Al(OH)3 Remark:– HI > HBr > HCl > HF HClO3 > HClO2 > HClO > HI > HBr > HCl > HNO3 > H2SO4 > CH3COOH Note:– KOH > NaOH > Ca(OH)2 > Mg(OH)2 > NH3 > NH3OH pH (pH Scale) (Power of Hydrogen)  pH Scale  pH Scale 0 14  7 pH  7 pH  7 pH  pH 7 Note:– pH 5.5 * pH – 2.2 – 2.4 – 2.8 – 4 – 5 – 5.5 – 5.5 – 7.5 – 6.5 – 7.5 – 7.4 – 6.4 0 Acid 7 Base 14 Note:– PH 7 PH 7 Buffer Solution PH  PH 7 26 26 KHAN G. S. RESEARCH CENTRE Q. PH (OH–) 10–4 PH+ + POH– = 14 PH+ = 14 – POH– = 14 – (– log [OH–]) = 14 – (– log 10–4) = 14 – (+ 4) = 10 Ans. Q. 10–13 PH POH– = –log [OH–] = –log 10–13 = –(–13) = 13 pH+ + POH– = 14 N Q. HCl pH 1000 1 3 [H+] = 10 100 pH = – log [H+] = – log [10–3] = – (–3) = 3 (SALT) HCl + NaOH —— NaCl + H2O Eg:– 1. (Normal Salt):– Eg:– Na2SO4, NaCl, CuSO4 2. (Acidic Salt):– Eg:– NaHCO3 3. (Basic Salt):– OH Eg:– Ca(OH) Cl 4. (Dboule Salt):– Eg:– K2SO4.Al2 (SO4)3. 24 H2O 5. (Mixed Salt):– 27 27 KHAN G. S. RESEARCH CENTRE Eg:– Ca(OCl) Cl or CaOCl2 6. (Complex Salt):– Eg:– Ag Na(N)2 – K4 Fe(CN)6 – (Radio Activity)           – – + U 92  – 2– 4 92 U 235 90 231 92 U 238 90 A 234 88 B230 86 C 226 84 D 222  – 92 U 235 93 235 92 U 235 93 A 235 94 B235 95 C 235 96 D 235  – 28 28 KHAN G. S. RESEARCH CENTRE –,  92 U 235 93 235 Ratio - Active Decay Radio Active  (Half life Period):– Q. 140 30 gm 15 gm Ans. 140 Q. 1600 12 gm 3 gm Ans. 3200 yrs. Q. 1600 125% Ans. 4800 yrs. 1 Q. 1600 16 Ans. 6400 yrs. Q. Active 2 4 Ta 1.44 t ½ Ta = t½ = Q. Active 100 Ta = 1.44  t½ = 1.44  100 = 144 Days Q. Radio Active 2.88 Ta t½ = 1.44 2.88 = 2 Days 1.44  Radio Activity  Radio Activity  Radio Active (G. M. Counter) 29 29 KHAN G. S. RESEARCH CENTRE  Radio Active Neutron 1.5 Radio Active Pr oton  Radio Active (82Pb206) (Proof) 206 82Pb P = 82 n = 206-82 = 124 n 124 = 1.5 p 82 Radio Active    1. 2. (2He4) (He2+) 3.  4. 1 9 10 10 30 30 KHAN G. S. RESEARCH CENTRE  (Nuclear Fusion):–     (Fission)    (Atom Bomb)   18 May 1974  11 13 May 1998 98  Yellow Cake   (BARC Bhabha Atomic Research Centre) 1956  (Safety wall) :– 6-10 m 31 31 KHAN G. S. RESEARCH CENTRE (Moderator) (D2O) (Moderator) Remark:– (Coolant):– Remark:– (Zr) 1972 (USA ) (CATALYST)  Eg:– MnO2, Fe, V2O5 Eg:– C2 H5OH, Eg:– (Mo) 32 32 KHAN G. S. RESEARCH CENTRE (Activation Energy):– 1. 2. (HNO3) H2SO4 3. 4. 5. 6. 7. 8. Acid  (Physical Change) Eg:– etc. (Chemical Change) Eg:– etc. (Reactant) (Reactant) (Product) Remark:– Na + Cl —— NaCl Reactant Product 33 33 KHAN G. S. RESEARCH CENTRE 1. (Exothermic Reaction):– (System) Eg:– 1. CH 4 2O2 CO2 2H 2 O Heat 2. 3. CO2 4. 2. (Endothermal Reaction):– System Eg:– 1. N 2 O2 2NO 2. 3. 3. (Combination Reaction):– CaO H 2O Ca(OH) 2 C O2 CO2 2Ca O 2 2CaO 4. (Discomposition Reaction):– Eg:– CaCO 3 CaO CO 2 2H 2 O 2H 2 O2 5. (Displacement Reaction):– Zn + H 2SO 4 ZnSO 4 H2 Zn + CaSO 4 ZnSO 4 Ca Fe + CaSO 4 FeSO 4 Ca 6. (Double Displacement Reaction):– AgCl + NaNO3 —— AgNO3 + NaCl 34 34 KHAN G. S. RESEARCH CENTRE 7. (Reverssible Reaction):– Eg:–  H2 + I2  2HI N2 + O2  2 NO 8. (Irrevessible Reaction):– Eg:– C O2 CO 2 2Mg O 2 2MgO 9. (Nutrification Reaction):– Eg:– HCl NaOH NaCl H 2 O H 2SO 4 2KOH K 2SO 4 2H 2 O (Periodic Table)   Triad  Newland Note:–  (PT) 7 9 P.T. 1. 2. P.T. 3. P.T.  P.T.   35 35 KHAN G. S. RESEARCH CENTRE (PERIOD):– 7 1 2 3 4 5 6 7 (Group):– (Vertical) 18 BLOCK:– 4–Block s, p, d f Block 1–2 13 – 18 Block 3 – 12 s p d f  36 36 KHAN G. S. RESEARCH CENTRE   3–12 d-Block Block (Transition Metal) Block etc.  13 – 18 p - Block Block  16 Eg:– Sulfer  17 Eg:– Cl, F, I  18 Eg:– He, Ne, Ar, Kr, Xn, Rn  He:– He  Ne:– Ne  Ar:– Ar  Xn:– (Xn) Xn Stranger Gas  Rn:– Note:– (Flash light) Mg F-Block:– 1. Lenthanide Series:– 14 58 – 72 2. Actinide Series:– 90 - 103 14 f-Block (Inner Transition Metal) f-Block etc   11Na 17 Cl 37 37 KHAN G. S. RESEARCH CENTRE Q. 22 Period Group Block Period = 4 Group = 22 – 19 = 3 (3 + 1) = 4 Block = d-Block Q. 26, 29, 40, 46, 43, 56, 60 Periodic Table 26 29 40 Period = 4 Period = 4 Period = 5 Group = (26 - 19) Group = (29 - 19) Group = (40 - 37) = 7+1=8 = 10 + 1 = 11 = 3+1=4 Block = d-Block Block = d-Block Block = d-Bock 46 43 56 Period = 5 Period = 5 Period = 6 Group = (46 - 37) Group = (43 - 37) Group = (56 - 55) = 9 + 1 = 10 = 6+1=7 = 1+1=2 Block = d-Block Block = d-Block Block = s-Bock 1. Green = Cr, Fe++ (Crow Eg:– FeSO4.7H2O 2. Blue = Cup (cup = Copper) Eg:– CuSO4.5H2O 3. = Mn Eg:– KMnO4 4. = Fe+++ (Gasses Law) (Intermolecular Force) * 1. (Mono-Atomic Gas):– Eg:– He, Ne, Ar etc. 2. (DI-Atomic Gas):– Eg:– O2, Cl2, H2, CO etc. 38 38 KHAN G. S. RESEARCH CENTRE 3. (Tri-Atomic Gas):– Eg:– CO2, SO2, NO2, O3 4. (Poly-Atomic Gas):– Eg:– CH4, NH4 * (IDEAL GAS):– Remark:– CO2, H2 N2 * (REAL GAS):–  1. (Pressure) 'P’ 2. (Temperature) 'T' 3. (Volume) 'V' Trick:– T.V. T V T = V  constant T = Constant V T1 T2 V1 T2 Q. 15ºC 360 ml 400 ml T1 T2  V1 T2 15 273 T2 360 400 32 10 288 400 T2 360 9 39 39 KHAN G. S. RESEARCH CENTRE T2 = 320 K T2 = (320 – 273)ºC = 47ºC T2 47º C Q. 27ºC 200 ml Oº T1 T2 V1 V2 27 273 0 273 200 V2 91 273 200 V2 = = 91  2 = 182 Ans. 300 Trick:– VIP Boy P 1 V P V Constant V P V  P = Constant V1 P1 V2 P2 Q. 700 mm 500 ml 100 ml P1V1 = P2V2 700 500 = P2 100 P2 = 3500 Pressure = 3500 – 700 = 2800 mm Q. 750 mm 120 ml 760 mm P1 = 750 mm 40 40 KHAN G. S. RESEARCH CENTRE V1 = 120 ml P1 V1 = P2 V2 P2 = 760 V2 = ? 3 19 6 38 750 120 = 760 x 2250 x 118.42 19 V2 = 118.42 ml PT P = T  Constant P Cons tan t T P1 P2 T1 T2 Q. 0ºC 120 mm 27ºC P1 = 120 mm P2 = ? T1 = 0ºC = 273K T2 = 27ºC = 273 + 27 = 300 P1 P2 T1 T2 120 P1 273 300 40 120 300 12000 P2 131.86 mm 273 91 91 * P1V1 P2 V2 T1 T2 Q. 27ºC 760 mm 50 ml 207ºC 25 ml 41 41 KHAN G. S. RESEARCH CENTRE Vn V = n  Constant V constant n V1 V2 n1 n2 Q. 200 ml 40 molls 60 Soln.: V1 = 200 ml n1 = 40, n2 = 60 V2 = ? V1 V2 n1 = n2 * PV nRT R= n- R = 8.314 Joule / Mole-Kelvin Note:– * STP (Standered Temprature & Pressure) STP 0ºC 1 atm * NTP (Normal Temperature & Pressure) NTP 20ºC 1 atm STP 56 gm CO STP P = 1 atm = 56 gm T = 0ºC = 273 K 56 56 56 Mole 2m CO 12 16 28 PV = nRT 1V = 2  8.314  273 Q. NTP 132 gm CO2 NTP P = 1 atm T = 20ºC = (273 + 20) = 293 K 42 42 KHAN G. S. RESEARCH CENTRE 132 132 Mole 3 CO 2 44 Mole = 3 PV = nRT nRT V = P 3 8.314 293 = 7308.006 Ans. 1 * (Diffusion):– * 1 V M V1 M2 V2 M1 Q. 400 ml H2 Pipe O2 * (Mole Concept) 1 mole 1 Mole 22.4 l 1 Mole 6.022 10 23 ( ) 1 Mole Q. 64 gm O2 litre  = 64 cm = 16  2 = 32 43 43 KHAN G. S. RESEARCH CENTRE 64 Mole = = 2 32 1. Mole = 2 2. = 2  1 Mole (Mole  No.) = 2  6.022  1023 = 12.044  1023 3. = 222.4 = 44.8 l Q. 67 l (N2) N2 = 214 = 28 = 67 l 1 mole = 22.4l  22.4 l 1 mole 1 67 67 l 3 mole 22.4 Mole (1) = Mole  = 3  28 = 84 gm Ans. (2) = Mole  = 3  6.022  1023 = 18.066  1023 Ans. Q. 45 l 12 gm 1 = 22.4 l 45l —— 2 mole 12 6 Ans. 2 Q. 12  1023 4 gm Q. 196 gm H2SO4 OXIDATION & REDUCTION Oxidation:– Eg:– (1) Na Na (2) Mg M 44 44 KHAN G. S. RESEARCH CENTRE Reduction:– Reduction Eg:– (1) O O (2) Cl Cl Trick:– LEO GER Loss of e – Gain of e – is oxidation is Reduction * Oxidation Numbers:– 0.N Na 0 Cl 0 Cl– –1 H+ +1 NaCl 0 CaCO3– – –2 * (1) +1 –1 (2) (0) (3) –2 (4) 0. N. (5) 0.N. –½ Eg:– KO2 (6) IA +1 (7) IIA 0.N. +2 (8) VIIA 0.N. –1 OXIDATION Eg:– (1) 2Ca + O2 ——— 2CaO  Ca 45 45 KHAN G. S. RESEARCH CENTRE (1) 2Na + Cl2 ——— 2 NaCl  Na  Oxidation No. o o + – 2Na + Cl2 ———— 2NaCl Oxidation REDUCTION 1. Reduction 17 Cl Cl – 2. Reduction 2H 2 O2 2H 2 O Re duction 3. Reduction o – 2Na + Cl2 ———— 2NaCl Reduction Redox Reaction:– Oxidation Reduction Redox Reaction o o + – 2Na + Cl ———— 2NaCl Reduction Oxidation o o + – H2 + O2 ———— 2H2O Reduction Oxidation (Oxidising Agent) : Electron Oxidation 46 46 KHAN G. S. RESEARCH CENTRE Note:– Oxidising Agent Eg:– Na+ Cl– 2, 8, 1 2, 8, 7 Oxidation Oxidising Agent (Reducting Agent):– electron Reducing Agent Note:– Reducing Agent Eg:– K+ Cl– 2, 8, 8, 1 2, 8, 7 Reducing Agent Reduction (WATER) H2O (Soft) (Hard) H2O H2O (Soft Water):– (Hard Water):– 1. Ca Mg Bicarbonate 2. Ca Mg Note:– (Na2CO3) Note:– (D2O) 20 (Heavy Water) (Distilled Water) 0 P.P.M. (Partical Per Million) 300 – 600 PPM 47 47 KHAN G. S. RESEARCH CENTRE Note:– (Vapourisation):– (Condensation):– (Distallation Process):– Vote:– Reverse Osmosis (Fractional Distallation):– (Sublimation):– Note:– (ORES & METALLURGY) (Mineral):– (Ores):– Remark:– 48 48 KHAN G. S. RESEARCH CENTRE (ROASTING) (Calcination):– MATRIX / GANG Gang Flux Flux (Slag):– (Gang) = Gang + Flux Si SiO2 + CaO —————— SiCaO3 (Flux) 49 49 KHAN G. S. RESEARCH CENTRE (A) (1) (Al2O3.2H2O) (2) (Al2O3.H2O) (3) (Al2O3) Note:– (4) (Na3AlF6) Note:– (1) (2) Al (3) Al (B) (1) (Cu FeS2) (2) (Cu2S) (3) (Cu2O) (4) (Cu(CO)3)2. Cu(OH)2 Cu CO3 Cu (OH)2 Note:– (C) (1) (CaSO4.2H2O) (2) (CaSO4.½H2O) (D) (1) (MgSO4.7H2O) (E) (1) (KCl) (2) (KNO3) (F) (1) (NaCl) (2) (Na2CO3.10H2O) (3) (NaNO3) (4) (Na2B4O7.10H2O) (G) (OP) (1) (PbS) (H) (1) (Ag2S) (I) (1) (ZnS) (2) (ZnO) (J) (1) (HgS) (K) (1) (Fe3O4) (2) (Fe2O3) (3) (FeCO3) (4) (FeS2) (L) (BaCO3) (M) Note:– Hope Metal Note:– 50 50 KHAN G. S. RESEARCH CENTRE (ALLOY)      (1) (Brass) / Cu (70%) + Zn (30%) (2) (Bronze) Cu (90%) + Sn (10%) (3) Cu (60%) + Zn (20%) + Ni (20%) (4) Cu (90%) + Al (10%) (5) Cu (90%) + Zn (2%) + Sn (8%) (6) Pb (82%) + Sb (15%) + Sn (3%) (7) Solder Sn (67%) + Pb (33%) (8) (99%) + C (1%) STEEL:–  0.5 – 1.5%   Alloy Steel   Fe, Cr, Ni C (Fe Cr Nic)   (1) LD Process (2) Open Hearth Process (3) Process (Vital Force Theory):– 51 51 KHAN G. S. RESEARCH CENTRE 1828 NH 4CNO NH 2CONH 2 4 (Allotrope):– (1) SP3 (1) SP2 (2) (2) (3) (2) Note:– (ISOMERISM)     (C4H10) ISO – (C4H10) CH3 – CH2 – CH2 – CH3 CH3 – CH – CH3  CH 3 (Hydro-Carbon):– CH4, (C2H2), (C6H14) 52 52 KHAN G. S. RESEARCH CENTRE Alkane CnH2n+2 Alkene Alkyne CnH2n CnH2n–2 Single Bond Oil CnH2n+2 'n' (n) Alkane (CnH2n+2) 1. CH4 2. C2H6 3. C3H8 4. C4H10 5. C5H12 6. C6H14 10. C10H22   LPG (Liquified Petrolium Gas):– CNG (Comporessed Natural Gas):– 85% Eco-friendly Double Bond or Triple Bond (Alkene):– Hydrocarbon Double Bond 53 53 KHAN G. S. RESEARCH CENTRE (Alkone) CnH2n n=1 Alkane (CnH2n) n=1 CH2 n=2 C2H4 n=3 C3H6 n=4 C4H8 n=4 C5H10 n = 6. C6H12 n=7 C7H14 n=8 C8H16 n=9 C9H18 n = 10 C10H20 (Alkyne):– Hydro-Carbon Triple Bond Akyne (CnH2n–2) n=1 Alkane (CnH2n–2) n=1 CH0 n=2 C2H2 n=3 C3H4 n=4 C4H6 n=5 C5H8 n = 6. C6H10 n=7 C7H12 n=8 C8H14 n=9 C9H16 n = 10 C10H18 (FUEL) (Ignesion Temperature) :–  CNG, LPG, Petrol,  54 54 KHAN G. S. RESEARCH CENTRE 90–95% = CO + N2 Water Gas (Coal Gas):– 50% (COAL) 1. 90–95% 2. 88% 3. 78% 4. 50 – 60%     Crude Oil   1 159 litre 55 55 KHAN G. S. RESEARCH CENTRE * (Calorific Value):– 17 kg j/g 30 K-J/g 33 48 50 LPG 50 Methane (CNG) 55 Hydrozen 150 Note:– Hydrozen (–256ºC)  Hydrozen (Knocking):– TEL:– Anti-knocking Agent 1. (Thrmo Plastic):– Thrmo Plastic Eg:– PVC (Poly Venile Cloride) Polythine Polystrin, Teflan Teflon:– (Non-Stick) 56 56 KHAN G. S. RESEARCH CENTRE * Cover etc. * Thrmo Setting Plastic Eg:– Note:– Remark:–      10%   Note:–  [(KOH) ] 57 57

Use Quizgecko on...
Browser
Browser