APPENDIX I Elements, their Atomic Number and Molar Mass

1 261 AppendixElementSymbolAtomicMolarNumberma ss/ (g mol 1) (243) (247) (264) (263) (252) ( ) (223) (269) ( ) (268) ( ) (259) (244) (145) (226)RadonRn86(222) (261) (266) ( ) (277)UnunniliumUun110(269)UnununiumUuu11 1(272) (g mol 1)The value given in parenthesis is the Molar mass of the isotope of largest known , their Atomic Number and Molar MassAPPENDIX I NCERTnot to be republished262 ChemistryCommon Unit of mass and Weight1 pound = grams1 pound = grams = kilogram1 kilogram = 1000 grams = pounds1 gram = 10 decigrams = 100 centigrams = 1000 milligrams1 gram = 1023 Atomic mass units or u1 Atomic mass unit = 10 24 gram1 metric tonne = 1000 kilograms = 2205 poundsCommon Unit of Volume1 quart = litre1 litre = quarts1 litre = 1 cubic decimetre = 1000 cubiccentimetres = cubic metre1 millilitre = 1 cubic centimetre = litre = 10-3 quart1 cubic

2 Foot = litres = quarts = gallonsCommon Units of Energy1 joule = 1 107 ergs1 thermochemical calorie**= joules= 107 ergs= 10 2 litre-atmospheres= 1019 electron volts1 ergs = 1 10 7 joule = 10 8 calorie1 electron volt = 10 19 joule = 10 12 erg = kJ/mol 1 litre-atmosphere = calories = joules = 109 ergs1 British thermal unit = joules= 1010 ergs= caloriesCommon Units of Length1 inch = centimetres (exactly)1 mile = 5280 feet = kilometres1 yard = 36 inches = metre1 metre = 100 centimetres = inches= feet= yards1 kilometre = 1000 metres = 1094 yards= mile1 Angstrom = 10 8 centimetre = nanometre = 10 10 metre = 10 9 inchCommon Units of Force* and Pressure1 atmosphere = 760 millimetres of mercury = 105 pascals = pounds per square inch1 bar = 105 pascals1 torr = 1 millimetre of mercury1 pascal = 1 kg/ms2 = 1 N/m2 TemperatureSI Base Unit: Kelvin (K)K = CK = C + F = ( C) + 32F =*Force.

3 1 newton (N) = 1 kg m/s2, ,the force that, when applied for 1 second, gives a1-kilogram mass a velocity of 1 metre per second.**The amount of heat required to raise the temperature of one gram of water from to Note that the other units are per particle and must be multiplied by 1023 to be Useful Conversion FactorsAPPENDIX II NCERTnot to be republished263 AppendixReduction half-reactionE /VH4 XeO6 + 2H+ + 2e XeO3 + 3H2O+ + 2e 2F + + 2H+ + 2e O2 + H2O+ 8 + 2e 2SO2 4+ + + e Ag++ + + e Co2++ + 2H+ + 2e 2H2O+ + + e Au+ + + 2e Pb2++ + 2H+ + 2e Cl2 + 2H2O+ + + e Ce3++ + 2H+ + 2e Br2 + 2H2O+ 4 + 8H+ + 5e Mn2+ + 4H2O+

4 + + e Mn2++ + + 3e Au+ + 2e 2Cl + 7 + 14H+ + 6e 2Cr3+ + 7H2O+ + H2O + 2e O2 + 2OH + + 4H+ + 4e 2H2O+ 4 + 2H+ +2e ClO 3 + 2H2O+ + 4H+ + 2e Mn2+ + 2H2O+ + + 2e Pt+ + 2e 2Br + + + e Pu3++ 3 + 4H+ + 3e NO + 2H2O+ + + 2e Hg2+2+ + H2O + 2e Cl + 2OH + + + 2e Hg+ 3 + 2H+ + e NO2 + H2O+ + + e Ag+ +2 +2e 2Hg+ + + e Fe2++ + H2O + 2e Br + 2OH + +2e 2Hg + SO2 4+ 4 + 2H2O + 2e MnO2 + 4OH + 4 + e MnO2 4+ + 2e 2I + 3 + 2e 3I + half-reactionE /VCu+ + e Cu+ + H2O + e Ni(OH)2 + OH + + 2e 2Ag + CrO2 4+ + 2H2O + 4e 4OH + 4 + H2O + 2e ClO 3 + 2OH + [Fe(CN)6]3 + e [Fe(CN)]

5 6]4 + + + 2e Cu+ + 2e 2Hg + 2Cl + + e Ag + Cl + + + 3e Bi+ + 4H+ + 2e H2SO3 + H2O+ + + e Cu++ + + 2e Sn2++ + e Ag + Br + + + e Ti3+ + + 2e H2 bydefinitionFe3+ + 3e Fe + H2O + 2e HO 2 + OH + + 2e Pb + + e In + + 2e Sn + e Ag + I + + 2e Ni + + e V2+ + + 2e Co + + 3e In + + e Tl + 2e Pb + SO2 4 + + e Ti2+ + + 2e Cd + + e In+ + + e Cr2+ + + 2e Fe + + 2e In+ + 2e S2 + + e In2+ + + e U3+ + + 3e Cr + + 2e Zn (continued)Standard potentials at 298 K in electrochemical orderAPPENDIX III NCERTnot to be republished264 ChemistryReduction half-reactionE /VCd(OH)

6 2 + 2e Cd + 2OH + 2e H2 + 2OH + + 2e Cr + + 2e Mn + + 2e V + + 2e Ti + + 3e Al + + 3e U + + 3e Sc + + 2e Mg + + 3e Ce half-reactionE /VLa3+ + 3e La + + e Na + + 2e Ca + + 2e Sr + + 2e Ba + + 2e Ra + + e Cs + + e Rb + +e K + + e Li III CONTINUED NCERTnot to be republished265 AppendixSometimes, a numerical expression may involve multiplication, division or rational powers of largenumbers. For such calculations, logarithms are very useful. They help us in making difficult calculationseasy.

7 In Chemistry, logarithm values are required in solving problems of chemical kinetics, thermodynamics,electrochemistry, etc. We shall first introduce this concept, and discuss the laws, which will have to befollowed in working with logarithms, and then apply this technique to a Number of problems to showhow it makes difficult calculations know that23 = 8, 32 = 9, 53 = 125, 70 = 1In general, for a positive real Number a, and a rational Number m, let am = b,where b is a real Number . In other wordsthe mth power of base a is way of stating the same fact islogarithm of b to base a is for a positive real Number a, a 1am = b,we say that m is the logarithm of b to the base write this as balo gm ,= log being the abbreviation of the word logarithm.

8 Thus, we have3223307log 83,Since 28log 92,Since 39125log3,Since 51255log 1 0,Since 71========Laws of LogarithmsIn the following discussion, we shall take logarithms to any base a, (a > 0 and a 1)First Law: loga (mn) = logam + loganProof: Suppose that logam = x and logan = yThen ax= m, ay = nHence mn = = ax+yIt now follows from the definition of logarithms thatloga (mn) = x + y = loga m loga nSecond Law: loga mn = loga m loganProof: Let logam = x, logan = yLogarithmsAPPENDIX IV NCERTnot to be republished266 ChemistryThen ax = m, ay = nHence xx yyamana ==Thereforeaaamlogxylog m log nn= = Third Law : loga(mn) = n logamProof : As before, if logam = x, then ax = mThen ()nnxnxmaa==giving loga(mn) = nx = n loga mThus according to First Law: the log of the product of two numbers is equal to the sum of their , the Second Law says: the log of the ratio of two numbers is the difference of their logs.

9 Thus,the use of these laws converts a problem of multiplication / division into a problem of addition/subtraction, which are far easier to perform than multiplication/division. That is why logarithms are souseful in all numerical to Base 10 Because Number 10 is the base of writing numbers, it is very convenient to use logarithms to the base10. Some examples are:log10 10 = 1,since 101 = 10log10 100 = 2,since 102 = 100log10 10000 = 4,since 104 = 10000log10 = 2,since 10 2 = = 3,since 10 3 = log101 = 0since 100 = 1 The above results indicate that if n is an integral power of 10, , 1 followed by several zeros or1 preceded by several zeros immediately to the right of the decimal point, then log n can be easily n is not an integral power of 10, then it is not easy to calculate log n.

10 But mathematicians havemade tables from which we can read off approximate value of the logarithm of any positive numberbetween 1 and 10. And these are sufficient for us to calculate the logarithm of any Number expressedin decimal form. For this purpose, we always express the given decimal as the product of an integralpower of 10 and a Number between 1 and Form of DecimalWe can express any Number in decimal form, as the product of (i) an integral power of 10, and (ii)a Number between 1 and 10.