Antilog

Excess of solid AgCl is added to a 0.1 M solution of Br^(c-) ions. E^(c-) for half cell is : AgBr+e^(-) rarr Ag+Br^(c-),E^(c-)=0.095V AgCl+e^(-) rarr Ag+Cl^(c-), E^(c-)=0.222V The value of [Br^(c-)] ion at equilibrium is : [ Given : Antilog (2.152)=142]

The first order rate constant for the decomposition of ethyl iodide by the reaction. C_(2)H_(5)(g) to C_(2)H_(4)(g) + HI(g) at 600 K is 160 xx 10^(-5)s^(-1) , T_(1)= 600K , T_(2)=700K, E_(a) = 209 kJmol^(-1) . logk_(2)= log(1.60 xx 10^(-5)s^(-1)) + (209000Jmol^(-1))/(2.303 xx 8.314J mol^(-1)K^(-1))[1/(600K)-1/(700K)] log k_(2)=(-5 + 0.2041) + 2.5989 = -2.197 = bar3.8080 k_(2)="Antilog" bar3.8030 = 6.36 xx 10^(-3)s^(-1)

Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

Let L denote antilog 320.6 and M denote the number of positive integers which have the characteristic 4, when the base of log is 5, and N denote the value of 49^((1-log_(7)2))+5^(-log_(5)4) Find the value of (LM)/(N) .

Let 'L' denotes the antilog of 0.4 to the base 1024. and 'M' denotes the number of digits in 6^(10) (Given log,02-03 and 'N' denotes the number of positive integers which have the characteristic 2, when base of the logarithm is 6. Find the value of LMN

Consider the atomization of Br_(2)(g) according to the reaction : Br_(2)(g)hArr 2Br(g)" "{"Given antilog "(-7.02)=9xx10^(-4)} If the heat of atomization of bromine gas is 82 Kj/mol, standard molar entropies of Br(g) and Br_(2)(g) are 175 and "245.4 JK"^(-1) respectively, calculate degree of dissociation when the total pressure is 40 atm at 500 K. (assume alpha ltlt1 in your calculation)

At Tl^(+) |Tl couple was prepared by saturating 0.1 M KBr with TlBr and allowing the Tl^(+) from the relatively insoluble bromide to equilibrate. This couple was observed to have a potential of -0443 V with respect to Pb6(2+)|Pb couple in which Pb6(2+) was 0.1 M. Waht is K_(sp) of TlBr? (Report answer in multiplication of 10^(-8) ) (E_(Pb^(2+0/Pb)^(@) = -0.126 , E_(Tl^(+)//Tl)^(@) = -0.336V) (Take antilog(0.5509) = 3.55, (2.303RT)/(F) = 0.059) [Hint: Take Pb as anode]