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Why is that for both very small as well ...

Why is that for both very small as well as very large (in addition to ordinary) quantities, the physical measurements are usually expressed in scientific notation, with powers of ten ?

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To denote the precision of the measurement.
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Stabilities of alkanes can be compared by converting these compounds to a common product and comparing the amount of the heat given off. One possiblitiy would be to measure the heat of combustion from converting alkenes to xo_(2) and H_(2)O . The heats of combustion are of large values and measuring small difference in these large numbers is difficult. Alkene of the lowest heat of combustion among isomeric alkenes is of the lowest energy and is most stable. Th stability of alkenes is often compared by meansuring the ehat of hydrogenation 9 heat given off, Delta H_(h)^(@) during catalytic hydrogenation. The heat of hydrogenation is in smal number, which provides more accurate energy difference. For a compound containing more than one double bond, Delta_(h)^(@) is the sum of heat of hydrogenation of individual double bonds. For non - conjugated diens, this additive relatioship is found to hold. For conjugated dienes, however, the measured value is slightly lower than expected. Cumulated dienes, which are even less stable than non - conjugated dienes. The more stable is the alkene, lower is the heat of combustion and heat of hydrogenation. More highly substituted double bonds are usually more stable. In case of cyclokanes, compounds having higher angle strin are less stable. If the heat of hydrogenation of cyclooctene is about 23 k cal mo^(-1) , what would be the probable DeltaH_(b) or 1,3,5,7- cyclooctatetraene ?

The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of 10^(-3) to 10^(-4) M.At concentration of 10^(-3) mol dm^(-3) , the magnitude of osmotic pressure of 300 K is : P=10^(-3)xx0.082xx300=0.0246 atm or 0.0246xx1.01325xx10^5=2492.595 Pa At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally. Further polymers have following two types of molar masses : (A) Number average molar mass (barM_n) , which is given by (undersetisumN_iM_i)/(undersetisumN_i) where N_i is the number of molecules having molar mass M_i . (B) Molar average molar mass (barM_m) , which is given by (undersetisumN_iM_i^2)/(undersetisumN_iM_i) Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined. One gram each of polymer A (molar mass=2000) and B(molar mass=6000) is dissolved in water to form one litre solution at 27^@C .The osmotic pressure of this solution will be :

A metal foil is at a certain distance from an isotropic point source that emits energy at the rate P. Let us assume the classical physics to be applicable. The incident light energy will be absorbed continuously and smoothly. The electrons present in the foil soak up the energy incident on them. For simplicity , we can assume that the energy incident on a circular path of the foil with radius 5xx10^(-11) m (about that of a typical atom ) is absorbed by a single electron. The electron absorbs sufficient energy to break through the binding forces and comes out from the foil. By knowing the work function, we can calculate the time taken by an electron to come out i.e., we can find out the time taken by photoelectric emission to start. As you will see in the following questions, the time decay comes out to be large, which is not practically observed. The time lag is very small. Apparently, the electron does not have to soak up energy . It absorbs energy all at once in a single photon electron interaction. If power of the source P is 1.5 W and distance of the foil from the source is 3.5 m, the energy received by an electron per second is

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