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yes i m very much sure.

Ka for NH₃⁺ is 10^-10 whereas for -COOH its 10^-3

NH₃⁺CHRCOOH------>NH₃⁺CHRCOO^- + H⁺

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this is becoz we can only calculate ph frm 1 to 14 practically by using conventional ph meter.

ans is a

becoz -COOH is acidic group it has to be fst then Z which is close to acidic part then y

yes it is.

bonding in this involves 5pz of xe and 2pz orbital of the 2 f atoms . these three orbital combine to give three molecular orbital, one bonding , one nonbonding and one antibonding. thus 2 electron occupy the bonding molecular orbital and this pair of electron is responsible for binding all the three atoms . the remaining 2 electron goes to the nonbonding MO which is situated mainly on f and confer to ionic character

Theoretically, value of pH could range from - infinity to + infinity but practically the limits in aqueous solutions are from -1.2 to +15

pH is basically not equals to -log[H+] bt it actually depends upon the thermodynamics activities rather than [H+] ie on log aH+ where aH+ =[H+]F+ . in dilute solution F+ is near enough to unity and thus aH+ =[H+]. thus ph of 10m solution can't be calculated n its near to zero

a control is a standard condition which includes all the factor thought to be significant, except one, so that details of its effect may be discovered by comparison.

Effects of pH

Enzymes are affected by changes in pH. The most favorable pH value - the point where the enzyme is most active - is known as the optimum pH. This is graphically illustrated in Figure 14.

Extremely high or low pH values generally result in complete loss of activity for most enzymes. pH is also a factor in the stability of enzymes. As with activity, for each enzyme there is also a region of pH optimal stability.

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This Post 5 points (1 Olaaa!! Perrrfect answer.

in 1 votes ) [?]

The enzyme used in this lab exercise is amylase, which is commonly found in saliva and germinating seeds. It catalyzes the breakdown of starch. When amylase reacts with starch, it cuts off the disaccharide maltose (two glucose molecules linked together). As the reaction progresses, less starch will be present and more sugar (maltose) will be present. In lab, the activity of amylase was observed by using iodine. Remember, iodine reacts with starch to form a dark brown/purple color. As amylase breaks down starch, less and less starch will be present and the color of the solution (if iodine is added) will become lighter and lighter. The color change was observed using spot-plates as illustrated on the diagram below.

jus revise wat ever u have done till now..n don't loose your cool in the exam

well try my technique...do inorganic in the form of questions...

pick -up ne of the objective books like krishna,op agarwal , or dinesh...read all the questions for 3 to 4 times..n u 'll c that u r through with inorganic..

in sulphonation , the energy barrier on either side of the carbocation ArH(S0₃^-)

must be roughly of same height some ion go on one way and some ion go on other way. giving the reversible nature to the reaction

dear your question is not complete. bt i knw that this is rorkee 1992 problem.

H₂S--------> H⁺ + HS^- K_a1= 10^-7

HS---------> H⁺ + S^2- K_a2 =1.3 x10^-13

HCl----> H⁺ + Cl^-

now due to common ion effect the dissociation of H₂S is suppressed and the [H⁺] in the solution is due to HCl

K_a1 = [H⁺][HS^-]

---------------

[H₂S]

10^-7 =[0.3][HS^-]/[0.1]

Therefore [HS^-]= 3.3 x10^-8 M

Ka₂ =[H⁺] [S^2-]/[HS^-]

K_a1 x K_a2=[H⁺]²[S²]/[H₂S]

Placing Ka1 and Ka2 in the obove expression we get [S^2-] as 1.4x10^-20 M

ans ) fst ans is benzaldehyde1705cm- is a peak for C=O stretch.

2820 and 2710 is aldehydic C-H stretch doublet due to fermi resonance.

above 3000 aromatic C-H stretch

refer spectroscopy by ps kalsi or ELIEL.

well this is not a part of jee syllabus.

2-bromo anisole ----- NaNH2/NH3------>

it w'll give 2,3 dehydroanisole and then m-anisidine