94420 + Interview Questions in Chemistry in Class 12

1.	Carbon cannot be used in the reduction of Al_2 O_3 because
Answer» it is an EXPENSIVE PROPOSITION the ENTHALPY of FORMATION of `CO_2` is more than that of` Al_2 O_3` pure carbon is not easily available the enthalpy of formation of `Al_2 O_3` Answer :D

Discussion

2.	Carbon can reduce ferric oxide to iron at a temperature above 983 K because
Answer» carbon monoxide formed is thermodynamically less satble than ferric oxide carbon has a HIGHER affinity TOWARDS oxidation than iron. free energy change for the formation of carbon dioxide is less NEGATIVE than that for ferric oxide. iron has a higher affinity towards oxygen than carbon. Answer :B

Discussion

3.	Carbon burns in air and forms two oxides CO and CO_2.This shows that carbon has :
Answer» Two ALLOTROPIC forms Two oxidation states Two isotopes 4 ELECTRONS in VALENCE shell Answer :B

Discussion

4.	Carbon black, which is used in making printer's ink is obtained by the decomposition of
Answer» Acetylene Carbon-tetrachloride Benzene Methane Answer :D

Discussion

5.	Carbon black, which is used in making printer's ink, is obtained by decomposition of
Answer» Acetylene Benzene Carbon tetrachloride Methane Solution :`CH_4+O_2underset"Limited supply of air"oversetDeltatoC+2H_2O` It CONTAINS 98-99% carbon. It is USED in MAKING black ink, PAINTS and shoe polishes

Discussion

6.	Carbonblackis obtained whenmethaneis :
Answer» heatedin absenceof air Heated in PRESENCEOF NITROGEN heatedin presenceofammonia heatedwithsteam ANSWER :A

Discussion

7.	Carbon atoms in the compound (CN)_(4)C_(2) are
Answer» SP HYBRIDIZED `sp^(2)` hybridized sp & `sp^(2)` hybridized sp, `sp^(2)` & `sp^(3)` hybridized Solution :

Discussion

8.	Carbon atoms in fullerene with formula C_(60) have
Answer» `SP^(3)` hybridised sp hybridised `sp^(2)` hybridised partially `sp^(2)` and partially `sp^(3)` hybridised Answer :B::C::D

Discussion

9.	Carbon atoms in diamond are bonded with each other in configuration :
Answer» planer linear octahedral tetrahedral Answer :D

Discussion

10.	Carbon atoms in fullerence with formula C_(60) have………
Answer» `SP^(3)` hybridised sp hybridised `sp^(2)` hybridised partialy `sp^(2)` and PARTIALLY `sp^(3)` hybridised Answer :C

Discussion

11.	Carbon atoms in diamond are bonded to each other in configuration which is :
Answer» Planar Linear Tetrahedral Octahedral Answer :C

Discussion

12.	Carbon atoms bearing the -OH group in phenols and CH_(2)=CH-OH are
Answer» both hybridised `SP^(2)` both hybridised `sp^(3)` `sp^(2) and sp^(3)` hybridised RESPECTIVELY. `sp^(3) and sp^(@)` hybridised respectively. ANSWER :A

Discussion

13.	Carbon atom of carbonyl gp. In aldehyde is of :
Answer» `1^@` `2^@` `3^@` None Answer :A

Discussion

14.	Carbon atom of carbonyl group In a ketone is of :
Answer» `1^@` `2^@` `3^@` None Answer :B

Discussion

15.	Carbon atom of carbonyl group In aldehyde is of :
Answer» `1^@` `2^@` `3^@` None Answer :A

Discussion

16.	Carbon atom in methyl carbocation contains how many pairs of electrons?
Answer» 8 4 3 5 Answer :C

Discussion

17.	Carbon atom in carboxyl group is
Answer» SP-hybridised state `sp^(2)` - hybridised state `sp^(3)` - hybridised state `sp^(3)` -d- hybridised state Answer :B

Discussion

18.	Carbon atom holding halogen in aryl halides is:
Answer» `sp^2-Hybridized` `sp^3-Hybridized` `sp-Hybridized` `sp^3d-Hybridized` ANSWER :A

Discussion

19.	Carbon atom holding halogen in aryl halides is :
Answer» `SP^(2)` HYBRIDIZED `sp^(3)` hybridized sp-hybridized `sp^(3)` d-hybridized Answer :A

Discussion

20.	Carbon and oxygen forms two compound . Carbon content in one of them is 42.9% while in the other is 27.3% . Thegiven data is in support with
Answer» LAW of definite proportions law of RECIPROCAL proportions law of MULTIPLE proportions law of CONSERVATION of mass Answer :C

Discussion

21.	Carbon and hydrogen are estimated by :
Answer» Liebig's METHOD Carius method Duma's method NONE of the above Answer :A

Discussion

22.	Carbon and CO gas are used to reduce which of the following pairs of metal oxides for extraction of metals?
Answer» `FEO, SNO` `SnO, ZNO ` `BAO, Na_2 O_2` `FeO,ZnO ` SOLUTION :`FeO,ZnO `

Discussion

23.	Carbon and CO gas are used to reduce which of the following pairs of metal oxides for extraction of metals ?
Answer» FEO, SnO SnO, ZNO `BAO, Na_(2)O_(2)` `FeO_(4)ZnO` Solution :

Discussion

24.	Carbon and carbon monoxide burn in oxygen to form carbon dioxide according to the following reactionsC+O_(2)rarrCO_(2), DeltaH=-394 " kJ mol"^(-1)2CO+O_(2)rarr2CO_(2),DeltaH=-569 " kJ mol"^(-1)The heat of formation of 1 mol of carbon monoxide is thus
Answer» `-219.0 kJ mol^(-1)` `-109.5 kJ mol^(-1)` `-175.0 kJ mol^(-1)` `-87.5 kJ mol^(-1)` Solution :`C+O_(2)rarrCO_(2)+394 kJ "….(i)"` `2C+2O_(2)rarr2CO_(2)+788 kJ "….(ii)"` `2CO_(2)+O_(2)rarr2CO_(2)+569 kJ "….(iii)"` `2CO_(2)rarr2CO+O_(2)-569 kJ "….(IV)"` EQ. (ii) +eq. (iv) = - 109.5 kJ`mol^(-1)`.

Discussion

25.	Carbon-14 used to determine the age of organic material. The procedure is absed on the formation of C^(14) byneutron capture iin the upper atmosphere. ._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1) C^(14) is absorbed by living organisms during photosynthesis. The C^(14) content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of C^(14) in the dead being falls due to the decay, which C^(14) undergoes. ._(6)C^(14)rarr ._(7)N^(14)+beta^(c-) The half- life period of C^(14) is 5770 year. The decay constant (lambda) can be calculated by using the following formuls : lambda=(0.693)/(t_(1//2)) The comparison of the beta^(c-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of C^(14) to C^(12) in living matter is 1:10^(12). A nuclear explosion has taken place leading to an increase in the concentration of C^(14) in nearby areas. C^(14) concentration is C_(1) in nearby areas and C_(2) in areas far away. If the age of the fossil is determined to be T_(1) and T_(2) at the places , respectively, then
Answer» The age of the fossil will increase at the place where explosion has taken place and`T_(1)-T_(2)=(1)/(lambda)ln ``(C_(1))/(C_(2))` The age of the fossil will decrease at the place where explosion has taken place and`T_(1)-T_(2)=(1)/(lambda)ln``(C_(1))/(C_(2))` The age of fossile will be DETERMINED to be same. `(T_(1))/(T_(2))=(C_(1))/(C_(2))` Solution :All radioactive decays are examples of first`-` order KINETICS. So decay constant`lambda=(1)/(T_(1)-T_(2))ln``(C_(1))/(C_(2))` `C_(1)` is the concentration at `T_(1)` time. `C_(2)` is the concentration at `T_(2)` time. So, `T_(1)-T_(2)=(1)/(lambda)ln ``(C_(1))/(C_(2))`

Discussion

26.	Carbon-14 used to determine the age of organic material. The procedure is absed on the formation of C^(14) byneutron capture iin the upper atmosphere. ._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1) C^(14) is absorbed by living organisms during photosynthesis. The C^(14) content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of C^(14) in the dead being falls due to the decay, which C^(14) undergoes. ._(6)C^(14)rarr ._(7)N^(14)+beta^(c-) The half- life period of C^(14) is 5770 year. The decay constant (lambda) can be calculated by using the following formuls : lambda=(0.693)/(t_(1//2)) The comparison of the beta^(c-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of C^(14) to C^(12) in living matter is 1:10^(12). What should be the age of fossil for meaningful determination of its age ?
Answer» 6 years 6000 years 60000years It can be used to CALCULATE any age Solution :Radiacarbon DATING method ceases to be accurate over PERIODS longer than 30000 years.

Discussion

27.	Carbon-14 used to determine the age of organic material. The procedure is absed on the formation of C^(14) byneutron capture iin the upper atmosphere. ._(7)N^(14)+ ._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1) C^(14) is absorbed by living organisms during photosynthesis. The C^(14) content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of C^(14) in the dead being falls due to the decay, which C^(14) undergoes. ._(6)C^(14)rarr._(7)N^(14)+beta^(c-) The half- life period of C^(14) is 5770 year. The decay constant (lambda) can be calculated by using the following formuls : lambda=(0.693)/(t_(1//2)) The comparison of the beta^(c-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of C^(14) to C^(12) in living matter is 1:10^(12). Which of the following options is correct ?
Answer» In LIVING organisms, circulation of `.^(14)C` from the atmosphere is high, so the carbon content is constant in organism. Carbond DATING can be used to find out the age of earth CRUST and rocks. Radioactive ABSORPTION due to cosmic radiation is equal to the rate of radioactive decay. Hence, the carbon content remains constant in living organisms. Carbon dating cannot be used to determineconcentration of `C^(14)` in dead beings. SOLUTION :Radioactive absorption due to cosmic radiation is equal to the rate of radiactive decay, hence the carbon content as the ratio of `C^(14)` remains constant in living organism.

Discussion

28.	Carbon - 14 is used to determine the age of organic material . The procedure is based on the formation of ""^(14) C by neutron capture in the atmosphere . T_(1) - T_(2) = (1)/(lambda) "log" (C_(1))/(C_(2)) ""^(14) C is absorbed by living organisms during photosynthesis . The ""^(14) C content is content is constant in living organism once the plant or animal dies , the uptake of carbon dioxide by it increases and the level of ""^(14) C in the dead being falls due to the decay which C^(14) undergoes . ""_(6)^(14) C to""_(7)^(14) N + beta^(-) (""_(-1)^(0)e) . The half- life period of ""^(14) C is 5770 years . The decay constant (lambda) can be calculated using the formula lambda = (0.693)/(t_(1//2)). The comparison of the b^(-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle . the method , however , ceases to be accurate over period longer than 30,000 years . The proportion of ""^(14) C to ""^(12) C in living matter is 1 : 10^(2) A nuclear explosion has taken place leading toincrease in concentration of C^(14) is C_(3) nearby areas and C_(2) in areas far away . If the age of the fossil is determined to be T_(1) and T_(2) at the places respectively then
Answer» the age of the fossil will INCREASE at the place where explosion has taken place and `T_(1)- T_(2) = (1)/(LAMBDA) "log" (C_(1))/(C_(2))` the age of the fossil will DECREASES at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lambda) "log" (C_(1))/(C_(2))` the age of fossil will be determined to be same `(T_(1))/(T_(2)) = (C_(1))/(C_(2))` Solution :`lambda T_(1) = 1_n ((c_(1))/(c_(2))) …… (1) lambda T_(2) = 1_(n) ((c_(2))/(c_(1)))….. (2)` `lambda (T_(1) - T_(2)) = 1_(n) ((c_1)/(c_1)) - 1_n ((c_2)/(c_1)) , lambda (T_(1) -T_(2)) = 1_(n) ((c_(1))/(c_(2)) xx (c_(1))/(c_(2))) , (T_(1) - T_(2)) = (1)/(lambda) xx 1_n ((c_1)/(c_2))`

Discussion

29.	Carbon - 14 is used to determine the age of organic material . The procedure is based on the formation of ""^(14) C by neutron capture in the atmosphere . T_(1) - T_(2) = (1)/(lambda) "log" (C_(1))/(C_(2)) ""^(14) C is absorbed by living organisms during photosynthesis . The ""^(14) C content is constant in living organism once the plant or animal dies , the uptake of carbon dioxide by it increases and the level of ""^(14) C in the dead being falls due to the decay which C^(14) undergoes . ""_(6)^(14) C to""_(7)^(14) N + beta^(-) (""_(-1)^(0)e) . The half- life period of ""^(14) C is 5770 years . The decay constant (lambda) can be calculated using the formula lambda = (0.693)/(t_(1//2)). The comparison of the b^(-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle . the method , however , ceases to be accurate over period longer than 30,000 years . The proportion of ""^(14) C to ""^(12) C in living matter is 1 : 10^(2) Which of the following options is correct ?
Answer» In living organisms , circulation of `""^(14) C` from atmosphere is high so the carbon content is constant in organism Carbon DATING can be USED to find out the age of earth crust of ROCKS Radioactive material absorption is equal to its rate of decay , hence the carbon content remains constant in living organism. Carbon dating cannot be used to determine concentration of `""^(14) C` in dead being . Solution :living organism `((C^(14))/(C^(12)))` is maintained as constant VOLUME .

Discussion

30.	Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of .^(14)C by neutron capture in the upper atmosphere. ._(7)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1) .^(14)C is absorbed by living organisms during photosynthesis. The .^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of .^(14)C in the dead being, falls due to the decay which .^(14)C undergoes. ._(6)^(14)C rarr ._(7)^(14)N + beta^(-) The half-life period of .^(14)C is 5770 years. The decay constant (lamda) can be calculated by using the following formula lamda = (0.693)/(t_(1//2)). The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of .^(14)C " to " .^(12)C in living matter is 1 : 10^(12).A nuclear explosion has taken place leading to increase in concentration of .^(14)C in nearby areas. C^(14) concentrations is C_(1) in nearby areas and C_(2) in areas far away. If the age of the fossil is determined to be T_(1) and T_(2) at the places respectively then
Answer» The age of the fossil will increase at the place where EXPLOSION has taken and `T_(1) - T_(2) = (1)/(LAMDA "LN" (C_(1))/(C_(2))` The age of the fossil will decrease at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lamda) "ln" (C_(1))/(C_(2))` The age of fossil will be determined to be same `(T_(1))/(T_(2)) = (C_(1))/(C_(2))` SOLUTION :All radioactive decays are the examples of first order kinetics So decay constant `lamda = (1)/(T_(1) - T_(2)) "ln" (C_(1))/(C_(2))` `C_(1)` is the concentration at `T_(1)` times `C_(2)` is the concentration at `T_(2)` time So `T_(1) - T_(2) = (1)/(lamda) ln (C_(1))/(C_(2))`

Discussion

31.	Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of .^(14)C by neutron capture in the upper atmosphere. ._(7)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1) .^(14)C is absorbed by living organisms during photosynthesis. The .^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of .^(14)C in the dead being, falls due to the decay which .^(14)C undergoes. ._(6)^(14)C rarr ._(7)^(14)N + beta^(-) The half-life period of .^(14)C is 5770 years. The decay constant (lamda) can be calculated by using the following formula lamda = (0.693)/(t_(1//2)). The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of .^(14)C " to " .^(12)C in living matter is 1 : 10^(12). What should be the age of fossil for meaningful determination of its age
Answer» 6 YEARS 6000 years 60,000 years It can be used to calculate any age Solution :RADIO carbon dating method CEASES to be accurate over periods longer than 30,000 years

Discussion

32.	Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of .^(14)C by neutron capture in the upper atmosphere. ._(7)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1) .^(14)C is absorbed by living organisms during photosynthesis. The .^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of .^(14)C in the dead being, falls due to the decay which .^(14)C undergoes. ._(6)^(14)C rarr ._(7)^(14)N + beta^(-) The half-life period of .^(14)C is 5770 years. The decay constant (lamda) can be calculated by using the following formula lamda = (0.693)/(t_(1//2)). The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of .^(14)C " to " .^(12)C in living matter is 1 : 10^(12). Which of the following option is correct
Answer» In living organisms, circulation of `.^(14)C` from atmosphere is high so the carbon content is constant in organism Carbon dating can be used to find out the age of earth crust and rocks RADIOACTIVE absorption due to COSMIC RADIATION is equal to the rate of radioactive decay, hence the carbon content remains constant in living organisms Carbon dating can not be used to determine concentration of `.^(14)C` in dead beings. Solution :Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content as the RATIO of `C^(14) " to " C^(12)` remains constant in living organism.

Discussion

33.	Carbon -14 is used to determine the age of organic material. The procedure, is based on the formation of .^(14)C by neutron capture in the upper atmosphere. ""_(7)^(14)N+_(0)n^(1) to _(6)^(14)C+_(1)n^(1). .^(14)C is absorbed by living organisms during photosynthesis. The .^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of .^(14)C in the dead being, falls due to the decay.which C^(14) undergoes ""_(6)^(14)C to _(7)^(14)N+beta^(-) The half life period of ^(14)C is 5770 years. The decay constant (lambda) can be calculated by using the following formula lambda=(0.693)/(t_(1//2)) The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of .^(14)C to .^(12)C in livingmatter is 1 : 10^(12). What should be the age of fossil for meaningful determination of its age ?
Answer» `6` years `6000` years `60,000` years It can be used to calculate any age Answer :B

Discussion

34.	Carbon -14 is used to determine the age of organic material. The procedure, is based on the formation of .^(14)C by neutron capture in the upper atmosphere. ""_(7)^(14)N+_(0)n^(1) to ""_(6)^(14)C+""_(1)n^(1). ""^(14)C is absorbed by living organisms during photosynthesis. The .^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ""^(14)C in the dead being, falls due to the decay.which C^(14) undergoes ""_(6)^(14)C to ""_(7)^(14)N+beta^(-) The half life period of ""^(14)C is 5770 years. The decay constant (lambda) can be calculated by using the following formula lambda=(0.693)/(t_(1//2)) The comparison of the beta^(-) activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ""^(14)C to ""^(12)C in livingmatter is 1 : 10^(12). Which of the following option is correct ?
Answer» In living organisms, CIRCULATION of `.^(14)C` from atmosphere is HIGH so the carbon content is constant in organism Carbon dating can be USED to FIND out the age of earth crust and rocks Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism Carbon dating can not be used to DETERMINE concentration of `.^(14)C` in dead beings Answer :C

Discussion

35.	Carbon 14 is used to determine the age of organic material. The procedure is based on the formation of 14C by neutron capture in the upper atmosphere ""_(7)N^(14) + _(0)n^(1)rarr ""_(6)C^(14)+ ""_(1)H^(1) ""^(14)C is absorbed by living organism during photosynthesis. The ""^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ""^(14)C in the dead being falls due to the decay which C^(14) undergoes. ""_(6)C^(14)rarr ""_(7)N^(14) + _(-1)e^(0) The half life period of ""^(14)C is 5770 years. The decay constant (lambda) can be calculated by using the following formulalambda = (0.693)/t_(1/2)The comparison of the beta –activity of the dead matter with that of carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ""^(14)C to ""^(12)C in living matter is 1 : 10^(12 ) A nuclear explosion has taken place leading to increase in concentration of C^(14) in nearby areas. C^(14) concentration is C_(1) in nearby areas and C_(2) in areas far away. If the age of the fossil is determined to be t_(1) and t_(2) at the places respectively, then
Answer» The age of the fossil will increase at the place where EXPLOSION has taken place and `t_(1) - t_(2) =1/lambda LN frac (C_(1))(C_(2))` The age of the fossil will decrease at the place where explosion has taken place and `t_(1) - t_(2) =1/lambda ln frac (C_(1))(C_(2))` The age of fossil will be determined to be the same `t_(1)/t_(2) =frac (C_(1))(C_(2))` Answer :A

Discussion

36.	Carbon 14 is used to determine the age of organic material. The procedure is based on the formation of 14C by neutron capture in the upper atmosphere ""_(7)N^(14) + _(0)n^(1)rarr ""_(6)C^(14)+ ""_(1)H^(1) ""^(14)C is absorbed by living organism during photosynthesis. The ""^(14)C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ""^(14)C in the dead being falls due to the decay which C^(14) undergoes. ""_(6)C^(14)rarr ""_(7)N^(14) + _(-1)e^(0) The half life period of ""^(14)C is 5770 years. The decay constant (lambda) can be calculated by using the following formulalambda = (0.693)/t_(1/2)The comparison of the beta –activity of the dead matter with that of carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ""^(14)C to ""^(12)C in living matter is 1 : 10^(12 ) Which of the following option is correct ?
Answer» In LIVING organisms, circulation of `""^(14)C` from atmosphere is high so the carbon content is constant in organism Carbon dating can be used to FIND out the age of earth CRUST and rocks Radioactive absorption due to cosmic radiation is equal to the RATE of radioactive decay, hence the carbon content REMAINS constant in living organism Carbon dating cannot be used to determine concentration of `""^(14)C` in dead beings. Answer :C

Discussion

37.	Carbon 14 dating method is based on the fact that
Answer» CARBON -14 FRACTION is the same in all objects Carbon-14 is highly insoluble Ratio of carbon -14 and carbon -12 is constant All of these Solution :A basis for the `C -14` dating TECHNIQUE

Discussion

38.	Carbon-14 dating method is that the :
Answer» Carbon-14 FRACTION is the same in all objects Carbon-14 is HIGHLY insoluble Ratio of carbon-14 and carbon-12 are constant All the above. ANSWER :C

Discussion

39.	Carbolic acid is reacted with conc. H_2SO_4 at 300 K gives,
Answer» 2-hydroxybenzenesulphonic acid 3-hydroxybenzenesulphonic acid 4-hydroxybenzenesulphonic acid 2 and 4-phenol sulphonic acid Answer :A

Discussion

40.	Carbolic acid is obtained from oxidation of
Answer» SODIUM salicylate salicylic acid toluene cumene Answer :D

Discussion

41.	Carbolic acid is …………………. .
Answer» PHENOL Pieric ACID BENZOIC acid pheny lacetic acid SOLUTION :Phenol

Discussion

42.	Carbohydrates which differ in configuration at the glycosidic carbon (i.e. C_(1) in aldose and C_(2) in ketoses) are called
Answer» Anomers Epimers Diastereomers Enantiomers Answer :A

Discussion

43.	Carbohydrates which give two molecules of mono-saccharides are called disaccharides. These are:
Answer» MALTOSE CELLULOSE MALTASE LACTOSE ANSWER :A::D

Discussion

44.	Carbohydrates that produce two to ten monosaccharide units on hydrolysis are called________.
Answer» SOLUTION :OLIGOSACCHARIDES

Discussion

45.	Carbohydrates may be regarded as
Answer» AROMATIC compounds alicyclic compounds polyfunctional compounds all of these Answer :C

Discussion

46.	Carbohydrates , proteins and facts are the thre main components of our food. For a normal person, a balanced diet which contains all these three components in right proportions is recommended for good health. Arjun , a class XII students , however, pointed out that this is not right always. Under special situations, it is not the balanced diet but it is the protein rich diet which is often recommended. Read the above passage and answer the following questions: (i) Why a protein rich diet is recommended for HIV/AIDS patients ? (ii) Why children and expectant mothers need a protein rich diet ? (iii) Person who wish to reduce body weight and heart patients are also recommended protein rich diet. Why ?
Answer» Solution :(i) The immune system of HIV/AIDS PATIENTS is largely/ completely distroyed. Since REBUILDING immune system besides medication also requires antibodies which are proteins. Therefore, to rebuilt/strengthen immune systems, protein rich diet is recommended. (ii) Children for their growth and expectant mothers for the growth of their expected infants require building of muscles which requires protein rich foods. (iii) Protein increases satiety, i.e., you feel fail full for longer periods, thereby reducing the total number of calories eaten per DAY. Further , it increases protein metabolism thereby maintaining lean muscles which is required for weight loss. A protein rich diet helps to REDUCES the blood pressure as well as the CHOLESTEROL level in the blood. Both these factors are beneficial for heart patients.

Discussion

47.	Carbohydrates in which large number of monosaccharide units linked to each other by oxides bridges are
Answer» non-sugars POLYSACCHARIDES tetrasaccharides boht (a) and (B) SOLUTION :boht (a) and (b)

Discussion

48.	Carbohydrates may be :
Answer» sugars starch POLYHYDROXY aldehyde/ketones Compounds that can be HYDROLYSED to sugar Solution :N//A

Discussion

49.	Carbohydrates have
Answer» BITTER taste sour taste SWEET taste some have sweet TEST and some are tasteless Answer :D

Discussion

50.	Carbohydrates have the general formula C_X(H_2O)_Y. Which of the following is not a carbohydrate:
Answer» `C_6H_16O_6` `C_6H_10O_5)_n` `C_12H_22O_11` `C_2H_4O_2` ANSWER :D

Discussion

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Carbon cannot be used in the reduction of Al_2 O_3 because

Carbon can reduce ferric oxide to iron at a temperature above 983 K because

Carbon burns in air and forms two oxides CO and CO_2.This shows that carbon has :

Carbon black, which is used in making printer's ink is obtained by the decomposition of

Carbon black, which is used in making printer's ink, is obtained by decomposition of

Carbonblackis obtained whenmethaneis :

Carbon atoms in the compound (CN)_(4)C_(2) are

Carbon atoms in fullerene with formula C_(60) have

Carbon atoms in diamond are bonded with each other in configuration :

Carbon atoms in fullerence with formula C_(60) have………

Carbon atoms in diamond are bonded to each other in configuration which is :

Carbon atoms bearing the -OH group in phenols and CH_(2)=CH-OH are

Carbon atom of carbonyl gp. In aldehyde is of :

Carbon atom of carbonyl group In a ketone is of :

Carbon atom of carbonyl group In aldehyde is of :

Carbon atom in methyl carbocation contains how many pairs of electrons?

Carbon atom in carboxyl group is

Carbon atom holding halogen in aryl halides is:

Carbon atom holding halogen in aryl halides is :

Carbon and oxygen forms two compound . Carbon content in one of them is 42.9% while in the other is 27.3% . Thegiven data is in support with

Carbon and hydrogen are estimated by :

Carbon and CO gas are used to reduce which of the following pairs of metal oxides for extraction of metals?

Carbon and CO gas are used to reduce which of the following pairs of metal oxides for extraction of metals ?

Carbon and carbon monoxide burn in oxygen to form carbon dioxide according to the following reactionsC+O_(2)rarrCO_(2), DeltaH=-394 " kJ mol"^(-1)2CO+O_(2)rarr2CO_(2),DeltaH=-569 " kJ mol"^(-1)The heat of formation of 1 mol of carbon monoxide is thus

Carbon 14 dating method is based on the fact that

Carbon-14 dating method is that the :

Carbolic acid is reacted with conc. H_2SO_4 at 300 K gives,

Carbolic acid is obtained from oxidation of

Carbolic acid is …………………. .

Carbohydrates which differ in configuration at the glycosidic carbon (i.e. C_(1) in aldose and C_(2) in ketoses) are called

Carbohydrates which give two molecules of mono-saccharides are called disaccharides. These are:

Carbohydrates that produce two to ten monosaccharide units on hydrolysis are called________.

Carbohydrates may be regarded as

Carbohydrates in which large number of monosaccharide units linked to each other by oxides bridges are

Carbohydrates may be :

Carbohydrates have

Carbohydrates have the general formula C_X(H_2O)_Y. Which of the following is not a carbohydrate: