94420 + Interview Questions in Chemistry in Class 12

1.	b) Atoms of element B form hep lattice and those of element A occupies 2//3^(rd) of tetrahedral voids. Calculate the formula of the compound formed by A and B.
Answer» SOLUTION :The hcp lattice is formed by the atoms of the element B. Here, the number of TETRAHEDRAL voids generated is equal to TWICE the number of atoms of the element B. The atoms of element A occupy `(2)/(3)` rd of the tetrahedral voids. Therefore, the number of atoms of A is equal to `2 xx (2)/(3) = (4)/(3)` rd of the number of atoms of B. Therefore, ratio of the number of atoms of A to that of B is `A : B = (4)/(3) : 1` = 4 : 3 Thus the formula of the compound ia `A_(4) B_(3)`.

Discussion

2.	Atoms of element B form hcp lattice and those of the element A occupy 2//3^(rd) of tetrahedral voids. What is the formula of the compound formed by the elements A and B ?
Answer» Solution :The ATOMS of element B is forming close packed STRUCTURE. Let number of B atoms = n The total number of tetrahedral voids = 2n Atoms A OCCUPY `(2//3)^(rd)` of tetrahedral voids. Hence, the total number of Atoms = `2n xx 2/3 =(4n)/3` So the formula of a compound A`((4n)/3),B_n` Taking n = 3, we GET `A_4B_3`.

Discussion

3.	Atoms of element B form hcp lattice and those of the element A occupy 1/(3^(rd)) of tetrahedral voids. What is the formula of the compound formed by the elements A and B ?
Answer» `A_(2)B_(3)` `A_(4)B_(3)` `A_(3)B_(2)` AB Answer :A

Discussion

4.	Atoms of an element A occupy(2)/(3)tetrahedral volds in the hcp, formed by the elements, 'B' . The formula of compound is
Answer» `A_(2)B_(3)` `A_(2)B` `AB_(2)` `A_(4)B_(3)` SOLUTION :No . Ofoctahedral voids FORMED`=2xx`No . Of ATOMS of B . Only`(2)/(3)`rdof these are occupied by atoms of A `:.`No .of atoms A : No. of atoms of B `=2xx(2)/(3):1=4:3` `:.`FORMULA is `A_(4)B_(3)`

Discussion

5.	Atoms of an element 'A' occupy 2/3 tetrahedral voids in the hexagonal close packed (hep) unit cell lattice formed by the element 'B'. The B' is formula of the compound formed by 'A' and
Answer» `a_(2)b` `ab_(2)` `a_(4)b_(3)` `a_(2)b_(3)` ANSWER :C

Discussion

6.	Atoms may be regarded as comprising protons, neutrons and electrons. If the mass of a neutron were halved and that of electron was doubled , the atomic mass of ""_(6)C^(12) would :
Answer» REMAIN APPROXIMATELY the same be approximately doubled be approximately halved be REDUCED approximately by 25% ANSWER :D

Discussion

7.	Atoms in hydrogen have abundance of :
Answer» `._(1)H^(1)` atoms `._(1)D^(2)` atoms `._(1)T^(3)` atoms All THREE are in EQUAL proportions Answer :A

Discussion

8.	Atoms in a P_(4) molecule of white phosphorous are arranged regularly in the following way:
Answer» At the CORNERS of the cube At the corners of an octahedron At the corners of a tertrahedron At the CENTRE and corners of a tetrahedron. Answer :C

Discussion

9.	Atoms C and Dformfcc crystalline structure. Atom C ispresent at thecorners of thecube and D is at theface centres of thecube. What is theformula of thecompound ?
Answer» SOLUTION :Given : Crystal has fcc structure . AtomsC areat 8cornerswhileatoms D are at 6 FACECENTRES of thecubicunitcell. At thecorner .`(1)/(8)` th ofeach Catomis presentwhile ateachfacecentrehalfof each D ATOMIS present . NUMBER of THEC atoms `= (1)/(8) xx 8=1 ` Number of Datoms `= (1)/(2)xx 6 = 3` Thusunitcellcontainsone C atomand three D atoms Hencetheformula of thecompoundis CD

Discussion

10.	Atomicity of sulphur in orthorhombic ( alpha sulphur) is
Answer» 1 2 8 6 Solution :ORTHORHOMBIC SULPHUR is OCTAHEDRAL `(S_8)`

Discussion

11.	Atomicity of phosphorus is:
Answer» 1 2 3 4 Answer :D

Discussion

12.	Atomicity of dimeric phosphorous pentoxide is 'x' and the number of shared electron pairs is 'y'. Then a) y-x=6""b)2x=y+8 c) 10x-7y=0
Answer» Only 'a' is CORRECT Only 'B' is correct Only 'C' is correct All the above are correct ANSWER :4

Discussion

13.	Atomic weights of practically all the elements are very nearly whole numbers and not always whole numbers. This is because :
Answer» of EXISTENCE of ALLOTROPIC forms Valency of ELEMENTS have frictional values Equivalent WEIGHTS are not whole numbers Of the existence of ISOTOPES for all the elements Answer :D

Discussion

14.	Atomic weight of the isotope of hydrogen which contains 2 neutrons in the nucleus would be
Answer» 2 3 1 4 Solution :`._(1)T^(3)` - TRITIUM has 2 NEUTRON and 1 proton

Discussion

15.	Atomic sizes generally increase as we come down a group of the periodic table. But in the 4th group of the periodic table, Zr and Hf have almost the same atomic sizes. Why?
Answer» Solution :This is a CONSEQUENCE of LANTHANOID contraction. Along the lanthanoids there is progressive decrease in size with INCREASE in atomic number which is known as lanthanoid contraction.Since comes IMMEDIATELY after the lanthanoids, its size, is smaller than expected"and it is almost the same as that of Zr.

Discussion

16.	Atomic radius of Cu is greater than that of Cr but ionic ofCr^(2+)is greater than that of Cu^(2+). Give suitable explanation.
Answer» Solution :In Cu, all the d-electronsarepaired `(3d^(10)4s^(1))`. In CR , all the d-ELECTRONS are unpaired`( 3d^(5) 4s^(1))`. Hence, d-delectron repulsion in Cu are much greater than those in Cr. Therefore ,Cu atomis larger size than Cr. In `Cu^(2+) (3d^(9))`, d-d electron repulsions decrease due to presenceofone unpaired d-electron. MOREOVER, the electrons are attracted by 29 PROTONS of the nucleus whereas in `Cr^(2+)`, three unpaired electrons are still presenct but they are attractedby only 24 protons of the nucleus. Thus , `Cu^(2+)`issmaller in size than`Cr^(2+)`.

Discussion

17.	Atomic size of 3d-series elements from chromium to copper is almost the same. Give reason.
Answer» Solution :The increase in the nuclear charge from CR to Cu is COMPENSATED by the SCREENING EFFECT of 3d-electrons.

Discussion

18.	Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. In the homonuclear diatomic molecule which of the following sets of M.O. orbitals are grade or un-grade
Answer» `sigma_(2S) , pi_(2p_x)` `sigma_(2s)^() , pi_(2p_x)^()` `sigma_(2s)^() , pi_(2p_x)` `sigma_(2p_x) , pi_(2p_x)^()` ANSWER :D

Discussion

19.	Atomic orbitals of bonded atoms combine to form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals taking part in the bond formation. When two atomic orbitals combine, two molecular orbitals are formed one of which has lower energy than the combining orbitals and is called bonding Molecular Orbital (MO). Whereas the other having higher energy than the two combining atomic orbitals is called Anti Bonding Molecular orbitals (ABMO) The two combining atomic orbitals must have comparable energies and should be properly oriented to allow considerable overlapping. If the overlapping is end to end along internuclear axis, the molecular orbital is called sigma and if the overlapping is lateral 1.e., sidewise the molecular orbital is called pie. Just like atomic orbitals, the molecular orbitals also have varying energy levels. Filling of electrons in molecular orbitals takes place following the same rules as followed for filing of atomic orbitals. The order of filling may not be same for all the molecules or their ions. Bond order is a useful parameter for comparing the various characteristics of molecules. Which of the following combinations is not allowed (assume z-axis as the internuclear axis ) ?
Answer» 2S and 2s `2p_x and 2p_x` `2p_z and 2p_z` `2p_x and 2p_y` ANSWER :D

Discussion

20.	Atomic numbers of Ar and Xe respectively are
Answer» 36 and 54 18 and 54 10 and 36 10 and 54 SOLUTION :18 and 54

Discussion

21.	Atomic number of N is 7. the atomic number of IIIrd member of nitrogen familyis
Answer» 23 15 33 43 Solution :`{:(,"I","II","III","IV","V"),("ELEMENT-",N,P,As,SB,Bi),("ATOMIC no.",7,15,33,51,83):}`

Discussion

22.	Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons ?
Answer» `[MnCl_(6)]^(3-)` `[FeF_(6)]^(3-)` `[CoF_(6)]^(3-)` `[Ni(NH_(3))_(6)]^(2+)` Solution :(a) and (C ) are `d^(4),d^(6)` respectively (HIGH spin)

Discussion

23.	Atomic number of Mn, Fe, and Co are 25, 26, 27 and 28 respectively. Which of the following inner orbital octahedral compexes ions are diamagnetic ?
Answer» `[CO(NH_(3))_(6)]^(3+)` `[MN(CN)_(6)]^(4-)` `[Fe(CN)_(6)]^(4-)` `[Fe(CN)_(6)]^(3-)` Solution :(a) and (C ) are `d^(6)` (spin PAIRED)

Discussion

24.	Atomic number of Mn. Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic ?
Answer» `[CO(NH_(3))_(6)]^(3+)` `[Mn(CN)_(6)]^(3-)` `[Fe(CN)_(6)]^(4-)` `[Fe(CN)_(6)]^(3-)` ANSWER :A::C

Discussion

25.	Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral compelxes have same number of unpaired electrons ?
Answer» `[MnCl_(6)]^(3-)` `[FeF_(6)]^(3-)` `[CoF_(6)]^(3-)` `[NI(NH_(3))_(6)]^(2+)` SOLUTION :Proceed as in

Discussion

26.	Atomic number of Mn, Fe and Co and 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic ?
Answer» `[Co(NH_(3))_(6)]^(3+)` `[Mn(CN)_(6)]^(3-)` `[FE(CN)_(6)]^(4-)` `[Fe(CN)_(6)]^(3-)` Solution :In (a), `Co^(3+)=3d^(6)=d^(2)d^(1)d^(1)d^(1)d^(1)`. In presence of `NH_(3)`, it becomes `d^(2)d^(2)d^(2)d^(0)d^(0)` In (b), `Mn^(2+)=3d^(5)=d^(1)d^(1)d^(1)d^(1)d^(1)`. In presence of `CN^(-)`, it is `d^(2)d^(2)d^(1)d^(0)d^(0)` In (C ), `Fe^(2+)=3d^(6)`. Similar to (a) In (d), `Fe^(3+)=3d^(5)`. Similar to (b).

Discussion

27.	Atomic number of Cr is 24, then Cr^(3+) will be :
Answer» Diamagnetic Paramagnetic Ferromagnetic None of these Answer :B

Discussion

28.	Atomic number of At is ..........
Answer» 117 85 53 167 Answer :B

Discussion

29.	Atomic number of Ag is 47. In the same group the atomic numbers of elements placed and below Ag in long form of periodic table will be x and y respectively. Given the value fo (x+y)//12.
Answer» SOLUTION :Atomic number of `CU` is `29 = X` Atomic number of `AU` is `79 = y` `x +y = 108` `(x+y)/(12) = (108)/(12) = 9`.

Discussion

30.	Atomic number and electronic configuration of Cerium are respectively
Answer» `59, 4F^(1)5d^(1)6S^(2)` `58, 4f^(1)5d^(2)6s^(1)` `58, 4f^(1)5d^(1)6s^(2)` `59, 4f^(1)5d^(2)6s^(1)` Answer :C

Discussion

31.	Atomic number ""_(64)Gd, ""_(66)Dy, ""_(69)Tm, ""_(71)Lu. which of the following does not have unpaired electron ?
Answer» `Gd^(3+)` `Dy^(3+)` `TM^(3+)` `Lu^(3+)` Solution :We know that, `Gd ^(3+)` has `4F^(7)` configuration containing 7 UNPAIRED electrons. `Dy ^(3+)` has `4f ^(9)` congiguration, containing five unpaired electrons. `Tm^(3+)` has `4f ^(12)` configuration, containing two unpaired electrons. `Lu^(3+)` has `4f^(14)` configuration does not CONTAINS unpaired electrons.

Discussion

32.	Atomic number 64 will have electronic configuration
Answer» `[Xe]_(54), 5s^(2) 4f^(8)` `[Xe]_(54), 6S ^(2), 4f^(2), 5D^(1)` `[Xe]_(54), 4t^(10)` `[Xe]_(54), 6s^(2), 4f^(7), 5p^(1)` Answer :B

Discussion

33.	Atomic no. of N is 7, the atomic no. of IVth member of nitrogen family will be
Answer» 23 51 33 43 Solution :`IV^(th)` MEMBER will be in 5th PERIOD. `therefore` Atomic number `=overset"II"7+overset"III"8+overset"IV"18+overset"V"18 =51`

Discussion

34.	Atomic mass of chlorine is 35.5. it has two isotopes of atomic mass 35 and 37. the percentage of heavier isotope is:
Answer» 10 15 20 25 Answer :D

Discussion

35.	Atomic mass of boron is 10.81. it has two isotopes with 80% and 20% abundance respectively. The atomic mass of the isotope having 80% abundance is 11.01. the atomic mass of the other isotope is:
Answer» 1081 11.01 10.01 21.82 Answer :C

Discussion

36.	Atomic mass of an element is
Answer» Actual mass of one ATOM of the element Relative mass of an atom of the element Average relative mass of DIFFERENT ATOMS of the element Always a whole number Answer :C

Discussion

37.	Atomic hydrogen produces formaldehyde when it reacts with:
Answer» `CO_2` CO `O_2` `C_2H_2` ANSWER :B

Discussion

38.	Atomic hydrogen is obtained by :
Answer» Electrolysis of HEAVY WATER Reaction of water with heavy metals Thermal decomposition of water Passing SILENT ELECTRIC DISCHARGE through hydrogen at low pressure Answer :D

Discussion

39.	Atom that requires high energy of excitation
Answer» F Cl Br I Answer :A

Discussion

40.	Atom A possesses higher values of packing fraction than atom B. The relative stabilities of A and B are
Answer» A is more STABLE than B B is more stable than A A and B both are EQUALLY stable Stability does not depend on PACKING FRACTION Solution :The atom which has lower value of packing fraction is stable

Discussion

41.	Atom bomb is based on the principle of
Answer» Nuclear fusion Nuclear FISSION Radioactivity Fusion and fission both Solution :Atom BOMB is based on the PRINCIPLE of nuclear fission

Discussion

42.	For the reaction, following data is given: AtoB,K_(1)=10^(15)"exp"^(((-2000)/T)),CtoD,K_(2)=10^(14)"exp"^(((-1000)/T)) The temperature at which K_(1)=K_(2) is
Answer» `1000K` `2000K` `2000/2.303K` 1000/2.303K Solution :`K_(A)=K_(C),10^(15),E^(((-2000)/T))=10^(14).e^(((-1000)/T))` `10=(e^(-1000+2000)/T),2.303xxlog_(10)^(10)=1000/T,T=(1000/2.303)`Kelving

Discussion

43.	(A)The reaction of oxalic acid with acidified KMnO_(4) first slow and then proceeds with faster speed(R ) Acidified KMnO_(4) is a strong oxidizing agent
Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) Both (A)and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false ANSWER :B

Discussion

44.	(a)Tin stone-Oxides ore (b)Copper pyrite-Oxide ore (c )Zincite-Oxide ore (d)bauxite-Oxides ore
Answer» SOLUTION :(B)COPPER pyrite-Oxides ORE

Discussion

45.	(a)The rate constants of a reaction at 500 K and 700 K are 0.02 s^(-1) and 0.07 s^(-1) respectively. Calculate the value of Ea. (b) Under what condition a bimolecular reaction behaves kinetically first order reaction ?
Answer» SOLUTION :(a)Accordingtoarrheniusequation `log(K_2) /(k_1) =(E_(Act ))/( 2.303 R) [(T_2 -T_1)/(T_1 T_2)]` `(K_2)/(K_1) = ( 0.07 )/( 0.02)=7/2` `T_2 = 70 K` `T_1 = 500 K` `R=8.314JK^(-1) Mol^(-1)` ` thereforelog (7)/(2)= (E_(a))/(2.303 xx (8.314 )/( 1000 ) )[ (700 -500 )/( 700 xx 500 )]` `E_a= log(7 )/(2)xx ( 2.303 xx 8.314 )/( 1000 ) xx (7000 xx 500 )/( 200 )` `=( 0.5441 xx 2.303 xx 8.314 xx 700 xx 500)/( 1000 xx 200 )` `=18.23 KJ mol^(-1)` (b)Thisis possibleonlywhen theone of the reactionis PRESENT inlargeexcess .

Discussion

46.	[A]:The Arrhenius equation k=Ae^(-E_(a)//RT) gives the relation between rate constant and temperature. [R]:The graph of log kto(1)/(T) is linear and with the help of this calculation of energy of activation can be possible.
Answer» ASSERTION [A] and reason [R] both are correct and [R] gives correct EXPLANATION of [A] Assertion [A] and reason [R] both are correct but [R] does not give correct explanation of [A] Assertion [A] is wrong but Reason [R] is wrong Assertion [A] is wrong but Reason [R]is correct Solution :By taking log both side of Atthenious EQUATION, In k=In A`-(E_(a))/(RT)` and log k=log A`-(E_(a))/(2.303R)((1)/(T))` The equation is obtain.This euqation is equal to y=mc+c.So,in graph log `kto(1)/(T)`,the valie of slope is `(E_(a))/(2.303R)` So,the value of `E_(a)` is CALCULATED by getting of slope and the explanation of Arrhenius equation is obtain from graph.

Discussion

47.	[A]:The catalyst increases the rate of reaction. [R]:The catalyst decreases the enthalpy change (DeltaH)
Answer» Assertion [A] and reason [R] both are correct and [R] gives correct explanation of [A] Assertion [A] and reason [R] both are correct but [R] does not give correct explanation of [A] Assertion [A] is wrong but Reason [R] is wrong Assertion [A] is wrong but Reason [R]is correct Solution :In presence of catalyst the path of REACTION is change and LESS ACTIVATION energy is REQUIRED . because of that the rate of reaction increase. In presence of catalyst the energy of product or reactant does not change so,the ENTHALPY change for reaction remain constant and does not change.

Discussion

48.	Atetrapeptide on complete hydrolysis gave Alanine, Glycine, Leucine and valine . It never gave Ala-Gly and Val-Ala on partial hydrolysis. How many such tetrapeptides are possible ?
Answer» SOLUTION :N//A

Discussion

49.	[A]:Temperature of reaction increases by 10^(@)C the collision frequency of molecule increase by 2% to 3% [B]:By increases temperature 10^(@)C the rate of reaction is 200% from 100%
Answer» Assertion [A] and reason [R] both are correct and [R] gives correct explanation of [A] Assertion [A] and reason [R] both are correct but [R] does not GIVE correct explanation of [A] Assertion [A] is wrong but Reason [R] is wrong Assertion [A] is wrong but Reason [R]is correct Solution :By increases the `10^(@)` temperature the frequency of collision of MOLECULE increases .Due to which reason theis frequency increases ?Its explanation ISNOT given by [R], because [R] is a result of statement [A],but its not a EXPLANTION of statement[A].

Discussion

50.	At^(3)Y, minimum numbers of elecrolyte required for a compound to obtained its conductivity is
Answer» 2 3 4 1 Answer :A

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

b) Atoms of element B form hep lattice and those of element A occupies 2//3^(rd) of tetrahedral voids. Calculate the formula of the compound formed by A and B.

Atoms of element B form hcp lattice and those of the element A occupy 2//3^(rd) of tetrahedral voids. What is the formula of the compound formed by the elements A and B ?

Atoms of element B form hcp lattice and those of the element A occupy 1/(3^(rd)) of tetrahedral voids. What is the formula of the compound formed by the elements A and B ?

Atoms of an element A occupy(2)/(3)tetrahedral volds in the hcp, formed by the elements, 'B' . The formula of compound is

Atoms of an element 'A' occupy 2/3 tetrahedral voids in the hexagonal close packed (hep) unit cell lattice formed by the element 'B'. The B' is formula of the compound formed by 'A' and

Atoms may be regarded as comprising protons, neutrons and electrons. If the mass of a neutron were halved and that of electron was doubled , the atomic mass of ""_(6)C^(12) would :

Atoms in hydrogen have abundance of :

Atoms in a P_(4) molecule of white phosphorous are arranged regularly in the following way:

Atoms C and Dformfcc crystalline structure. Atom C ispresent at thecorners of thecube and D is at theface centres of thecube. What is theformula of thecompound ?

Atomicity of sulphur in orthorhombic ( alpha sulphur) is

Atomicity of phosphorus is:

Atomicity of dimeric phosphorous pentoxide is 'x' and the number of shared electron pairs is 'y'. Then a) y-x=6""b)2x=y+8 c) 10x-7y=0

Atomic weights of practically all the elements are very nearly whole numbers and not always whole numbers. This is because :

Atomic weight of the isotope of hydrogen which contains 2 neutrons in the nucleus would be

Atomic sizes generally increase as we come down a group of the periodic table. But in the 4th group of the periodic table, Zr and Hf have almost the same atomic sizes. Why?

Atomic radius of Cu is greater than that of Cr but ionic ofCr^(2+)is greater than that of Cu^(2+). Give suitable explanation.

Atomic size of 3d-series elements from chromium to copper is almost the same. Give reason.

Atomic numbers of Ar and Xe respectively are

Atomic number of N is 7. the atomic number of IIIrd member of nitrogen familyis

Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons ?

Atomic number of Mn, Fe, and Co are 25, 26, 27 and 28 respectively. Which of the following inner orbital octahedral compexes ions are diamagnetic ?

Atomic number of Mn. Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic ?

Atomic number of Mn, Fe, Co and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral compelxes have same number of unpaired electrons ?

Atomic number of Mn, Fe and Co and 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic ?

Atomic number of Cr is 24, then Cr^(3+) will be :

Atomic number of At is ..........

Atomic number of Ag is 47. In the same group the atomic numbers of elements placed and below Ag in long form of periodic table will be x and y respectively. Given the value fo (x+y)//12.

Atomic number and electronic configuration of Cerium are respectively

Atomic number ""_(64)Gd, ""_(66)Dy, ""_(69)Tm, ""_(71)Lu. which of the following does not have unpaired electron ?

Atomic number 64 will have electronic configuration

Atomic no. of N is 7, the atomic no. of IVth member of nitrogen family will be

Atomic mass of chlorine is 35.5. it has two isotopes of atomic mass 35 and 37. the percentage of heavier isotope is:

Atomic mass of boron is 10.81. it has two isotopes with 80% and 20% abundance respectively. The atomic mass of the isotope having 80% abundance is 11.01. the atomic mass of the other isotope is:

Atomic mass of an element is

Atomic hydrogen produces formaldehyde when it reacts with:

Atomic hydrogen is obtained by :

Atom that requires high energy of excitation

Atom A possesses higher values of packing fraction than atom B. The relative stabilities of A and B are

Atom bomb is based on the principle of

For the reaction, following data is given: AtoB,K_(1)=10^(15)"exp"^(((-2000)/T)),CtoD,K_(2)=10^(14)"exp"^(((-1000)/T)) The temperature at which K_(1)=K_(2) is

(A)The reaction of oxalic acid with acidified KMnO_(4) first slow and then proceeds with faster speed(R ) Acidified KMnO_(4) is a strong oxidizing agent

(a)Tin stone-Oxides ore (b)Copper pyrite-Oxide ore (c )Zincite-Oxide ore (d)bauxite-Oxides ore

(a)The rate constants of a reaction at 500 K and 700 K are 0.02 s^(-1) and 0.07 s^(-1) respectively. Calculate the value of Ea. (b) Under what condition a bimolecular reaction behaves kinetically first order reaction ?

[A]:The Arrhenius equation k=Ae^(-E_(a)//RT) gives the relation between rate constant and temperature. [R]:The graph of log kto(1)/(T) is linear and with the help of this calculation of energy of activation can be possible.

[A]:The catalyst increases the rate of reaction. [R]:The catalyst decreases the enthalpy change (DeltaH)

Atetrapeptide on complete hydrolysis gave Alanine, Glycine, Leucine and valine . It never gave Ala-Gly and Val-Ala on partial hydrolysis. How many such tetrapeptides are possible ?

[A]:Temperature of reaction increases by 10^(@)C the collision frequency of molecule increase by 2% to 3% [B]:By increases temperature 10^(@)C the rate of reaction is 200% from 100%

At^(3)Y, minimum numbers of elecrolyte required for a compound to obtained its conductivity is