94420 + Interview Questions in Chemistry in Class 12

1.	If one strand of DNA has the sequence 5' -G-G-A-C-T-A-C-T-3', what is the sequence of bases in the complementary strand ?
Answer» SOLUTION :3.-C-C-T-G-A-T-G-A-5.

Discussion

2.	If one mole of monoatomic ideal gas expanded from 2 atm to 0.5 atm at 27°C, then the entropy change will be
Answer» R In 2 4R In 2 3 RIN 2 2R In 2 Answer :1

Discussion

3.	If one mole of an ideal gas (C_(p.m) = (5)/(2)R) is expanded isothermally at 300 K until it's volume is tripled , then change in entropy of gas is:
Answer» zero INFINITY `(5)/(2)R In 3` `R In 3` SOLUTION :N//A

Discussion

4.	If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas , then
Answer» `DeltaHltDeltaU` there is no relationship `DeltaHgtDeltaU` `DeltaH=DeltaU` Solution :`NH_(3(G))+HCl_((g))toNH_(4)Cl_((g))` `DELTAN=` no . Ofmoles of product - no of moles of REACTANT `=1-2` `Deltan=1` Now , `DeltaH=DeltaE+DeltanRT` `DeltaH=DeltaE-RTorDeltaE=DeltaH+RT` `thereforeDeltaEgtDeltaH(DeltaE=DeltaU)`=change in internal energy.

Discussion

5.	If one mole of AgCZ is dopped with 10^(-5) mole of CaCl_(2) then number of Ag^(+) ions lost from the lattice
Answer» `10^(-5)` `6 xx 10^18` `1.2 xx 10^19` `3 xx 10^18` Answer :C

Discussion

6.	If one mole of AgCl is dopped with 10^(-5)mole of AICI_3, the number of cation vacancies created are
Answer» `10^(-5)` `6 XX 10^18` `12 xx 10^18` `12 xx 10^19` ANSWER :C

Discussion

7.	If onemole monoatmic ideal gas was taken through process AB as shown is figure , then select correcgtoption(s). Given: In 1.5 = 0.4
Answer» `W_(AB)= - 1496.52 J` `q_(AB)=5237.82 J` `DeltaH_(AB) = 3741.3 J` `DeltaS_(AB) "is + ve"` Solution :(a)`"y = mx" + C` `10 = mm xx 300 + C` `15 = mxx 600+ C` `5 = 300 m` `m = (1)/(60), "" c= 5` `V=(1)/(60) T + 5""w = - int PDeltaV` `P = 60 R - (300R)/(V)` ` w = -.^(15)int_(10) (60R-(300R)/(V))DV =- 1496.52J` (b)`DELTAU= q + w or q= DeltaU-w` `DeltaU= C_(V) DeltaT =374 .13` `q = 374.13 + 1496.525 = 5237. 82 J` (c)`Deltah = (5)/(2) xx Rxx 300 ` (d) `DeltaS = + ve`

Discussion

8.	If one million atoms of silver weight 1.79xx10^(-16)gm the gram atomic mass of silver is:
Answer» 107 g/mol 107.2 g/mol 107.8 g/mol 108.2 g/mol Answer :C

Discussion

9.	If one litre of air is passed repeatedly over heated copper and magnesium till no further reduction in volume takes place, the volume finally obtained is
Answer» 800ML 990 ML 10 ml 100 ml Answer :C

Discussion

10.	If one first order reaction is completed by 50% in 1.26xx10^(14) second then how much time will be taken by this reaction to complete 100%?
Answer» `1.26xx10^(15)` SECOND `2.52xx10^(14)` second `2.52xx10^(25)`second INFINITE time Solution :100% reaction will COMPLETED when it runs for infinite time and hence,time can not be calculated.

Discussion

11.	If one end of a piece of a metal is heated the other end becomes hot after some time. This is due to
Answer» ENERGISED ELECTRONS MOVING to the other part of the metal resistance of the metal mobility of atoms, in the metal minor PERTURBATION in the energy of atoms Solution :It is due to MOVEMENT of energised electrons. `(KE propT)`.

Discussion

12.	If one end of a piece of a metal is heated, the other end becomes hot after some time. This is due to
Answer» ENERGISED electrons MOVING to the other part of the metal Resistance of the metal Mobility of ATOMS in the metal Minor perturbation in the energy of atoms Answer :A

Discussion

13.	If one cell has standard electrode potential of 0.0295V and n=2 then calculate its equilibrium constant at 298K temperature
Answer» ANSWER :`K_(C)=1.0xx10^(10)`

Discussion

14.	Ifon performing theexperiment, 2 . 28 g of (NH_(4))_(2) PbCl_(6) was produced, calculatethe percentage yield of (NH_(4))_(2) Pb Cl_(6)
Answer» Solution :`% "yield " = ("amount produced experimentally")/("amountfor 100% yield") XX 100` `= (2 . 28)/( 4 . 56) xx 100 = 50%`

Discussion

15.	If one atom of hydrogen weighs 1.65 xx 10^(-24)g, then mass of one atom of carbon weighs
Answer» `1.98 xx 10^(-23) g ` `1.65 xx 10^(-24) g ` `1.37 xx 10^(-25) g ` `1.40 xx 10^(-23) g ` Solution :An atom of carbon is 12 times HEAVIER than an atom of HYDROGEN . THEREFORE , mass of one atom of carbon ` = 1.65 xx 10^(-24) xx 12 = 1.98 xx 10^(-23) g `

Discussion

16.	If OH-group in lactic acidis replaced by hydrogen atom, which will happened ?
Answer» OPTICAL ACTIVITY is retained optical activity is lost form recemic mixutre PRODUCE non-superimposable mirror image Answer :B

Discussion

17.	If O_(2) gas in bubbled through water at 293 K, how many millimoles of O_(2) gas would dissove in 1 littre of water ? Assume that O_(2) exerts a partial pressure of 0.987 bar and k_(H) for O_(2)=34.86 k bar.
Answer» Solution :Step I. Calculation of `x_(O_(2))` `x_(O_(2))=rho_(O_(2))/k_(H)=((0.98"bar"))/((43860"bar"))=2.83xx10^(-5)` Step II. Calculation os milli-MOLES of `O_(2)` If N moles os `O_(2)` are present in 1 L of water (or 1000 g of water) `X_(N_(2))=n/(n+55.5)=n/(55.5)` `n=X_(N_(2))xx55.5=2.83xx10^(-5)xx55.5` `=157.06xx10^(-5)mol=157.06xx10^(-2)`milli-moles =1.57 milli-moles

Discussion

18.	If observed molar mass of a solute is more than calculated molar mass, then the solute undergoes .............. in the solvent.
Answer» SOLUTION :ASSOCIATION

Discussion

19.	If [OH^-] is 1 xx 10^-8 ion/litre.Is pH is:
Answer» 6 7 5 8 Answer :A

Discussion

20.	If N_(x) is the number of bonding orbitals of an atom and N_y is the number of antibonding orbitals, then the molecule/atom will be stable if
Answer» `N_x ltN_y` `N_x = N_y` `N_x gtN_y` `N_x leN_y` ANSWER :C

Discussion

21.	If n_(t) number of radioatoms are present at time t, the following expression will be a constant
Answer» `n_(t)//t` ln `n_(t)//t` `d ln n_(t)//DT` `t.n_(t)` Solution :`- (dn_(t))/(dt) = LAMDA n_(t) RARR - (1)/(n_(1)) (dn_(t))/(dt) = lamda` `rArr (d)/(dt) (ln, n_(t)) = - lamda`, a CONSTANT

Discussion

22.	If n_t number of radiatoms are present at time t, the following expression will be a constant.
Answer» `n_t//t` `LN n_t//t` `d ln n_t//dt` `t cdot n_t` Solution :`-(dn_t)/(dt) = LAMBDA n_t implies - 1/(n_t) (dn_t)/(dt) = lambda` `implies d/(dt) (ln n_t) = -lambda`, a constant.

Discussion

23.	If no catalyst (H^(+)) is present in acid hydrolysis of ester (in above question) then constant K is :
Answer» `(2.303)/(t)"LOG"(V_(0))/(V_(t)-V_(oo))` `(2.303)/(t)"log"(V_(oo))/(V_(oo)-V_(t))` `(2.303)/(t)"log"(V_(0))/(V_(t))` `(2.303)/(t)"log"(V_(oo))/(V_(t)-V_(oo))` ANSWER :B

Discussion

24.	If non ideal solution is prepared by mixing 30 mL CHCl_(3) and 50 mL Acetone, then volume of solution is ……..
Answer» `GT` 80 ML `lt` 80 mL `= 80` mL `ge` 80 mL Solution :DUE to formation of H - bond between `CHCl_(3)` and acetone A-B attraction is more than A-A and B-B, so this solution shows negative deviation from Raoult.s law. Means,`DELTA V_(mix)-ve, Delta H_(mix)=-ve` So, total volume `= lt` 80 mL.

Discussion

25.	If NH_4OH is added to the (PtCl_2)^(2-) ion, the complexformed represents:
Answer» ZERO dipole Finite dipole Infinite-dipole All of the above Answer :B

Discussion

26.	If N_(B) is the number of bonding electron and N_(A) is the number of antibonding electrons of a molecules. Then choose the incorrect statement(s) for the relationship , N_(B)gtN_(A)"
Answer» Molecule may be STABLE or unstable Molecule may have any integral, frational or zero VALUE of bond order Molecule is only PARAMAGNETIC species Molecule does not EXIST Solution :Molecule may have only integral/ fractional value of bond order Molecule may or not be paramagnetic molecule will exist

Discussion

27.	If NaOH is added to an aqueous solution of zinc ions, a white precipitate appears ad on adding excess NaOH, the precipitate dissolves. In this solution zinc exists in the
Answer» Cationic part ANIONIC part Both in cationic and anionic parts There is not ZINC in the solution Solution :`ZN^(2+)+2NaOHtoZn(OH)_(2)+2Na^(+)` `Zn(OH)_(2)+2NaOHtoNa_(2)ZnO_(2)+2H_(2)O`.

Discussion

28.	If NaOH is added to an aqueous solution of zinc ions a white precipitate appears and on adding excess of NaOH, the precipitate dissolves. In this solution, zinc exists in the
Answer» CATIONIC part Anionic part Both in the cationic and anionic parts there is no zinc ION in the SOLUTION Solution :`Zn^(2+)+2NaOH to Zn(OH)_(2) +2Na^(+)` `Zn(OH)_(2) +2NaOH to Na_(2)ZnO_(2)+2H_(2)O`

Discussion

29.	If NaCl is doped with 10^(4) mol % of SrCl_(2) the concentration of cation vacancies will be (N_(A) = 6.02 xx 10^(23) mol^(-1))
Answer» `6.02 xx10^(16) MOL^(-1)` `6.02 xx10^(17) mol^(-1)` `6.02 xx10^(14) mol^(-1)` `6.02 xx10^(15) mol^(-1)` Answer :B

Discussion

30.	If NaCl is doped with 10^(-4) mol % of SrCl_(2) , the concentration of cation vacancies will be ( N_(A) = 6.022 xx 10^(23) mol^(-1))
Answer» `6.02 xx 10^(18) MOL^(-1)` `6.02 xx 10^(17) mol^(-1)` `6.02 xx 10^(14) mol^(-1)` `6.02 xx 10^(15) mol^(-1)` Solution :ONE cation of `Sr^(2+)` would create one cation vacancy in NaCl. Therefore , the number of cation vacancies created in the LATTICE of NaCl is equal to the number of divalent `Sr^(2+)` ions added. No. of moles of cationic vacancies ` = ( 10^(-4))/( 10^(2)) = 10^(-6)` mol No. of cation vacancies ` = 10^(-6) xx 6.02 xx 10^(23) = 6.02 xx 10^(17)`

Discussion

31.	If NaCl is doped with 10^(-3) mole percent SrCl_(2), what will be the concentration of cation vacancies ? (N_(A) = 6.02 xx 10^(23) mol^(-1))
Answer» Solution :Every `Sr^(2+)` ion causes ONE cation vacancy (because two `Na^(+)` ions are REPLACED by one `Sr^(2+)`) Therefore, introduction of `10^(-3)` moles of `SrCl_(2)` per 100 moles of NaCl would introduce `10^(-3)` mole cation VACANCIES in 100 moles of NaCl. No. of vacancies per mole of NaCl `= (10^(-3))/(10000) xx 6.02 xx 10^(23) = 6.02 xx 10^(18)` vacancies.

Discussion

32.	If NaCl is doped with 10^(-3) mol % SrCl_2, what is the concentration of cation vacancies?
Answer» Solution :Doping of NaCl with `10^(-3) mol % SrCl_2`, means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_2`. `:.` 1 mole of NaCl is doped with `SrCl_2 =(10^(-3))/(100) = 10^(-5)` mol As each `SR^(2+)` ION introduces one cation vacancy, THEREFORE concentration of cation vacancies = `10^(-5) mol//mol` of NaCl = `10^(-5) xx 6.02 xx 10^(23)` cation vacancies `mol^(-1)` = `6.02 xx 10^(18)` cation vacancies `mol^(-1)`

Discussion

33.	If NaCl is doped with 10^(-3) mol % of SrCl_(2) what is the concentration of cation vacancy?
Answer» Solution :ONE cation of `SR^(2+)` would create one cation vacancy in NaCl. `:.`Concentration of cation vacancy on being doped with `10^(-3)` mol% `SrCl_(2)` `= 10 ^(-3) mol%` `=(10 ^(-3))/(100) "mol"` `=(10 ^(-3))/(100) XX 6.02 xx 10^(23)` cation vacancies PER mole `= 6.02 xx10^(18)`cation vacancies per mole.

Discussion

34.	If NaCl is doped with 10^(-3) mol % of SrCl_2, what is the concentration of cation vacancies ?
Answer» SOLUTION :Let moles of NaCl in the crystal = 100 Moles of `SrCl_2` doped = `10^(-3)` `therefore` No. of `NA^(+)` ions replaced by each `SR^(2+)` ions = 1 No. of CATIONIC vacancies formed by `Sr^(2+)` ions = 1 `therefore` Number of cationic vacancies = `10^(-3)` formed by `10^(-3)` mol `SrCl_2` `therefore` Vacancy per mole of `NaCl =(10^(-3))/(100) = 10^(-5)` `therefore` Number of vacancies per mole of NaCl ` = 6.022 xx 10^23 X 10^(-5)` ` = 6.022 xx 10^18`.

Discussion

35.	If NaCl is doped with 10^(-3) mol % of SrCl_(2), what is the concentration of cation vacancies ?
Answer» Solution :Let us take one MOLE of `NaCl`. For every `Sr^(2+)` ion INTRODUCED, one `Na^(+)` ion will be removed and one vacancy will be created. Amount of `SrCl_(2)=(10^(-3))/(100)" moles"` `=10^(-5)" moles "=10^(-5)xx6.02xx10^(23)" molecules "=6.02xx10^(18)" molecules"` As each `Sr^(2+)` ion intrduced creates one cation vacancy, CONCENTRATION of cation vacancies `=6.02xx10^(18)` PER mole.

Discussion

36.	If NaCl is doped with 10^(-2) mol % SrCl_2, what is the concentration of cation vacancies?
Answer» SOLUTION :Doping of NaCl with `10^(-2)` mole % of `SrCl_2` means that 100 MOLES of NaCl are doped with 10 mole of `SrCl_2`. `:.` 1 mole of NaCl is doped with `SrCl_2=(10^(-2))/(100)= 10^(-4)` MOL As each `Sr^(2+)` ion introduces one cation vacancy, therefore, concentration of cation vacancies = `10^(-4) mol//mol` of NaCl = `10^(-4) xx 6.02 xx 10^(23)` cation vacancies `mol^(-1) = 6.02 xx 10^(19)` cation vacancies `mol^(-1)`.

Discussion

37.	If NaCl crystals are doped with 2 xx 10^(-3) mol per cent of SrCl_(2), calculate the cation vacancies per mole.
Answer» Solution :Doping of `NaCl` with `2xx10^(-3)` mol per cent of `SrCl_(2)` means 100 MOLES of `NaCl` are dropped with `2xx10^(-3)` moles of `SrCl_(2)` or 1 mole of `NaCl` is doped with `=2xx10^(-5)` mole of `SrCl_(2)`. Each `Sr^(2+)` will occupy the place of `Na^(+)` and displace one `Na^(+)` from crystal lattice to create cation vacancies. Cation vacancies = Number of `Sr^(2+)` ion added `=2xx10^(-5)mol=2xx10^(-5)xx6.023xx10^(23)=12.046xx10^(18)=1.2046xx10^(19)mol^(-1)`.

Discussion

38.	If NaCl is doped with 10^(-2) mol percentage of strontium chloride, what is the concentration of cation vacancy ?
Answer» Solution :We KNOW that two `Na^(+)`ions are replaced by each of the `Sr^(2+)` ions while `SrCl_(2)` is doped with NaCl. But in this case, only one lattice POINT is occupied by each of the `Sr^(2+)` ions and PRODUCE one cation vacancy. Here `10^(-2)` mole of `SrCI_(2)` is doped with 100 moles of NaCl. Thus, cation vacancies produced by NaCl = `10^(-2)` mol. Since, 100 moles of NaCl produces cation vacancies after doping = `10^(-2)` mol. Therefore, 1 mole of NaCl will produce cation vacancies after doping `=(10^(-2))/(100)=10^(-4)"mol"` `therefore"Total cationic vacanices,"` `=10^(-4)xx"Avogadro.s NUMBER"` `=10^(-4)xx6.023xx10^(23)` `=6.023xx10^(19)" vacancies"`

Discussion

39.	If Na_2SO_3 is left open in air, we get:
Answer» `Na_2S` `Na_2SO_4` `Na_2HSO_4` `Na_2HSO_3` ANSWER :B

Discussion

40.	If N_A is Avogadro's number, then the number of hydrogen atoms in one g-equivalent of hydrogen is :
Answer» `N_A` `N_A//2` `N_A//4` `2N_A` ANSWER :B

Discussion

41.	If Na^(+) ion and S^(2-) ion is larger than Cl^(-) ion, which of the following will b least soluble in water
Answer» `MgS` `NACL` `Na_(2)S` `MgCl_(2)` Solution :`UNDERSET("Solubility DECREASING order")OVERSET(NaCl gt Mg Cl_(2) gt Na_(2)S gt MgS)to`

Discussion

42.	If n_A and n_Brepresent the number of moles of components A and B, the mole fraction of A(x_A) is given by
Answer» `n_A//n_B` `n_B // n_A` `n_A // n_A + n_B` `n_A + n_B // n_A` ANSWER :C

Discussion

43.	If n=6, the correct sequence for filling of electrons will be:
Answer» `NSTO(N-2)F to(n-1)d tonp` `nsto(n-1)d to (n-2)f to np` `nsto(n-2)f to n p to (n-1)d` `nstonpto(n-1)d to (n-2)f` Solution :For n=6 `6s t o4f to 5d to 6P`

Discussion

44.	If N_(2) gas is bubbled through water at 293 K, how many millimoles of N_(2) gas would dissolve in 1 litre of water ? Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 293 K is 76.48 k bar.
Answer» Solution :The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry.s law. Thus : x (Nitrogen) `= ("p (nitrogen)")/(K_(H))` `=(0.987" bar")/(76480" bar")=1.29xx10^(-5)` As 1 LITRE of water contains 55.5 MOL of it, therefore if n represents number of MOLES of `N_(2)` in solution, x(Nitrogen) `= ("n mol")/(n mol +55.5 mol)` `= (n)/(55.5)=1.29xx10^(-5)` (n in denominator is neglected as it is `LT lt 55.5`) Thus, `n = 1.29xx10^(-5)xx55.5 mol = 7.16xx10^(-4)` mol `= (7.16xx10^(-4)mol xx 1000 m mol)/(1 mol)` = 0.716 m mol

Discussion

45.	If N_(2) gas is bubbled through water at 293 K, how many of millimoles of N_(2) gas would dissolve in 1 litre of water 7 Assume that N_(1) exerls a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 293 K is 76.48 K bar.
Answer» Solution :ACCORDING to Henry.s LAW x(Nitrogen ) = `(p("Nitrogen"))/(K_(H))` = `(0.987)/(74,480("bar"))` = `1.29 xx 10^(-5)` As 1 litre of water contains 55.5 mole of it, therefore if n represents number of moles of `N_(2)` in solution. x (Nitrogen ) = `(n"mole")/("n mole + 55.5 mol")` = `(n)/(55.5) = 1.29 xx 10^(-5) = 1.29 xx 10^(-5)` (n in denominator is neglecti!d as it is `lt lt `55.5) So, n= 1.29` xx 10^(-5) xx 55.5 `mol = `7.16 xx 10^(-4)` mol `= (7.16 xx 10^(-4)"mol" xx 1000 "mol")/(1 "mol")` = 0.716 m/mol

Discussion

46.	If N_2 gas is bubbled through water at 293 K , how many millimoles of N_2 gas would dissolve in 1 litre of water? Assume that N_2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N_2 at 293 K is 76.48 k bar.
Answer» ANSWER :0.716 MOL

Discussion

47.	If N_(2) gas is bubbled through water at 293 K, how many millimoles of N_(2) gas would dissolve in 1 litre of water ? Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 298 K is 76.48 kbar.
Answer» <P> Solution :`"According to Henry's law, "p_(N_(2))=K_(H)xxx_(N_(2)), thereforex_(N_(2))=(p_(N_(2)))/(K_(H))=("0.987 bar")/("76480 bar")=1.29xx10^(-5)` If N moles of `N_(2)` are prsent in 1 L of water (i.e., 55.5 moles), `x_(N_(2))=(n)/(n+55.5)~=(n)/(55.5)("as "ltlt 55.5)` `therefore""(n)/(55.5)=1.29xx10^(-5)"or"n=1.29xx10^(-5) xx 55.5" moles"=71.595xx10^(-5)" moles = 0.716 MILLIMOLES"`

Discussion

48.	If N_2 + 3H_2 hArr 2NH_3 has equilibrium constant K and 2N_2 + 6H_2 hArr 4NH_3 has equilibrium constant k', then k' =
Answer» `K^2` sqrtK 1/sqrtK 1/K^2 Answer :A

Discussion

49.	If N_1,N_2,N_3…….. Are the number of molecules with molecular masses M_1,M_2,M_3………. Respectively, than mass average molecular mass is expressed:
Answer» `(sumN_iM_i^2)/(sumN_iM_i)` `(sumN_iM_i)/(sumN_i)` `(sumM_i^2)/(sumN_i)` `(sumN_iM_i^2)/(sumM_i)` ANSWER :A

Discussion

50.	If n_(1) is the number of radio-atoms present at tiem 't' the following expression will be a constant,
Answer» `(n_(t))/(t)` `(In n_(t))/(t)` `(d In n_(t))/(dt)` `t n_(t)` SOLUTION :`(d n_(t))/(dt) = LAMBDA n_(t)` `(1)/(n_(t)) (d n_(t))/(dt) = lambda` `(d)/(dt) (l_(n) n_(t)) = -lambda` constant 1

Discussion

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If N_A is Avogadro's number, then the number of hydrogen atoms in one g-equivalent of hydrogen is :

If Na^(+) ion and S^(2-) ion is larger than Cl^(-) ion, which of the following will b least soluble in water

If n_A and n_Brepresent the number of moles of components A and B, the mole fraction of A(x_A) is given by

If n=6, the correct sequence for filling of electrons will be:

If N_(2) gas is bubbled through water at 293 K, how many millimoles of N_(2) gas would dissolve in 1 litre of water ? Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 293 K is 76.48 k bar.

If N_(2) gas is bubbled through water at 293 K, how many of millimoles of N_(2) gas would dissolve in 1 litre of water 7 Assume that N_(1) exerls a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 293 K is 76.48 K bar.

If N_2 gas is bubbled through water at 293 K , how many millimoles of N_2 gas would dissolve in 1 litre of water? Assume that N_2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N_2 at 293 K is 76.48 k bar.

If N_(2) gas is bubbled through water at 293 K, how many millimoles of N_(2) gas would dissolve in 1 litre of water ? Assume that N_(2) exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N_(2) at 298 K is 76.48 kbar.

If N_2 + 3H_2 hArr 2NH_3 has equilibrium constant K and 2N_2 + 6H_2 hArr 4NH_3 has equilibrium constant k', then k' =

If N_1,N_2,N_3…….. Are the number of molecules with molecular masses M_1,M_2,M_3………. Respectively, than mass average molecular mass is expressed:

If n_(1) is the number of radio-atoms present at tiem 't' the following expression will be a constant,