52671 + Interview Questions in Chemistry in Class 11

1.	Conpound obtained by passing CO_(2) through BaO_(2) in water is
Answer» `CO` `Ba(OH)_(2)` `H_(2)O_(2)` `O_(2)` Answer :C

Discussion

2.	Consdier the following energy level diagram: In the given diagram z refer to
Answer» `6 XX Delta H_(f(CO_(2)))^(0)` `Delta H_(f (C_(6)H_(12)O_(6)))^(0)` `Delta H_("Combustion" (C_(6)H_(12)O_(6)))^(0)` `Delta H_("combustion" (C_(6)H_(12)O_(6)))^(0) + Delta H_(f(H_(2)O_((L))))^(0)` SOLUTION :`C_(6)H_(12) O_(6) + 6O_(2) rarr 6CO_(2) + 6H_(2)O, Delta H= -z`

Discussion

3.	Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne's test is separately done.
Answer» `NH_(2)NH_(2)HCl` and `ClCH_(2)-CHO` `NH_(2)CS NH_(2)` and `CH_(3)-CH_(2)Cl` `NH_(2)CH_(2) COOH` and `NH_(2) CONH_(2)` `C_(6)H_(5)NH_(2)` and `ClCH_(2)-CHO` Answer :D

Discussion

4.	Conjugate base of H_(2) is
Answer» `H_(3)^(+)` `H_(3)^(-)` `H^(+)` `H^(-)` Solution :CONJUGATE base = Conjugate ACID `-H^(+)=H_(2)-H^(+)=H^(-)`

Discussion

5.	Conjugate base of [Cu(NH_(3))_(6)]^(2+) is
Answer» `[Cu(NH_3) _3NH_2]^(-) ` ` [Cu(NH_3)_5NH_2]^(+) ` `[Cu(NH_3) _4NH_2]^(+) ` `[Cu(NH_3)_4NH_2]^(2+) ` SOLUTION :` [Cu(NH_3)_6 ]^(2+)to [Cu (NH_3) _5 (NH_2)]^(+)+H^(+) `

Discussion

6.	Conjugate base of a strong acid is
Answer» a WEAK base a STRONG base neutral a weak acid. Answer :A

Discussion

7.	Conjugate acid of a weak base isalways stronger. What will be the decreasing order of basic strength of the following conjugate bases ? OH^(_), RO^(-), CH_(3)CO O^(-), Cl^(-)
Answer» SOLUTION :Conjugate acids of given bases are : `H_(2)O, ROH, CH_(3)CO OH, HCl`. Their acidic strength is in the order `HCl GT CH_(3)CO OH gt H_(2)O gt ROH`. HENCE, their conjugate bases have strength in the order : `CL^(-) lt CH_(3)CO O^(-) lt OH^(-) lt RO^(-)` or `RO^(-) gt OH^(-) gt CH_(3)CO O^(-) Cl^(-)`.

Discussion

8.	Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases ? OH^(-), RO^(-) , CH_3COO^(-) , Cl^(-)
Answer» SOLUTION :Conjugate acid of the given BASES are `H_2O, ROH, CH_3COOH` and HCL. Order of their acidic strength is `HCl gt CH_3COOH gt H_2O gt ROH` HENCE, order of basic strength of their conjugate bases is `Cl lt CH_3COO lt OH^(-) lt RO^-`

Discussion

9.	conisder the followingdiseases from which human beings are suffered. (I) Asthma(II) Dyspepsia (III) Bronchitis (IV) Emphysema Diseases due to SO_(2) are
Answer» I,III and IV II, III and IV I,II and III Both I and IV Solution :Even low concentration of `SO_(2)` causes RESPIRATORY diseases as asthma, bronchitis emphysema in HUMAN beings.

Discussion

10.	Configuration that shows the highest energy released when an electron is added to the atom
Answer» `1s^2 2s^2 2p^3` `1s^2 2s^2 2p^4` `1s^2 2s^2 2p^5` `1s^2 2s^2 2p^6` ANSWER :C

Discussion

11.	Configuration that does not denote a transition element
Answer» `3d^14s^2` `3d^10 4s^1` `3d^10 4s^2 4p^2` `3d^8 4s^2` ANSWER :C

Discussion

12.	Configuration of the element with the highest ionisation energy is
Answer» `[N E] 3s^1` `[N e] 3s^2 3p^3` `[N e] 3d^(10) 4s^2 4p^3` `[N e] 3s^2 3p^4` ANSWER :B

Discussion

13.	Configuration of a reactive metal
Answer» `2s^2 2p^5` `3s^2 3p^6` `4s^2 3d^10` `5s^1` ANSWER :D

Discussion

14.	Configuration of
Answer» R S D L Solution :N//A

Discussion

15.	Configuration isomerism is shown by the compounds in which groups or atoms are arranged with rigid part like double bonded atoms, cycle asymmetric centre etc. Geometrical isomerism is possible in case of double bonded atoms as well as in cycle. The phytical. Properties of geometrical isomer differs which is observed fact. Which of the following is correct?
Answer» cis-1,4-dimethyl CYCLOHEXANE will contains dipole moment VALUE zero trans- 1,3-dimethyl cyclohexane will contain dipole moment value zero trans- 1,3 - dimethyl CYCLOBUTANE will contain dipole mament values zero trans - 1,2-dimethyl cyclopropane will contain dipole moment value zero SOLUTION :

Discussion

16.	Which of the following is an inner orbital complex as well as diamagnetic in nature ?
Answer» cis-1,4-DIMETHYL cyclohexane will contains dipole moment VALUE zero trans- 1,3-dimethyl cyclohexane will contain dipole moment value zero trans- 1,3 - dimethyl cyclobutane will contain dipole mament values zero trans - 1,2-dimethyl cyclopropane will contain dipole moment value zero Solution :

Discussion

17.	Configuration isomerism is shown by the compounds in which groups or atoms are arranged with rigid part like double bonded atoms, cycle asymmetric centre etc. Geometrical isomerism is possible in case of double bonded atoms as well as in cycle. The phytical. Properties of geometrical isomer differs which is observed fact. Which of the following is true for 2 - butene ?
Answer» cis-form-contain more b.p than trans form trans-form-contain less m.p than cis form cis-form is more stable than trans form refractive index of cis suriu is less than liaus funn Solution :BOLING Point `PROP mu`, therefore Boling Point of Cis `gt` tran

Discussion

18.	Conductivity water is
Answer» the WATER WHOSE own conductance is very very small highly PURIFIED water CONTAINING no suspended impurity the water whose conductance is QUITE high the water which has been sterilised. Answer :A

Discussion

19.	Conductivity (S m) is directly proportional to area of the plates and the concentration of solution ( mol m^(-3)) in it and is inversely proportional to the distance between plates then the unit of constant of proportionality is
Answer» `S m "mol"^(-1)` `S m^(2) "mol"^(-1)` `S^(-2) m^(2)` mol `S^(2) m^(2) "mol"^(-2)` Solution :Conductivity `prop ("Area" XX "CONCENTRATION")/("length")` `=(KAPPA xx "area" xx "concentration")/("length")` `therefore kappa=("conductivity" xx "length")/("area"xx"concentration")` Units of `kappa=(S m^(-) xx m)/(m^(2)xx"mol" m^(-3))` `= S m "mol"^(-1).`

Discussion

20.	Concentration of the Ag^(+) ions in a saturated solution of Ag_(2)C_(2)O_(4) is 2.2xx10^(-4) "mol" L^(-1).Solubilityproduct of Ag_(2)C_(2)O_(4) is
Answer» `2.42xx10^(-8)` `2.66xx10^(-12)` `4.5xx10^(-11)` `5.3xx10^(-12)` Solution :`Ag_(2)C_(2)O_(4)(s) HARR 2Ag^(+)(aq)+C_(2)O_(4)^(2-)(aq)`, `K_(sp)=[Ag^(+)]^(2)[C_(2)O_(4)^(2-)]` GIVEN `[Ag^(+)]=2.2xx10^(-4) "mol"L^(-1)` `:. [C_(2)O_(4)^(2-)]=(2.2xx10^(-4))/(2)=1.1xx10^(-4)M` `:. K_(sp)=(2.2xx10^(-4))^(2) (1.1xx10^(-4))` `=5.324xx10^(-12)`

Discussion

21.	Concentration of pentladite is carried out by using
Answer» GRAVITY process froth FLOATATION process electromagnetic process Any of these. Answer :B

Discussion

22.	Concentration of oxygen in cold water is 100 ppm.
Answer» Solution :FALSE statement (CONCENTRATION of oxygen in cold WATER is 10 PPM)

Discussion

23.	Concentration of H_3 O^(+)of aqueous barium hydroxide is 5 xx 10^(-9) mol L^(-1) What is the molar concentrationof barium hydroxide ?
Answer» Solution : IONIC product of water, `K_w = [H_3O^(+) ] [OH^(-)] = 1xx10^(-14)` Concentration of HYDRONIUM ion,` [H_3 O^(+) ]=5 xx 10^(-9) ` Concentration of hydroxyl ion, `[OH^(-)] = (K_w)/( [H_3 O^+])=(1 xx 10^(-14))/( 5 xx 10^(-9)) =2 xx 10^(-6) M` Hydroxyl ion concentration of base is given as, `[OH^(-) ]=2[Ba(OH)_2 ]` Concentration of` Ba(OH)_2= ( 2 xx 10^(-6) )/(2 )= 10^(-6)` M

Discussion

24.	Concentration of H_(2)O_(2) by vacuum distillation gives hydrogen peroxide
Answer» About 99% pure About 90% pure 30% pure About 50% pure Answer :B

Discussion

25.	Concentration of glucose (C_(6)H_(12)O_(6)) in normal blood is approximately 90 mg per 100 mL. What is the molarity of glucose in blood ?
Answer» Solution :Mass of glucose = 90 mg `= (90)/(1000) = 0.09 G` No. of moles of glucose `= ("Mass of glucose")/("Molar mass")=(("0.09 g"))/(("180 g MOL"^(-1)))=0.005` mol Volume of solution = 100 ML = 0.1 L Molarity of solution (M) `= ("No. of moles of glucose")/("Volume of solution in LITRES")=(("0.005 mol"))/(("0.1 L"))="0.005 mol L"^(-1)=0.005M`

Discussion

26.	Concentration of dissolved oxygen in water is ........ and concentration of dissolved oxygen in air is ........
Answer» SOLUTION :10 PPM, 2,00,000 ppm

Discussion

27.	Concentrated sulphuric acid is an excellent dehydrating agent and is widely used for drying such gases which do not react with it. But concentrated sulphuric acid cannot be used for drying dihydrogen gas. Why ?
Answer» Solution :Conc. `H_(2)SO_(4)` on absorbing `H_(2)O` from MOIST `H_(2)` PRODUCES so much heat that HYDROGEN catches fire.

Discussion

28.	Concentrated sulphuric acid has a density of 1.9 g/mL and is 99% H_2SO_4 by weight. Calculate the molarity of H_2SO_4 in this acid.
Answer» Solution :Suppose we have 1 litre (1000 mL) of the given acid. Mass of 1000 mL of this acid= DENSITY x Volume = `1.9 xx 1000 = 1900` g The given acid is 99% pure. `therefore`The mass of `H_2SO_4` present in 1900 g (1L) of the 99 sample = `99/100 xx 1900 = 1881 g` Moles of `H_(2)SO_(4)` present in 1 L `=1881/98 = 19.19` (`therefore` Molecular mass of `H_(2)SO_(4) = 98`) HENCE, the molarity of the given `H_2SO_4` sample is 19.19 M.

Discussion

29.	Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.50g mL^(-1) ?
Answer» Solution :`68%` NITRIC acid by MASS means that Mass of nitric acid = 68 g Mass of solution = 100 g Molar mass of `HNO_(3) = 63 g mol ^(-1)` `therefore 68g NHO_(3) = (68)/(63) ` mole `= 1. 079` mole Density of solution `= 1,. 504 g mL ^(-1)` `therefore ` Volume of solution `= (100)/(1.504) mL = 66.5 mL = 0.0665L` Molarity of the solution `= ("Moles of the SOLUTE")/("Volume of solution") ` in L `= (1.0-79)/(0.0665) M = 16.23 M`

Discussion

30.	Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of
Answer» NO `NO_(2)` `N_(2)O` `N_(2)O_(4)` SOLUTION :`4HNO_(5)to4NO_(2)+O_(2)+2H_(2)O`

Discussion

31.	Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23 g of HNO_3.Density of cone. HNO_3 solution is 1.41 g cm^(-3)?
Answer» ANSWER :`23.64 CM^(3)`

Discussion

32.	Concentrated nitric acid renders aluminium passive. Give reasons.
Answer» Solution :When conc. `HNO_(3)` is added to aluminium a thin protective layer of alumina `(Al_(2)O_(3))` is FORMED on its SURFACE which prevents further ACTION. As a result, Al BECOMES passive. `2Al+6HNO_(3)toAl_(2)O_(3)+6NO_(2)+3H_(2)O`

Discussion

33.	Conc. HNO_(3) on dehydration with P_(4)O_(10) gives…………
Answer» ANSWER :`N_(2)O_(5)`

Discussion

34.	Conc. HNO_(3) isstored in containers made up of
Answer» `CU` `Zn` `AL` `Sn`. Solution :Al is renderedpassiveby formation of a thin impervious LAYER of `Al_(2)O_(3)`on its surfacewhen broughtin contact with conc. `HNO_(3)`.

Discussion

35.	Conc. HNO_(3) can be stored in container of
Answer» FE Al Zn Sn SOLUTION : .Al. is passive towards concentrated HNO, due to formation of protective oxide layer `:. HNO_(3)`can be STORED in.Al. container

Discussion

36.	Compute the heat of formation of liquid methyl alcohol in kJ mol^(-1) , using the following : Heat of vaporisation of liquid methyl alcohol=38 kJ mol^(-1). Heat of formation of gaseous atoms from the elements in their standard states : H, 218 kJ//mol , C,715 kJ // mol, O, 249 kJ //mol. Averagebond energies : C-H, 415 kJ //mol ,C-O kJ //mol, O-H, 463 kJ //mol
Answer» Solution :We aim at `: C(s) +2H_(2)(g) +(1)/(2)O_(2) RARR CH_(3)OH(l), DeltaH =?` We are given`:(i) CH_(3)OH(l) rarr CH_(3) OH (g), DeltaH = 38 kJ mol^(-1)`(III) `C(s)rarr C(g), DeltaH =715 kJ mol^(-1)` (IV) `(1)/(2)O_(2)(g) rarrO(g) , DeltaaH -249 kJ mol^(-1)` Also fromthe givenbond energies, we have (V)`CH_(3)OH(g)(H- underset(H) underset(\|) overset(H)overset(\|) (C) -O-H) rarr C(g) + 4H(g)+O(g), DeltaH= 3 xx 415 +356 +463 = 2064kJ mol^(-1)` Eqn. (iii) `+4 ` Eqn. (ii) `+` Eqn. (iv)- Eqn.(i) - Eqn. (v) gives the requried RESULT i.e., `DeltaH = 715 + 4( 218) + 249-38 -2064 = -266 kJ mol^(-1)`

Discussion

37.	compuond with difference in their boiling point by about 3^(@)C can be separated by simple distillation.Reason: all liquid mixture can be separated by distillation method
Answer» If both ASSERTION and reason are correct and REASONIS crect EXPLANATION for assertion If both assertion and reason are correct but reasonis not COTTECT explanation for assertion If reason is correct but assertion is incorrect If both assertion and reason are incorrect Answer :c

Discussion

38.	Compressibility factor (Z = (PV)/(nRT))is plotted against pressure. What is the order of liquefiability of the gas,
Answer» `H_2 LT N_2 lt CH_4 lt CO_2` `CO_2 lt CH_4 lt N_2 lt CH_2` `H_2 lt CH_4 lt N_2 lt CO_2` `CH_4 lt H_2 lt N_2 lt CO_2` SOLUTION :`a UARR IMPLIES ` more deviation `implies` more liquid .

Discussion

39.	Compressibility factor, Z, of a gas is given as Z=(pV)/(nRT)(i) What is the value of Z for an ideal gas ?For real gas what will be the effect on value of Z above Boyle.s temperature ?
Answer» Solution :(i) For ideal gas, compressibility factor, Z = 1. (II) Above Boyle.s temperature, real gases show positive deviation. So, `Z gt 1` Note : FOLLOWING important points come out from the compressibility factor, `Z=(PV)/(nRT)` (i) For ideal Z = 1 at all TEMPERATURES and PRESSURE because pV = nRT. (ii) At very low pressure all gases shown have `Z ~=1` and behave as ideal gas. (iii) At high pressure, all the gases have `Z gt 1` These are more difficult to compress. (iv) At intermediate pressures most gases have `Z lt 1`.

Discussion

40.	Compressibility factor, Z, of a gas is given as Z=(pV)/(nRT) (i) What is the value of Z for an ideal gas ? (ii) For real gas what will be the effect on value of Z above Boyle's temperature ?
Answer» Solution :(i) For ideal GAS, Z=1. (II) For a real gas, above Boyle's temperature, gas SHOWS POSITIVE deviation and hence `Z gt 1`

Discussion

41.	Compressibility factor, Z of a gas is given as Z=(pV)/(nRT) (i) What is the value of Z for an ideal gas ? (ii) For real gas what will be the effect on value of Z above boyle's temperature ?
Answer» SOLUTION :`Z GT 1`

Discussion

42.	Compressibility factor for H2 behaving as real gas is
Answer» `1` `(1 - a/(RTV))` `(1 + (PB)/(RT))` `(RTV)/((1 - a))` ANSWER :C

Discussion

43.	Compressibility factor for CO_(2) at 400 K and 71.00 bar is 0.8697. The molar volume of CO_(2) under these conditions is. . . . . . . . . .
Answer» `22.04" dm"^(3)` `2.24" dm"^(3)` `0.41" dm"^(3)` `19.5" dm"^(3)` SOLUTION :Compressibility factor `(z)=(PV)/(NRT)` `V=(z XX nRT)/(P)=(0.8697xx 1xx8.314xx10^(-2)" bar dm"^(3)K^(-1)mol^(-1)xx400K)/("71 bar")` `V=0.41" dm"^(3)`

Discussion

44.	Compressibility factor for H_(2) behaving as real gas is
Answer» `1` `(1-(a)/(RTV))` `(1+(Pb)/(RT))` `(RTV)/((1-a))` ANSWER :C

Discussion

45.	Compreshension The first demontration of the stereochmistry the S_(N^(2)) reaction was carred out in 1935 by prof ,E,D Huges and his colleagues at the University of london , they allowed® -2-iodooctane to react with readioactiver iodide ion .^(I-). CH_(3)underset("2- iodooctane")underset(I)underset(\|)(CH)(CH_(2))_(5)CH_(3)+^()I^(-)hArrCH_(3)underset("radioactive ") underset("2-iodooctane")underset(^())underset(\|)(CH_(3))(CH_(2)_(5)Ch_(3)+I^(-) the rate of subsittuiton (rate constantK_(s) was determined by measuring the rate incoprotion of radioactvity into the alkyl halide the rate of loss of optical activity from the alky halide (rate constant K_(@)was alsodeterined under the same coditionsconsitions . what ratio K_(@)//K_(S) is prediected for each of hte following stereochemical scenarios : For inversion reaction : (A) (K_(@))/(K_(s))=1 (b) (K_(@))/(K_(S))lt1(c ) (K_(@))/(K_(s))gt1(d) can not be predicted B. For equal amounts of both retention and inversion ? (K_(@))/(K_(s))=1 (b) (K_(@))/(K_(S))lt1(c ) K_(@))/(K_(s))gt1(d) can not be predicted
Answer» SOLUTION :NA

Discussion

46.	Compressibility factor for CO_(2) at 400 K and 71.0 bar is 0.8697. The molar valume of CO_(2) under these conditions is
Answer» `22.04dm^(3)` `2.24dm^(3)` `0.41dm^(3)` `19.5dm^(3)` SOLUTION :`V_("IDEAL")` is EQUAL to nRT/P

Discussion

47.	Compressed oxygen is sold at a pressure of 100 atmosphere in a cylinder of 49 litre. The number of moles of oxygen in the cylinder is:
Answer» 400 100 300 200 Solution :ONE mole OCCUPIES a VOLUME approximately 24.5 LITRE under ORDINARY atmospheric conditions

Discussion

48.	Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The elements with atomic numbers 35, 53 and 85 are all .......
Answer» noble gases HALOGENS heavy metals light metals Solution :Each period in periodic table ends up with noble gas. The atomic number of noble gases are 2, 10, 18, 36, 54 and 86. (i.e., GROUP 18). ELEMENTS with having atomic numbers 35 (36 - 1), 53 (54 - 1) and 85 (86 - 1), lie in the group before noble gas, i.e., halogen family. THUS, atomic number 35, 53 and 85 all belongs to halogens.

Discussion

49.	Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table ?
Answer» 107 118 126 102 Solution :PERIODIC table contains ELEMENTS with atomic number from 1 to 118.

Discussion

50.	Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table. The electronic configuration of the element which is just above the element with atomic number 43 in the same group is .........
Answer» `1S^(2) 2s^(2) 2p^(6) 3s^(6) 3p^(6) 3d^(5) 4s^(2)` `1s^(2) 2s^(2) 2p^(6) 3s^(6) 3p^(6)3d^(5) 4s^(3)` `1s^(2) 2s^(2) 2p^(6) 3s^(6) 3p^(6) 3d^(6) 4s^(2)` `1s^(2) 2s^(2) 2p^(6) 3s^(6) 3p^(6) 3d^(7) 4s^(2)` Solution :Fifth period starts with RB (Z = 37) and ends at Xe (Z = 54). Thus, Z = 43 lies in the `5^(th)` period. As the `4^(th)` period has the 18 elements, therefore the atomic number of that element lies, above the element with atomic number 43 is 43 - 18 = 25. Electronic configuration of element z = 25 is, `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(2) ( :.i.e, Mn)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Conpound obtained by passing CO_(2) through BaO_(2) in water is

Consdier the following energy level diagram: In the given diagram z refer to

Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne's test is separately done.

Conjugate base of H_(2) is

Conjugate base of [Cu(NH_(3))_(6)]^(2+) is

Conjugate base of a strong acid is

Conjugate acid of a weak base isalways stronger. What will be the decreasing order of basic strength of the following conjugate bases ? OH^(_), RO^(-), CH_(3)CO O^(-), Cl^(-)

Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases ? OH^(-), RO^(-) , CH_3COO^(-) , Cl^(-)

conisder the followingdiseases from which human beings are suffered. (I) Asthma(II) Dyspepsia (III) Bronchitis (IV) Emphysema Diseases due to SO_(2) are

Configuration that shows the highest energy released when an electron is added to the atom

Configuration that does not denote a transition element

Configuration of the element with the highest ionisation energy is

Configuration of a reactive metal

Configuration of

Which of the following is an inner orbital complex as well as diamagnetic in nature ?

Conductivity water is

Conductivity (S m) is directly proportional to area of the plates and the concentration of solution ( mol m^(-3)) in it and is inversely proportional to the distance between plates then the unit of constant of proportionality is

Concentration of the Ag^(+) ions in a saturated solution of Ag_(2)C_(2)O_(4) is 2.2xx10^(-4) "mol" L^(-1).Solubilityproduct of Ag_(2)C_(2)O_(4) is

Concentration of pentladite is carried out by using

Concentration of oxygen in cold water is 100 ppm.

Concentration of H_3 O^(+)of aqueous barium hydroxide is 5 xx 10^(-9) mol L^(-1) What is the molar concentrationof barium hydroxide ?

Concentration of H_(2)O_(2) by vacuum distillation gives hydrogen peroxide

Concentration of glucose (C_(6)H_(12)O_(6)) in normal blood is approximately 90 mg per 100 mL. What is the molarity of glucose in blood ?

Concentration of dissolved oxygen in water is ........ and concentration of dissolved oxygen in air is ........

Concentrated sulphuric acid is an excellent dehydrating agent and is widely used for drying such gases which do not react with it. But concentrated sulphuric acid cannot be used for drying dihydrogen gas. Why ?

Concentrated sulphuric acid has a density of 1.9 g/mL and is 99% H_2SO_4 by weight. Calculate the molarity of H_2SO_4 in this acid.

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.50g mL^(-1) ?

Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of

Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23 g of HNO_3.Density of cone. HNO_3 solution is 1.41 g cm^(-3)?

Concentrated nitric acid renders aluminium passive. Give reasons.

Conc. HNO_(3) on dehydration with P_(4)O_(10) gives…………

Conc. HNO_(3) isstored in containers made up of

Conc. HNO_(3) can be stored in container of

compuond with difference in their boiling point by about 3^(@)C can be separated by simple distillation.Reason: all liquid mixture can be separated by distillation method

Compressibility factor (Z = (PV)/(nRT))is plotted against pressure. What is the order of liquefiability of the gas,

Compressibility factor, Z, of a gas is given as Z=(pV)/(nRT)(i) What is the value of Z for an ideal gas ?For real gas what will be the effect on value of Z above Boyle.s temperature ?

Compressibility factor, Z, of a gas is given as Z=(pV)/(nRT) (i) What is the value of Z for an ideal gas ? (ii) For real gas what will be the effect on value of Z above Boyle's temperature ?

Compressibility factor, Z of a gas is given as Z=(pV)/(nRT) (i) What is the value of Z for an ideal gas ? (ii) For real gas what will be the effect on value of Z above boyle's temperature ?

Compressibility factor for H2 behaving as real gas is

Compressibility factor for CO_(2) at 400 K and 71.00 bar is 0.8697. The molar volume of CO_(2) under these conditions is. . . . . . . . . .

Compressibility factor for H_(2) behaving as real gas is

Compressibility factor for CO_(2) at 400 K and 71.0 bar is 0.8697. The molar valume of CO_(2) under these conditions is

Compressed oxygen is sold at a pressure of 100 atmosphere in a cylinder of 49 litre. The number of moles of oxygen in the cylinder is: