52671 + Interview Questions in Chemistry in Class 11

1.	For the buffer solution containing NH_(4)OH and NH_(4) CI, P^(H) of the buffer solution can be increased by
Answer» Adding some more `NH_4 CL` Adding some more `NH_4 OH` Removing `NH_4 Cl` Both 2 and 3 SOLUTION :` POH =pK_b +log ""([S])/([B]) ` ` pH uparrow rArr pOH downarrow i.e. , ([S])/([B]) downarrow ` `therefore ` add base or remove salt

Discussion

2.	For the balanced chemical reaction HNO_(3) + H_(2)S rarr NO + H_(2)O + S. The correct statements are:-
Answer» a. The STOICHIOMETRIC coefficient of `HNO_3`is 2 b.The stoichiometric coefficient of `H_2S`is 3 c. The stoichiometric coefficient of `NO` is 2 d.The stoichiometric coeffecient of `H_2O`is 4 Solution :Change in OXIDATION number of N = 3 Change in oxidation number of S = 2 `2HNO_3 +3H_(2)Srarr 2NO +4H_(2)O +3S` `:.x = 2 , y = 3, a = 2, b = 4`

Discussion

3.	For the arrangement classified as ABAB type the correct combination is
Answer» (I) (ii) P (III) (i)Q (II) (iii)R (IV) (iv) S Answer :C

Discussion

4.	For testing the presence of sulphur by lead acetate test, the Lassaigne's extract is acidified with dilute acetic acid. Can we use dilute hydrochloric or sulphuric acid instead of acetic acid ?
Answer» Solution :For testing sulphur, the Lassaigne's extract is acidified with acetic because LEAD acetate is soluble and does not INTERFERE with the test. However, it HCL or `H_(2)SO_(4)` is used, lead acetate with REACT with it forming while ppt. of `PbCl_(2)` or `PbSO_(4)` and, therefore, it will interfere with the test. `{:((CH_(3)COO)_(2)Pb,+,2HCl,rarr,2CH_(3)COOH,+,PbCl_(2)),("Lead acetate",,,,"Acetic acid",,"(While ppt.)"),((CH_(3)COO)_(2)Pb,+,H_(2)SO_(4),rarr,2CH_(3)COOH,+,PbSO_(4)),(,,,,,,"(While ppt.)"):}`

Discussion

5.	For testing halogens in an organic compound with AgNO_(3) solution, sodium extract (Lassaigne's test) is acidified with dilute HNO_(3). What will happen if a student acidifies the extract with dilute H_(2)SO_(4) in place of dilute HNO_(3) ?
Answer» Solution :`AgNO_(3)` reacts with DILUTE `H_(2)SO_(4)` to form a white precipitate of silver SULPHATE. `2 AgNO_(3) (aq)+H_(2)SO_(4)(aq) rarr underset("(White ppt.)")(Ag_(2)SO_(4)(s)) +2 HNO_(3) (aq)` This precipitate INTERFERES with the precipitate of silver halide `(AgX)` which is formed in the Carius method for the estimation of halogens

Discussion

6.	For tartaric acid total number of possible stereo isomers (configurations only) and total no. of possible optically active isomers are.
Answer» 4, 4 3, 3 4, 2 3, 2 Solution :`HOOC-OVERSET(OH)overset(\|)CH-overset(OH)overset(\|)CH-COOH`

Discussion

7.	For spontaneous reaction what is the value of DeltaG?
Answer» For SPONTANEOUS REACTION what is the VALUE of Negative Zero Any Answer :B

Discussion

8.	For sharpening and grinding metals and other substances which substance is used ?
Answer» BERYLLIUM CARBIDE TUNGSTEN carbide Calcium carbide Carborendum Answer :D

Discussion

9.	Orbitals having equal energies are
Answer» the most probable DISTANCE increases with increase in ‘n’ the most probable distance DECREASES with increase in ‘n’ the most probable distance REMAINS CONSTANT with increase in .n. the most probable distance increases with increse in `.n^(2).` Solution :For similar ORBITALS most probable distance incresaes with .n.

Discussion

10.	The probability of occurrence of an event is 2/5 and the probability of non occurrence of another eventis 3/10 if these events are independent then the probability that only one of the two events occursis
Answer» ALSO INDEPENDENT of anlges spherically SYMMETRIC both (1) and (2) are correct both (1) and (2) are incorrect Solution :s-orbital spherically symmetic so PROBABILITY is independent of ANGLE of direction.

Discussion

11.	For real gases, van der Waals equation is written as (p+(an^(2))/(V^(2)))(V-nb)=nRT Where 'a' and 'b' are van der Waals constants. Two sets of gases are : (I) O_(2),CO_(2),H_(2) and He(II) CH_(4),O_(2) and H_(2) Thegases given in set-I in increasing order of 'b' and gases given in set-II in decreasing order of 'a' are arranged below. Select the correct order from the following :
Answer» `(I) H_(2) lt He lt O_(2) lt CO_(2) (II) CH_(4) gt O_(2) gt H_(2)` `(I) H_(2) lt O_(2) lt CO_(2) (II) O_(2) gt CH_(4) gt H_(2)` `(I) He lt H_(2) lt CO_(2) lt O_(2) (II) CH_(4) gt H_(2) gt O_(2)` `(I) O_(2) lt He lt H_(2) lt CO_(2) (II) H_(2) gt O_(2) gt CH_(4)` Solution :a' is related to intermolecular forces of attraction. Greater the intermolecular forces, higher is the critical TEMPERATURE and more easily the gas is LIQUIFIED. Thus, decreasing order of 'a' in II is `CH_(4) gt O_(2) gtH_(2)` (critical temperatures of `CH_(4),O_(2)` and `H_(2)` are 304 K, 154 K and 33 K respectively). 'B' is related to the size of the molecules. Hence, in set I, increasing order of 'b' is `H_(2)ltHeltCO_(2)`.

Discussion

12.	For real gases, van der Waal's equation is written as [p + (an^(2))/(V^(2))] (V - b) = nRT where a and b are van der Waal's constant. Two sets of gases are (I) O_(2), CO_(2), H_(2) " and " He (II) CH_(4), O_(2), H_(2). The gases given in set I in increasing order of b and gases given in set II in decreasing order of a are arranged below. Select the correct order from the following :
Answer» (I)` He lt H_(2) lt CO_92) lt O_(2)` (II) `CH_(4) gt H_(2) gt O_(2)` (I) `O_(2) lt He lt H_(2) lt CO_(2)` (II) `H_(2) gt O_(2) gt CH_(4)` (I) `H_(2) lt He lt O_(2) lt CO_(2)` (II) `CH_(4) gt O_(2) gt H_(2)` (I) `H_(2) lt O_(2) lt He lt CO_(2)` (II) `O_(2) gt CH_(4) gt H_(2)` SOLUTION :`a prop` intermolecular attractive forces `b prop` EFFECTIVE size of the molecules `:.` order should be (I) `H_(2), lt He lt O_(2) lr CO_(2)`, (II) `CH_(4) gt O_(2) gt H_(2)`

Discussion

13.	For real gases the relation between p, V and T is given by van der Waals equation : (p+(an^(2))/(V^(2)))(V-nb)=nRT where 'a' and 'b' are van der Waals constants, 'nb' is approximately equal to the total volume of the molecules of a gas, 'a' is the measure of magnitude of intermolecular attraction. (i) Arrange the following gases in the increasing order of 'b'. Give reason. O_(2),CO_(2),H_(2),He (ii) Arrange the following gases in the decreasing order of magnitude of 'a'. Give reason. CH_(4),O_(2),H_(2)
Answer» Solution :(i) As 'b' REPRESENTS molar volume occupied by the gas molecules, GREATER the size of the molecules, greater is the volume occupied by 1 mole of molecules. The size and hence the value of 'b' increase in the order : `H_(2) lt He lt O_(2) lt CO_(2)` (ii) As all the given molecules are non-polar, the magnitude of intermolecular attractions and hence the value of 'a' increase with the increase in numberof electrons in the MOLECULE, i.e., in the order : `CH_(4) gt O_(2) gt H_(2)` (Greater the number of electrons, greater is the size of electron cloud, greater is the polarisation of the molecule, greater is the attraction).

Discussion

14.	For real gases van der Waals equation is written as (P + (an^2)/(V^2))(V-nb )=nRTwhere a andb are van der Waals constants two sets of gases are (I)O_2 ,CO_2,H_2and He (II )CH_4,O_2and H_2 The gases given in set - I in increasing order of b and gases given in set - II in decreasing order of a, are arranged below. Select the correct order from the following
Answer» `(I) HELT H_2lt CO_2 LT O_2 (II)CH_4gt H_2gt O_2` `(I )O_2 ltHe lt H_2 lt CO_2 (II)H_2gtO_2 gtCH_4` `(I) HeltH_2 lt O_2lt CO_2 (II)CH_4gt O_2gt H_2 ` `(I)H_2 ltO_2lt Helt CO_2(II)O_2gtCH_4gt H_2 ` ANSWER :C

Discussion

15.	For real gases the relation between p, V and T is given by van der Waals equation :(p+(an^(2))/(V^(2)))(V-nb)=nRTWhere .a. and .b. are vasn der Waals constants, .nb. is approximately equal to the total volume of the molecules of a gas. .a. is the measure of magnitude of intermolecular attraction.(i) Arrange the following gases in the increasing order of .b.. Give reason.O_(2), CO_(2), H_(2), He(ii) Arrange the following gases in the decreasing order of magnitude of .a.. Give reason.CH_(4), O_(2), H_(2)
Answer» Solution :(i) Molar volume occupied by the gas molecules size of the molecules and VAN der Waals. CONSTANT .b. shows the molar volume of the gas molecules. Hence, value of .b. increases as `H_(2)lt He lt O_(2)lt CO_(2)` (ii) van der Waals. constasnt .a. is the measure of magnitude of intermolecular attraction. As the magnitude of intermolecular attractions increases there is a increae in size of electron cloud in molecule. Thus, for the gases magnitude of .a. decreases as `CH_(4)GT O_(2)gt H_(2)` Greater the size of electron cloud, greater is he DISPERSION forces or London forces and higher is the polarisability of molecule.

Discussion

16.	For reaction XeF_6+H_2O hArr XeOF_4+2HF, K_1=4 XeO_4+XeF_6 hArr XeOF_4 + XeO_3F_2 , K_2=100 Find equilibrium constant of1/2XeO_4+HF hArr 1/2XeO_3F_2+1/2H_2O
Answer» ANSWER :5

Discussion

17.	For reaction, PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)), K_c=1.79 "L mol"^(-1).Then at 500 K state the value of K_p with respect to R.
Answer» Solution :`K_p=K_c(RT)^(DELTAN)`, where `Deltan` =1+1-1=1 =1.79`[(R)(500)]^1` =895R

Discussion

18.	For reaction of (Propene + HBr)…. (i) 2-bromopropane is obtained on reaction of HBr with propene. (ii) 2-bromopropane is obtained as main product on reaction of HBr with propene in presence of peroxides. (iii) 1-bromopropane is obtained as main product on reaction of HBr with propene in presence of peroxides. (iv) Propene does not react with propene.
Answer» ANSWER :(i-T), (ii-F), (iii-T), (iv-F)

Discussion

19.	For a reaction, N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g), identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.
Answer» 14 G of `N_(2) + 4G` of `H_(2)` 28 g of `N_(2) + 6g` of `H_(2)` 56 g of `N_(2) + 10g` of `H_(2)` 35 g of `N_(2) + 8g` of `H_(2)` Answer :C

Discussion

20.	For reaction, assuming large volume of water. H_(2)O(l)hArrH_(2)O(g) , "at temp". T K Choose correct options:
Answer» On introduction of an inert gas at constant temperature pressure in the container remins same at equilibrium. For this system `%` relative humidity ALWAYS remains `100%` at constant temperature at equilibrium If steam at temperature `2T` is passed into given system. After equilibrium is attained relative humidity changes. This is a SPECIAL case of equilibrium where pressure of `H_(2)O(g)` remains same always due to UNIQUE structural feature of `H_(2)O`. Solution :N//A

Discussion

21.	For reaction, 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g) which statements are correct? (P) K_(c)=[SO_(2)][O_(2)]//[SO_(3)] (Q) Addition of O_(2)(g) tothe system at contant temperature and valume would decrease the value of K_(c).
Answer» <P>P only Q only Both P and Q Neither P nor Q Answer :D

Discussion

22.	For reaction, 2NOCl_((g))hArr2NO_((g))+Cl_(2(g)),K_(c) at 427^(@)C is 3xx10^(-6)L" "mol^(-1). The value of K_(p) is nearly,
Answer» `7.50xx10^(-5)` `2.50xx10^(-5)` `2.50xx10^(-4)` `1.75xx10^(-4)` ANSWER :D

Discussion

23.	For reacting with HCl, the alcohol which does not requrie ZnCl_(2) is_____
Answer» `CH_(3)CH_(2)OH` `CH_(3)-CH_(2)CH_(2)OH` `CH_(3)-CHOH` `(CH_(3))_(3)C-OH` SOLUTION :`(CH_(3))_(3)C-OH`

Discussion

24.	For reaction, 2A+BhArr2C,K=x. Equilibrium constant for ChArrA+1//2B will be
Answer» X `x/2` `(1)/(SQRT(x))` `sqrt(x)` ANSWER :C

Discussion

25.	Which of the following is/are correct about the radial probability curves ?
Answer» The number of MAXIMA in 2s orbital are two The number of spherical or radial nodes is equal to N -L-1 The number of ANGULAR nodes are .l.. `3d_x^2` has 3 angular nodes Solution :a,B,c are correct . d) `3d_2^2` has no angular nodes.

Discussion

26.	For process H_(2)o(l)(100^(@)C,1atm)hArrH_(2)O(g)(100^(@)C,1 atm) {:(,"Column-I",,"Column-II"),((a),DeltaU,(p),0),((b),DeltaS ("system+surrounding"),(q),DeltaH),((c),"work",(r),"positive"),((d),"Heat involved",(s),"Negative"):}
Answer» ANSWER :`(a-r);(b-p);(c-s);(d-q)`

Discussion

27.	For principal quantum number n = 4, the total number of orbitals having l = 3 is
Answer» 3 7 5 9 Solution :When `n = 4, l = 0, 1, 2, 3`, i.e., 4s, 4P, 4D 4f subshells. But `l = 3` is only for f-subshell which has 7 orbitals.

Discussion

28.	For principle quantum number n=4,the total number of orbitals having l=3 is
Answer» 3 7 5 9 Answer :B

Discussion

29.	What happens when Al_4C_3 reacts with H_2O ?
Answer»

Discussion

30.	For preparing monoalkyl benzene , acylation process is preferred than direct alkylation because
Answer» In ALKYLATION, a poisonous gas is evolved In alkylation, large AMOUNT of HEAT is evolved In alkylation, polyalkylated PROUDCT is formed Alkylation is very costly. Answer :C

Discussion

31.	For preparing an alkane a concentrated aqueous solution of sodium or potassium salt of a saturated caroboxylic acid is subjected to
Answer» Hydrolysis Oxidation Hydrogenation Electolysis Answer :D

Discussion

32.	For preparing 0.1 N solution ofa compound from the impure sample of which the percentage purity is known, the weight of the substance required will be
Answer» LESS than the THEORETICAL weight more than the theoretical weight same as the theoretical weight none of these Answer :B

Discussion

33.	For precipitation of sparingly soluble salt if, I_p lt K_(sp), then …..
Answer» nothing can be predicted sparingly soluble SALT will not get precipitated. solution will remain in saturated STATE. sparingly soluble salt gets precipitated. Answer :B

Discussion

34.	For PCl_(5 (g)) hArr PCl_(3 (g)) + Cl_(2 (g)) at equilibrium , K_(P)= P//3. Then degree of dissociation of PCl_(5) at that temperature is
Answer» `0.75` `0.25` `0.9` `0.5` ANSWER :D

Discussion

35.	For packing with minimum void space the true combination is (Consider 2D & 3D arrangements separately)
Answer» (I) (i) P (II) (i) Q (III) (iv) S (IV) (iii)R Answer :B

Discussion

36.	For particles having same kineticenergy , the deBrogle wavelenght is-
Answer» DIRECTLY propertioanl to its VELOCITY INVERSELY proportial to its velcotiy indepenedent of velocity and mass Unperdicatable SOLUTION :N//A

Discussion

37.	For oxidation of iron,, 4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3)(s),entropy change is - 549 .4 JK^(-1) mol^(-1) at 298 K . Inspite of the negativeentropy change of this reaction , why is the reaction spontaneous ? ( Delta H ^(@) for this reaction is - 1648 xx 10^(3) J mol^(-1))
Answer» SOLUTION :Proceed as in Solved PROBLEM on page `6 // 87` by either method (Prove that `DELTAG^(@) ` is `-ve` or `DeltaS_("total") ` is `+`ve).

Discussion

38.	For oxidation of iron, 4Fe_((s)) + 3O_(2(g)) to 2Fe_(2) O_(3(s)) entropy change is -549.4 "JK"^(-1) "mol"^(-1) at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous ? Delta_(r) H^( Theta ) for this reaction is 1648 xx 10^(3) "J mol"^(-1)
Answer» Solution :One decides the spontaneity of a reaction by considering : `DeltaS_("total")=(DeltaS_("sys")+ DeltaS_("SURR")).` For calculating. `DeltaS_("surr")` we have to consider the heat absorbed by the surroundings which is equal to `- Delta_(r) H^( Theta )`. At temperature T, entropy change of the surroundings is `DeltaS _("surr") =-(Deltar H^( Theta ) )/( T) ("at constant pressure")` `=- ((-1648 xx 10^(3) "J mol"^(-1) ))/(298 K)= 5530 "kJ mol"^(-1)` Thus, total entropy change for this reaction `Delta_("total") = 5530 "JK"^(-1) "mol"^(-1) + (-549 .4 "JK mol"^(-1) )` `=4980.6 "JK mol"^(-1)` This SHOWS that the above reaction is spontaneous.

Discussion

39.	For one solution, K_a = 1.0 xx 10^(-8) then what is the pK_a and PK_b of the solution ?
Answer» SOLUTION :`pK_a=-log K_a=-log (1XX10^(-8))=+8` `pK_b=14.0-pK_a=14.0-8`=6.0

Discussion

40.	For one mole of an ideal gas, if P = P_(0)/(1+(V/V_(0))^(2)) , where P_(0) and V_(0) are constants. Which of the following are true?
Answer» `P = P_0//2 "when " V = V_0` `V = 2 V_0` when `P = P_0` `T = P_0 V_0//2R ` when `V = V_0` `R = P/(TV_0) ` when `P = P_0` Solution :`P = (P_0)/(1 + (V/(V_0))^2)` a) `V = V_0 implies P = (P_0)/2` B) `P = P_0 implies 1 + (V/(V_0))^2 = 1 implies V/(V_0) = 0`. c) `V = V_0 implies P = (P_0)/(2) T = (PV)/(nR) = (P^@ xx V_0)/(2R)`.

Discussion

41.	For one mole of an ideal gas, increasing the temperature from 10^(@)C to 20^(@)C.
Answer» increases the average kinetic energy by two times increases the RMS by `sqrt(2)` times increased the rms by 2 times increases both the average kinetic energy and rms VELOCITY but not significantly Solution :`bar(KE)=(3)/(2)RT :. (bar(KE_(2)))/(bar(KE_(1)))=(T_(2))/(T_(1))=(293)/(283)=1.04` `:. bar(KE_(2))=1.04bar(KE_(1))` `u_(rms)=sqrt((3RT)/(M))` `:.(u_(2))/(u_(1))=sqrt((293)/(283))=sqrt(1.04)=1.02` `:. u_(2)=1.02" u"_(1)` Thus, both INCREASE but not significantly.

Discussion

42.	For one mole of gas the average kinetic energy is given as E. the U_("rms") of gas is:
Answer» `SQRT((2E)/(M))` `sqrt((3E)/(M))` `sqrt((2E)/(3M))` `sqrt((3E)/(2M))` Answer :a

Discussion

43.	For one mole of an ideal gas, ((del H)/(del T))_(P)-((del U)/(del T)) is equal to
Answer» 2.303 R `8.314 DM^(3)"atm mol"^(-1)` 0.0821 lit atm `mol^(-1) K^(-1)` 2.303 LOG R ANSWER :C

Discussion

44.	For oneatom the energydifferencebetweenexcitedstategroundstateis 4.4 xx 1 0^(-4)JThen what willbe wavelengthof photon ?
Answer» `2.26 xx 10^(12) m` `1.13 xx 10^(12) m` `4.25 xx 10^(12)m` `4.52 xx 10^(12)m` Solution :`DELTA E = (hc )/(LAMBDA)`

Discussion

45.	For O_(2) and heavier diatomic molecules orbital has maximum energy .
Answer» `pi_(2py)` `SIGMA _(2 Pz)` `sigma_(2PZ)^()` `pi_(2Py)^()` ANSWER :c

Discussion

46.	For NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) reaction started only with NH_(4)HS(s), the observed pressure for reaction mixture in equilibrium is 1.2 atm at 106^(@)C. What is the value of K_(p) for the reaction?
Answer» `1.44atm^(2)` `0.36atm^(2)` `0.16atm^(2)` `3.6atm^(2)` Solution :`NH_(4)HS(s)hArrNH_(3)(G)+H_(2)S(g)` `P P` `2P=1.2` `P=0.6` `K_(p)=P^(2)=(0.6)^(2)=0.36 ATM^(2)`

Discussion

47.	For nitrogen a = 1.4 L^(2) atm, mol^(-2), b = 0.04 L mol^(-1). Find out the diameter of nitrogen molecule.
Answer» Solution :`b = 4 xx (4)/(3) pir^(3) xx N` b = 0.04 L `mol^(-1) = 40 "CC" mol^(-1)` `40 = 4 xx (4)/(3) xx (22)/(7) xx r^(3) xx 6.023 xx 10^(23)` `r^(3) = 4 xx 10^(-24)` cm , `r = 1.58 A^(@)` Diameter of nitrogen MOLECULE ` = 1.58 xx2 = 3.16 A^(@)`

Discussion

48.	For Na^(+), Mg^(2+),F^(-) and O^(2-) , thecorrectorder ofincreasingionicradii is
Answer» `MG^(2+) lt Na^(+)lt F^(-)lt O^(2-)` `O^(2-)lt F^(-)lt Na^(+) lg Mg^(2+)` `Na^(+)lt Mg^(2+)lt F^(-)lt O^(2-)` `Mg^(2+)lt O^(2-)lt Na^(+) lt F^(-)` Solution :Allth SPEICES have10 elementseach andhenceare isoelectronicions . Amongisoelectronicions , ionicsizeincreasesas the + Vechargedecrease orthe - ve chargeincreases i.e.,OPTION( a) is CORRECT .

Discussion

49.	For N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g), show that K_(c)=K_(p)(RT)^(2).
Answer» Solution :For `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` But PV = NRT, `P=(n/v)RT`, i.e., P = CONCENTRATION `xx` RT `P_(NH_(3))=[NH_(3)]RT," "p_(n_(2))=[N_(2)]RT," "p_(H_(2))=[H_(2)]RT," "K_(p)=(p_(NH_(3))^(2))/(p_(N_(2))p_(H_(2))^(3))` `thereforeK_(p)=({[NH_(3)]RT}^(2))/({[N_(2)]RT}{[H_(2)]RT}^(3))=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))xx((RT)^(2))/((RT)(RT)^(3))` `K_(p)=K_(c)(RT)^(-2)""thereforeK_(c)=K_(p)(RT)^(2)`.

Discussion

50.	For N_(2) + 3 H_(2) hArr 2 NH_(3) , continous removal of NH_(3) maintains the following condition
Answer» `Q_(C)= K_(C)` `Q_(C) GT K_(C)` `Q_(C) lt K_(C)` `Q_(C) = K_(C) = 1` ANSWER :C

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For the buffer solution containing NH_(4)OH and NH_(4) CI, P^(H) of the buffer solution can be increased by

For the balanced chemical reaction HNO_(3) + H_(2)S rarr NO + H_(2)O + S. The correct statements are:-

For the arrangement classified as ABAB type the correct combination is

For testing the presence of sulphur by lead acetate test, the Lassaigne's extract is acidified with dilute acetic acid. Can we use dilute hydrochloric or sulphuric acid instead of acetic acid ?

For testing halogens in an organic compound with AgNO_(3) solution, sodium extract (Lassaigne's test) is acidified with dilute HNO_(3). What will happen if a student acidifies the extract with dilute H_(2)SO_(4) in place of dilute HNO_(3) ?

For tartaric acid total number of possible stereo isomers (configurations only) and total no. of possible optically active isomers are.

For spontaneous reaction what is the value of DeltaG?

For sharpening and grinding metals and other substances which substance is used ?

Orbitals having equal energies are

The probability of occurrence of an event is 2/5 and the probability of non occurrence of another eventis 3/10 if these events are independent then the probability that only one of the two events occursis

For reaction XeF_6+H_2O hArr XeOF_4+2HF, K_1=4 XeO_4+XeF_6 hArr XeOF_4 + XeO_3F_2 , K_2=100 Find equilibrium constant of1/2XeO_4+HF hArr 1/2XeO_3F_2+1/2H_2O

For reaction, PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)), K_c=1.79 "L mol"^(-1).Then at 500 K state the value of K_p with respect to R.

For a reaction, N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g), identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.

For reaction, assuming large volume of water. H_(2)O(l)hArrH_(2)O(g) , "at temp". T K Choose correct options:

For reaction, 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g) which statements are correct? (P) K_(c)=[SO_(2)][O_(2)]//[SO_(3)] (Q) Addition of O_(2)(g) tothe system at contant temperature and valume would decrease the value of K_(c).

For reaction, 2NOCl_((g))hArr2NO_((g))+Cl_(2(g)),K_(c) at 427^(@)C is 3xx10^(-6)L" "mol^(-1). The value of K_(p) is nearly,

For reacting with HCl, the alcohol which does not requrie ZnCl_(2) is_____

For reaction, 2A+BhArr2C,K=x. Equilibrium constant for ChArrA+1//2B will be

Which of the following is/are correct about the radial probability curves ?

For process H_(2)o(l)(100^(@)C,1atm)hArrH_(2)O(g)(100^(@)C,1 atm) {:(,"Column-I",,"Column-II"),((a),DeltaU,(p),0),((b),DeltaS ("system+surrounding"),(q),DeltaH),((c),"work",(r),"positive"),((d),"Heat involved",(s),"Negative"):}

For principal quantum number n = 4, the total number of orbitals having l = 3 is

For principle quantum number n=4,the total number of orbitals having l=3 is

What happens when Al_4C_3 reacts with H_2O ?

For preparing monoalkyl benzene , acylation process is preferred than direct alkylation because

For preparing an alkane a concentrated aqueous solution of sodium or potassium salt of a saturated caroboxylic acid is subjected to

For preparing 0.1 N solution ofa compound from the impure sample of which the percentage purity is known, the weight of the substance required will be

For precipitation of sparingly soluble salt if, I_p lt K_(sp), then …..

For PCl_(5 (g)) hArr PCl_(3 (g)) + Cl_(2 (g)) at equilibrium , K_(P)= P//3. Then degree of dissociation of PCl_(5) at that temperature is

For packing with minimum void space the true combination is (Consider 2D & 3D arrangements separately)

For particles having same kineticenergy , the deBrogle wavelenght is-

For oxidation of iron,, 4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3)(s),entropy change is - 549 .4 JK^(-1) mol^(-1) at 298 K . Inspite of the negativeentropy change of this reaction , why is the reaction spontaneous ? ( Delta H ^(@) for this reaction is - 1648 xx 10^(3) J mol^(-1))

For oxidation of iron, 4Fe_((s)) + 3O_(2(g)) to 2Fe_(2) O_(3(s)) entropy change is -549.4 "JK"^(-1) "mol"^(-1) at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous ? Delta_(r) H^( Theta ) for this reaction is 1648 xx 10^(3) "J mol"^(-1)

For one solution, K_a = 1.0 xx 10^(-8) then what is the pK_a and PK_b of the solution ?

For one mole of an ideal gas, if P = P_(0)/(1+(V/V_(0))^(2)) , where P_(0) and V_(0) are constants. Which of the following are true?

For one mole of an ideal gas, increasing the temperature from 10^(@)C to 20^(@)C.

For one mole of gas the average kinetic energy is given as E. the U_("rms") of gas is:

For one mole of an ideal gas, ((del H)/(del T))_(P)-((del U)/(del T)) is equal to

For oneatom the energydifferencebetweenexcitedstategroundstateis 4.4 xx 1 0^(-4)JThen what willbe wavelengthof photon ?

For O_(2) and heavier diatomic molecules orbital has maximum energy .

For NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) reaction started only with NH_(4)HS(s), the observed pressure for reaction mixture in equilibrium is 1.2 atm at 106^(@)C. What is the value of K_(p) for the reaction?

For nitrogen a = 1.4 L^(2) atm, mol^(-2), b = 0.04 L mol^(-1). Find out the diameter of nitrogen molecule.

For Na^(+), Mg^(2+),F^(-) and O^(2-) , thecorrectorder ofincreasingionicradii is

For N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g), show that K_(c)=K_(p)(RT)^(2).

For N_(2) + 3 H_(2) hArr 2 NH_(3) , continous removal of NH_(3) maintains the following condition