52671 + Interview Questions in Chemistry in Class 11

1.	Equilibrium constant for the reaction , N_(2) + 3 H_(2) hArr 2 NH_(3)is K, then equilibrium constant for the reaction, NH_(3) hArr 1/2 N_(2) + 3/2 H_(2)will be……….. .
Answer» ANSWER :`SQRT(1//K)`

Discussion

2.	Equilibrium constant for following reactions repectively K_(1),K_(2)"and"K_(3) N_(2)+3H_(2)hArr2NH_(3) K_(1) N_(2)+O_(2)hArr2NO_(3) K_(2) H_(2)+(1)/(2)O_(2)hArrH_(2)O K_(3) 2NH_(3)+(5)/(2)O_(2)hArr2NO+3H_(2)O K_(4) Which of the following relation is correct.
Answer» `K_(1)=K_(2)XX(K_(3)^(3))/(K_(4))` `K_(4)=K_(1)xxK_(2)//(K_(3))^(3)` `K_(2)=(K_(4)xx(K_(1))/((K_(3))^(3)` `K_(4)=(K_(2)xx(K_(3)^(3))/(K_(1)` Solution :N//A

Discussion

3.	Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium.Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g)=-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g)=-300kcal/mole DeltaG_(f)^(@)D(l)=-224.606 kcal/mole DeltaG_(f)^(@)D(g)= -226.9.9 kcal/ mole, All values at 500K Calculate equilibrium concentration of B(g) if A(g)at 10 bar, B(g) at 2 bar,C(g) at 20 bar is mixed with excess liquid D such that followingequilbrium gets established at 500K: A(g) B(g)iffC(g)+D(g)
Answer» 2M `(2)/(41.57)` `1M` `(1)/(41.57)` Answer :b

Discussion

4.	Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium.Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g)=-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g)=-300kcal/mole DeltaG_(f)^(@)D(l)=-224.606 kcal/mole DeltaG_(f)^(@)D(g)= -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)
Answer» `10 "BAR"^(-1) SEC^(-1)` `0.1 "bar"^(-1) sec^(-1)` `0.1 sec^(-1)` `10SEC^(-1)` Answer :C

Discussion

5.	Equilibrium constant changes with ………….
Answer» SOLUTION :Both TEMPERATURE and PRESSURE

Discussion

6.	Equilibrium can be attained in water and its vapour in open vessel ? Why ?
Answer» Solution :No EQUILIBRIUM is not attained in open vessel. As the total water will not CONVERT to VAPOUR TILL the vaporization CONTINUE but reverse reaction is not possible.

Discussion

7.	Equation for K_(sp) and its unit for the sparingly soluble salt Al(OH)_3 are …..
Answer» `4 S^4 , M^3` `4 S^3 , M^3` `27 S^4 , M^4` `S^2 , M^2` SOLUTION :Suppose the water solubility of `Al(OH)_3` is S MOLAR, so in solution at equilibrium, `underset"Concentration at equilibrium"(Al(OH)_(3(s))) HARR underset"SM"(Al_((aq))^(3+))+ underset"3SM"(3OH_((aq))^(-))` `K_(sp)=[Al^(3+)][OH^-]^3` =`[SM][3SM]^3=27S^4 , M^4`

Discussion

8.	Equation for Boyle's law is
Answer» `(DP)/(P)=(dV)/(V)` `(dP)/(P)=+(dV)/(V)` `(d^(23)P)/(P)=-(dV)/(dT)` `(d^(2)P)/(P)=+(d^(2)V)/(dT)` SOLUTION :By Boyle's law, PV=constant. DIFFERENTIATING this equation, we get PdV+VdP=0 or VdP=-PdV or`(dP)/(P)=-(dV)/(V)`

Discussion

9.	Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction .of total pressure exerted by oxygenis .....
Answer» <P>`1//3` `1//2` `2//3` `1//3 XX 273 xx 298` SOLUTION :`1//3` Massof METHANE= massof oxygen= a numberof molesofmethane`=(a)/(16)` numberof molesof Oxygen`= ((a)/(32))/( a/(16)+a/(32))` ` =((a)/(32))/( (3a)/(32)) =1/3` partialpressureofoxygen =MOLEFRACTION` xx`totalpressure`= 1/3` P

Discussion

10.	Equal weights of methane and oxygen are mixed in an empty container at 298K. The fraction of total pressure exerted by oxygen is
Answer» `(1)/(3)` `(1)/(2)` `(2)/(3)` `(1)/(3)xx273xx298` Solution :ACCORDING to Law of PARTIAL PRESSURE is EQUAL to mole FRACTION multiple total pressure

Discussion

11.	Equal weights of methane and oxygen are mixed in an empty container at 250C. The fraction of the total pressure exerted by oxygen is
Answer» `1//2` `2//3` 1/4 1/3 Answer :D

Discussion

12.	Equal volumes of two solutions withP^(H)=3 and P^(H)= 11are mixed . Then the P^(H)of resulting solution is
Answer» 8 7 6 0 Solution :` N_aV_a =10 ^(-3)xx V , N_b V_b = 10 ^(-3)xx V ` ` THEREFORE `exact NEUTRALIZATION

Discussion

13.	Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (C_p//C_v) of the mixture will be
Answer» `0.83` `1.50` `3.3` `1.67` ANSWER :D

Discussion

14.	Equal volumes of two jars contain HCI, NH_3 gases respectively at constant temperature and pressure P. When one of the jars is inverted over another jar so that they mix up, the pressure in either of the jars is
Answer» <P>1 atm Equal to P BECOMES Zero P+P = 2P SOLUTION :`NH_3 + HCl to NH_4Cl_((s)) implies P = 0`.

Discussion

15.	Equal volumes of two gases A and B diffuse through a porous pot in 20 and seconds respectvely.If the molar mass of A is 80, calculate the molar mass of B.
Answer» Solution :SUPPOSE the volume of each gas diffused =u c c. Then by Graham's law of DIFFUSION, `(r_(A))/(r_(B))=(v//20)/(v//20)=sqrt((M_(B))/(M_(A)))"or"(1)/(2)=sqrt((M_(B))/(80))"or"M_(B)=20 g mol^(-1)`

Discussion

16.	Equal volumes of the following Ca^(2+) and F^(-) solutions are mixed. In which of the solutions will precipitation occur ? (K_(sp) "of" CaF_(2)=1.7xx10^(-10)) (1) 10^(-2) M Ca^(2+) + 10^(-5) M F^(-) (2) 10^(-3) M Ca^(2+)+10^(-3)M F^(-) (3) 10^(-4) M Ca^(2+)+ 10^(-2)M F^(-) (4)10^(-2) M Ca^(2+) + 10^(-3) M F^(-) Select the correct answer using the codes given below:
Answer» `10^(-2) M Ca^(2+) + 10^(-5) M F^(-)` `10^(-3) M Ca^(2+) + 10^(-3) M F^(-)` `10^(-4) M Ca^(2+) + 10^(-2) M F^(-)` `10^(-2) M Ca^(2+) + 10^(-3) M F^(-)` SOLUTION :Ionic product of `CaF_92)=[Ca^(2+)][F^(-)](2)`. CONCENTRATION of ions will be halved after mixing. Thus, ionic products will be (1) `(10^(-2))/(2)XX((10^(-5))/(2))^(2)=(1)/(8)xx10^(-12)` (2) `(10^(-3))/(2)xx((10^(-3))/(2))^(2)=(1)/(8)xx10^(-9)` (3) `(10^(-4))/(2)xx((10^(-2))/(2))^(2)=(1)/(8)xx10^(-8)` (4) `(10^(-2))/(2)xx((10^(-3))/(2))^(2)=(1)/(8)xx10^(-8)` In (2), (3) and (4), ionic product `GT K_(sp)`. Hence, precipitation will OCCUR in (2), (3) and (4).

Discussion

17.	Equal volumes of solutions with pH = 4 and pH = 10 are mixed. Calculate the pH of the resulting solution ?
Answer» Solution :pH = 4 means `[H^(+)] = 10^(-4)` M and pH = 10 means `[H^(+)]=10^(-10)M or [OH^(-)]=10^(-4)M`. THUS, they will EXACTLY neutralise each other and pH of the resulting solution will be = 7 .

Discussion

18.	Equal volumes of one molar HCI and H_(2)SO_(4) are neutralised (separately) by dilute NaOH solution and x kcal and y kcal of heats are liberated.Which of hte following is true?
Answer» x = y x = 0.5 y `x = (1)/(2y)` None Solution :Heat EVOLVED for NEUTRALISATION of `H_(2)SO_(4)(2H^(+))` will be doubled the heat evolved for neutralisation of `HCI(H^(+))` i.e, y = 2X or x = 0.5 y

Discussion

19.	Equal volume of oxygen and ozone at a given temperature and pressure contain equal
Answer» NUMBER of moles masses number of gramatoms number of respective molecules Solution :OXYGEN `(48)/(16)` gram atoms Number of electrons = `(48)/(32)xx6.023xx10^(23)xx16` Ozone `(48)/(16)` grams atoms Number of ELECTRON = `(48)/(48)xx6.023xx10^(23)xx24`

Discussion

20.	Equal volumes of monoatomic and diatomic gases a same initial temperature and pressure are mixed.The ratio of specific heats of the mixture (C_(p)//C_(v))will be
Answer» `1` `2` `1.67` `1.5` Solution :`C_(v)=(3)/(2)RT,C_(P)=(5)/(2)RT` for monoatomic GAS `C_(v)=(5)/(2)RT,C_(P)=(7)/(2)RT` for monoatomic gas THUS for mixture of 1 mole each, `C_(v)=((3)/(2)RT(5)/(2)RT)/(2)` and `C_(p)=((5)/(2)RT(7)/(2)RT)/(2)` Therefore,`C_(p)//C_(v)=(3RT)/(2RT)=1.5` .

Discussion

21.	Equal volumes of H_(2)S & SO_(2) react at NTP to form H_(2)O and S. 2H_(2)S+SO_(2) to 2H_(2)O+3S. In this reaction, I. H_(2)S is the limiting reactant. II. SO_(2) is the limiting reactant. III. Sulphur formed is three times of SO_(2) reacted IV. sulphur formed is 1.5 times of H_(2)S reacted. Select the correct statements(s).
Answer» all EXCEPT I all except II all except I, III all except II, IV Answer :B

Discussion

22.	Equal volumes of equi molar HCl and H_(2)SO_(4) are separately neutralised by dilute NaOH solution, then heats liberated and x kCal and y kCal respectively. Which of the following is true.
Answer» `x=y` `x=y //2` `x=2y` `x=y//3` ANSWER :B

Discussion

23.	Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of atoms.
Answer» ANSWER :F

Discussion

24.	Equal volumes of 1M KMnO_4 and 1M K_2Cr_2O_7 solution are allowed to oxidise Fe^(2+) ions to Fe^(3+) ions in acidic medium. The number of moles of Fe^(2+) ions oxidised in the two cases are in the ratio:
Answer» `1:1` `3:1` `5:6` `6:5` Solution :Eq. mass of KMn04 in acidic medium =`("Mol. mass")/5` Eq. mass of `K_2Cr_2O_7` in acidic medium `=("Mol. mass")/56`. Equalvolumesof1M`KMnO_4`and1M `K_2Cr_2O_7`willhavegramequivalentsin the ratio of 5 : 6. Hence, number of GRAM equivalents (moles, as the eq. mass of `FE^(2+)` = mol. mass) of `Fe^(2+)` IONS oxidised in the TWO cases will also be in the ratio of `5 : 6`.

Discussion

25.	Equal volumes of 1 M each of KMnO_4 and K_2Cr_2O_7are used to oxidize Fe(II) solution in acidicmedium. The amcunt of Fe oxidized will be
Answer» more with `KMnO_4` more with `K_(2)Cr_(2) O_7` equal with both OXIDIZING agents cannot be determined SOLUTION : No of equivalents of `K_(2)Cr_(2)O_(7)=6` No. of equivalents of `KMnO_4` = 5

Discussion

26.	Equal volumes ofM NaOH and 0.3 M KOH are mixed in an experiment. Find the POH and pH of the resulting solution.
Answer» Solution :The concentration of HYDROXYL ions in the MIXTURE `N=(V_1N_1 +V_2 N_2)/( V_1 +V_2) = (V xx 0.6xx V xx 0.2 ) / (V +V ) = 0.4mol L^(-1)` POH of the mixture = -LOG(0.4) = 1-log 4 = 1-0.6=0.4. pH of the mixture = 14 -0.4 = 13.6.

Discussion

27.	Equal volumes of 0.1 M KCl and 0.1 M FeCl_(3) are mixed with no change in volume, which is/are correct?
Answer» `[Fe^(+3)]=0.05M` `[K^(+)]=0.05M` `[Cl^(-)]=0.2M` `[Cl^(-)]gt[K^(+)]` SOLUTION :Let volume of 0.1 M KCl and 0.1 `MFeCl_(3)=Vimplies(a),(b)` `[Fe^(+3)]=((0.1xxV))/(2V)=0.05M` C) `[Cl^(-)]=((0.1xxV)+0.1xx3xxV)/(2V)=(0.4)/(2)=0.2` d) `[Cl^(-)]gt[K^(+)]`

Discussion

28.	Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together . Will it lead to precipitation of copper iodate ? (For cupric iodate K_(sp)=7.4xx10^(-8) )
Answer» Solution :Sodium Iodate `(NaIO_3)` : It is completely ionized `{:(NaIO_3 to Na_((aq))^(+)+ IO_(3(aq))^(-)),(0.002M to 0.002 M + 0.002 M):}` `THEREFORE` Concentrationof iodate before mixing `[IO_3^-]`= 0.002 M "When equal volume of two solution mix together then the concentration of mixture is half" `therefore` In mixture `IO_3^(-)=("INITIAL" [IO_3^-])/2=0.002/2`=0.001 M By taking the complete ionization of Cupric chlorate `CU(ClO_3)_2` `{:(Cu(ClO_3)_2 hArr, Cu_((aq))^(2+) + , 2ClO_(3(aq))^(-)),(0.002 M, 0.002M , 2(0.002)M):}` `therefore [Cu^(2+)]`= 0.002 M before mixing "In equal volume of solutionwhen mix, the concentration become half " `therefore` (Concentration of copper ion in mixture ) = `1/2` (Initial concentration of `Cu^(2+)` ion ) `=1/2(0.002)`=0.001 M Equilibrium of `Cu(IO_3)_2` : `{:(Cu(IO_3)_2 hArr, Cu_((aq))^(2+) +, 2IO_(3)^(-)),("Concentration in mixture:", 0.001 M, 0.001 M):}` Thus, `Q_(SP)` of `Cu(IO_3)_2 = [Cu^(2+)][IO_3^-]^2` `=(0.001)(0.001)^2` `=1.0xx10^(-9)` `Cu(IO_3)_2` of `{:((Q_(sp)),(1.0xx10^9)) lt ((K_(sp)),(7.4xx10^(-8))):}` The value of `Q_(sp)` is less than `K_(sp)` so, precipitation of copper iodate does not occurs.

Discussion

29.	Equal volumes of 0.1 M potassium hydroxide and 0.1 M sulphuric acid are mixed . TheP^(H)of resultingsolution is
Answer» 7 0 less than 7 GREATER than 7 Solution :` N_aV_a =0.1 xx 2xx V, N_bV_b =0.1 xx V` ` rArr " ACID left"rArr PH lt 7`

Discussion

30.	Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to precipitationof copper iodate ? For copper iodate, K_(sp)=7.4xx10^(-8).
Answer» Solution :`2 NaIO_(3)+CuCrO_(4) rarr Na_(2)CrO_(4)+CU(IO_(3))_(2)` After mixing, `[NaIO_(3)]=[IO_(3)^(-)]=(2xx10^(3))/(2)=10^(-3)M` `[CuCrO_(4)]=[Cu^(2+)]=(2xx10^(-3))/(2) = 10^(-3)M` Ionic product of `Cu(IO_(3))_(2)=[Cu^(2+)][IO_(3)^(-)]^(2)=(10^(-3))(10^(-3))^(2)=10^(-9)` As ionic product is less than `K_(sp)`, no PRECIPITATION will OCCUR.

Discussion

31.	Equal volume of 0.1M HCl and 0.2M H_(2)SO_(4) are mixed. What is the molarity of proton in the mixture?
Answer» ANSWER :0.25mol`L^(-1)`

Discussion

32.	Equal quantities of zinc are separately treated with caustic soda solution and dilute sulphuric acid. Then
Answer» more HYDROGEN is liberated in the FIRST case more hydrogen is liberated in the second case equal AMOUNT of hydrogen is liberated in both cases no reaction TAKES places SOLUTION :`Zn+2NaOH rarrNa_(2)ZnO_2+H_2uarr` `Zn +H_(2)SO_(4)rarrZnSO_4+H_2uarr` equal amount of `H_2` is liberated in both cases.

Discussion

33.	Equal number of total nodes is present in
Answer» 5S 5P 50 50 Solution :NUMBER of nodes = n - L - 1

Discussion

34.	Equal moles of hydrogen and oxygen gases are placed in a container with a pin - hole through which both can escape. What fraction of the oxygen escapes in the time required for one - half of the hydrogen to escape ?
Answer» `(1)/(4)` `(3)/(8)` `(1)/(2)` `(1)/(8)` SOLUTION :`(tO_(2))/(tH_(2))=sqrt((MH_(2))/(MO_(2)))` `(tO_(2))/(0.5)=sqrt((2)/(32))` `tO_(2)=(1)/(8)`

Discussion

35.	Equal moles of water and urea taken in one flask. Give percentage weight of urea.
Answer» `23.077%` 0.77 0.7692 `7.7%` ANSWER :B::C

Discussion

36.	Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ?
Answer» `1 //8` `1//4` `3//8` `1//2` ANSWER :A

Discussion

37.	Equal moles of hydrogen and oxygen gas are placed ina container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half fo the hydrogen to escape ?
Answer» `3//8` `1//2` `1//8` `1//4` Solution :Suppose the number of moles of each gas taken =s moles `H_(2)` escaped in TIME `t=(X)/(2)` mole Suppose number of moles of `O_(2)` DIFFUSED in the same time=n TIMES Then `(r_(O_(2)))/(r_(H_(2)))=sqrt((M_(H_(2)))/(M_(O_(2))))` i.e., `(n//t)/((x//2)//t)=sqrt((2)/(32))"or" (2n)/(x)=(1)/(4)" or" n=(1)/(8)x` `:.` Fraction of `O_(2)` escaped`=(1)/(8)`.

Discussion

38.	Equal moles of F_(2(g))and Cl_(2(g)) are introduced into a sealed container and heated to temperature T to attain the following equilibria. Cl_(2(g)) + F_(2(g)) hArr 2C//F_((g)) , Kp_(I)= 3.2 , Cl_(2(g)) + 3 F_(2(g)) hArr 2 CIF_(2(g)) , Kp_(2) . If partial pressures of CIF and CIF_(3) at equilibria are 0.2 and 0.04 atm respectively. What is the value of Kp_(2)?
Answer» 14.66 `ATM^(-2)` 1.466 `atm^(-2)` 14.66 `atm^(-2)` 0.1466 `atm^(-2)` Solution : `:. 3.2=((0.2)^(2))/((P-0.12)(P-0.16)) IMPLIES P=0.2535` `:. 0.2535-0.12""0.2535-0.16""0.04` `=0.1335""=0.0935` `Kp_(2)=((0.04)^(2))/((0.1335)(0.0952)^(3))=14.66 atm^(-2)`

Discussion

39.	Equal moles of hydrogen and oxygen gas are placed in a container, with a pin-hole through which both can escape what fraction of oxygen esacpes in the time required for one-half of the hydrogen to escape.
Answer» `(3)/(8)` `(1)/(2)` `(1)/(8)` `(1)/(4)` SOLUTION :By Graham's law of diffusion `t_(2)//t_(1)` is equal to root of `M_(2)M_(1).t_(2)//t_(1)`

Discussion

40.	Equal masses of two samples of charcoal A and B are burntseparately and the resulting carbon dioxide is collected in two vessels. The radioacitivity of .^(14)C is measured for both the gas samples. The gas from the charcoalA gives1400 counts per week. Find teh age difference between the two samples. (Half-life .^(14)C = 5730yr)
Answer» ANSWER :`3353.16` YEAR

Discussion

41.	Equal masses of oxygen, hydrogen and methane are taken in a container in identical conditions. Find the ratio of the volumes of the gases.
Answer» SOLUTION :Suppose each gas has a mass of X g. HENCE, `O_(2) : H_(2) : CH_(4) = 1:16 : 2`

Discussion

42.	Equal masses of oxygen and ozone have equal
Answer» number of grammolecules number of GRAMATOMS volumes at STP number of electrons Solution :equal mass = x `{:(,O_(2),O_(3)),("moles =",(x)/(32),(x)/(48)),("volume =",(x)/(32)xx22.4,(x)/(48)xx22.4),("G atoms =",2XX(x)/(32),3XX(x)/(48)),(,=(x)/(16),=(x)/(16)):}` implies equal `e^(-)`

Discussion

43.	Equal masses of He, O_(2) and SO_(2) are taken in a closed container. The ratio of the partial pressures of gases He,O_(2) and SO_(2) would be
Answer» `1:2:8` `8:16:1` `16:2:1` `1:4:16` Solution :For a MIXTURE of non-reacting gases, PARTIAL pressure `prop` number of MOLES. If m is the mass of each gas taken, then `n_(He)=(m)/(4), n_(O_(2))=(m)/(32), n_(SO_(2))=(m)/(64)` `:. "Ratio " n_(He):n_(O_(2)):n_(SO_(2))=(1)/(4):(1)/(32):(1)/(64)=16:2:1`

Discussion

44.	Equal mass of methane and oxygen are mixed in an empty containerat 25^(@)C. The fraction of the total pressure exerted by oxygen is:
Answer» `2/3` `1/3 xx 273/298` `1/3` `1/2`. Solution :Suppose the equal mass of methane and oxygen =w=1 g MOLE FRACTION of oxygen (X)=`(W//32)/(W//32+W//16)` `(1/32)/(3/32)=1/3` Let the TOTAL pressure `(P_(total))` be equal to P. `:.` Pressure exerted by Oxygen (partial pressure) `=XO_2 xxP_(total)=Pxx1/3`.

Discussion

45.	Equal masses of H_(2)O_(2) and methane have been taken in a container of volume V at temperature 27^(@)C in identical conditions. The ratio of the volumes of gases H_(2) : O_(2) : methane would be
Answer» `8:16:1` `16:8:1` `16:1:2` `8:1:2` Solution :Suppose mass of each of `H_(2), O_(2)` and `CH_(4)` TAKEN =w G `w g H_(2)=(w)/(2)"mole",w g O_(2)=(w)/(32)"mole"`, `w g CH_(4)=(w)/(16)"mole"` As under identical conditions, VOLUME are in the ratio of moles. Hence, volume of `H_(2) : O_(2): CH_(4)=(w)/(2) : (w)/(32) : (w)/(16)` `=(1)/(2) : (1)/(32) : (1)/(16)=16 : 1 : 2`

Discussion

46.	Equal masses of H_(2), O_(2) and methane have been taken in a container of volume V at temperature 27""^(@)C in identical conditions. The ratio of the volumes of gases H_(2):O_(2):methane would be
Answer» `8:16:1` `16:8:1` `16:1:2` `8:1:2` SOLUTION :`H_(2)O_(2)H_(2)` `32:32:32` `16:1:2` MOLE RATIO

Discussion

47.	Equal masses of H_2, O_2, and methane have been taken in a container of volume V at temperature 27^@C in identical conditions. The ratio of the volumes of gases H_2 : O_2 : CH_4 would be
Answer» `8:16:1` `16:8:1` `16:1:2` `8:1:2`. SOLUTION :`16:1:2`

Discussion

48.	Equal masses of H_2, O_2 and methane have been taken in a container of volume V at temperature 27^@ C in identical conditions. The ratio of the volume of gasesH_2 :O_2 methane would be
Answer» `8:16:1` `16:8:1` `16:1:2` `8:1:2` ANSWER :C

Discussion

49.	Equal mass of KClO_(3) undergoes different reactions in two different containers: 2KClO_(3)(s) overset(Delta)to 2KCl(s) + 3O_(2)(g) ………………..(i) 4KClO_(3)(s) overset(Delta)to KCl(s) + 3KClO_(4)(s)………………..(ii) Mass ratio of KCl produced in respective reaction is n:1 then, value of n will be:
Answer» 4 2 74.5 90 Answer :A

Discussion

50.	Equal concentration possessing solutionCH_3COONa , HNO_3 , NH_4Cl , KI arrange in decreasing order of pH.
Answer» SOLUTION :`CH_3COONa, KI , NH_4Cl , HNO_3`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Equilibrium constant for the reaction , N_(2) + 3 H_(2) hArr 2 NH_(3)is K, then equilibrium constant for the reaction, NH_(3) hArr 1/2 N_(2) + 3/2 H_(2)will be……….. .

Equilibrium constant for following reactions repectively K_(1),K_(2)"and"K_(3) N_(2)+3H_(2)hArr2NH_(3) K_(1) N_(2)+O_(2)hArr2NO_(3) K_(2) H_(2)+(1)/(2)O_(2)hArrH_(2)O K_(3) 2NH_(3)+(5)/(2)O_(2)hArr2NO+3H_(2)O K_(4) Which of the following relation is correct.

Equilibrium constant changes with ………….

Equilibrium can be attained in water and its vapour in open vessel ? Why ?

Equation for K_(sp) and its unit for the sparingly soluble salt Al(OH)_3 are …..

Equation for Boyle's law is

Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction .of total pressure exerted by oxygenis .....

Equal weights of methane and oxygen are mixed in an empty container at 298K. The fraction of total pressure exerted by oxygen is

Equal weights of methane and oxygen are mixed in an empty container at 250C. The fraction of the total pressure exerted by oxygen is

Equal volumes of two solutions withP^(H)=3 and P^(H)= 11are mixed . Then the P^(H)of resulting solution is

Equal volumes of two monatomic gases, A and B at same temperature and pressure are mixed. The ratio of specific heats (C_p//C_v) of the mixture will be

Equal volumes of two jars contain HCI, NH_3 gases respectively at constant temperature and pressure P. When one of the jars is inverted over another jar so that they mix up, the pressure in either of the jars is

Equal volumes of two gases A and B diffuse through a porous pot in 20 and seconds respectvely.If the molar mass of A is 80, calculate the molar mass of B.

Equal volumes of solutions with pH = 4 and pH = 10 are mixed. Calculate the pH of the resulting solution ?

Equal volumes of one molar HCI and H_(2)SO_(4) are neutralised (separately) by dilute NaOH solution and x kcal and y kcal of heats are liberated.Which of hte following is true?

Equal volume of oxygen and ozone at a given temperature and pressure contain equal

Equal volumes of monoatomic and diatomic gases a same initial temperature and pressure are mixed.The ratio of specific heats of the mixture (C_(p)//C_(v))will be

Equal volumes of equi molar HCl and H_(2)SO_(4) are separately neutralised by dilute NaOH solution, then heats liberated and x kCal and y kCal respectively. Which of the following is true.

Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of atoms.

Equal volumes of 1M KMnO_4 and 1M K_2Cr_2O_7 solution are allowed to oxidise Fe^(2+) ions to Fe^(3+) ions in acidic medium. The number of moles of Fe^(2+) ions oxidised in the two cases are in the ratio:

Equal volumes of 1 M each of KMnO_4 and K_2Cr_2O_7are used to oxidize Fe(II) solution in acidicmedium. The amcunt of Fe oxidized will be

Equal volumes ofM NaOH and 0.3 M KOH are mixed in an experiment. Find the POH and pH of the resulting solution.

Equal volumes of 0.1 M KCl and 0.1 M FeCl_(3) are mixed with no change in volume, which is/are correct?

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together . Will it lead to precipitation of copper iodate ? (For cupric iodate K_(sp)=7.4xx10^(-8) )

Equal volumes of 0.1 M potassium hydroxide and 0.1 M sulphuric acid are mixed . TheP^(H)of resultingsolution is

Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to precipitationof copper iodate ? For copper iodate, K_(sp)=7.4xx10^(-8).

Equal volume of 0.1M HCl and 0.2M H_(2)SO_(4) are mixed. What is the molarity of proton in the mixture?

Equal quantities of zinc are separately treated with caustic soda solution and dilute sulphuric acid. Then

Equal number of total nodes is present in

Equal moles of hydrogen and oxygen gases are placed in a container with a pin - hole through which both can escape. What fraction of the oxygen escapes in the time required for one - half of the hydrogen to escape ?

Equal moles of water and urea taken in one flask. Give percentage weight of urea.

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape ?

Equal moles of hydrogen and oxygen gas are placed ina container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half fo the hydrogen to escape ?

Equal moles of hydrogen and oxygen gas are placed in a container, with a pin-hole through which both can escape what fraction of oxygen esacpes in the time required for one-half of the hydrogen to escape.

Equal masses of oxygen, hydrogen and methane are taken in a container in identical conditions. Find the ratio of the volumes of the gases.

Equal masses of oxygen and ozone have equal

Equal masses of He, O_(2) and SO_(2) are taken in a closed container. The ratio of the partial pressures of gases He,O_(2) and SO_(2) would be

Equal mass of methane and oxygen are mixed in an empty containerat 25^(@)C. The fraction of the total pressure exerted by oxygen is:

Equal masses of H_(2)O_(2) and methane have been taken in a container of volume V at temperature 27^(@)C in identical conditions. The ratio of the volumes of gases H_(2) : O_(2) : methane would be

Equal masses of H_(2), O_(2) and methane have been taken in a container of volume V at temperature 27""^(@)C in identical conditions. The ratio of the volumes of gases H_(2):O_(2):methane would be

Equal masses of H_2, O_2, and methane have been taken in a container of volume V at temperature 27^@C in identical conditions. The ratio of the volumes of gases H_2 : O_2 : CH_4 would be

Equal masses of H_2, O_2 and methane have been taken in a container of volume V at temperature 27^@ C in identical conditions. The ratio of the volume of gasesH_2 :O_2 methane would be

Equal concentration possessing solutionCH_3COONa , HNO_3 , NH_4Cl , KI arrange in decreasing order of pH.