94420 + Interview Questions in Chemistry in Class 12

1.	The Baeyer's angle strain is expected to be maximum in
Answer» Cyclodecane Cyclopentane Cyclobutane Cyclopropane Answer :D

Discussion

2.	The bakelite is made from phenol and fromaldehyde.The intial reaction between the two compounds is an example of:
Answer» AROMATIC ELECTROPHILIC substitution Aromatic NUCLEOPHILIC substitution Free RADICAL reaction Aldol reaction Answer :A

Discussion

3.	The bacterial growth follows the rate law,(dN)/(dt)=kN, where k is a constant and N is the number of bacterial cell at any time. If the population of bacteria (no. of cell) is doubled in 5 minutes then the time by which the population will be eight times of the initial one.
Answer» 15 mins 10 mins 12 mins 18 mins SOLUTION : We know that, `(dN)/(DT)`=kN `impliesN=N_0^(e^(KT))` In 5 min `impliesN=2N_0` `k=(2.303)/(t)log (N)/(N_0) = (2.303)/(5) log (2N_0)/(N_0)` `=(2.303)/(5) log 2`= 0.138 For 8 `N_0,t=(2.303)/(0.138) log (8N_0)/(N_0)`=15 min

Discussion

4.	The bad smelling substance formed by the action of alcoholic caustic potash on chloroform and aniline is :
Answer» PHENYL isocyanide Nitrobenzene Acetylene Chlorobenzene Answer :A

Discussion

5.	The bad smelling substance formed by the action of alcoholic caustic potash on chloroform and aniline is
Answer» Chlorobenzen Nitro benzene Acetylene Phenyl isocynide Answer :D

Discussion

6.	The bacteriostatic antibiotic among the following is :
Answer» Erythromycin Penicillin Aminoglycoside Ofloxacin. Answer :A

Discussion

7.	The BaCl_(2) ionises to an extent of 80% in aqueous solution, the value of van't Hoff factor is -
Answer» 2.6 0.4 0.8 2.4 Solution :`BaCl_(2) rarr Ba^(2+) + 2CL^(-) :. n=3` `ALPHA = (% alpha)/100 =80/100=0.8` `alpha = (i-1)/(n-1)` `:. 0.8 = (i-1)/(3-1)` ` :. 0.8 xx 2 = i - 1` 1.6 + 1 = I i=2.6`

Discussion

8.	the azocompoundsare colouredas theyabsorb thevisibleregionof electromagneticspectrumbecauseof presenceof
Answer» `-N^(+) -= NX^(-)` `-N^(+) = NH^(-)` conjugatesystemthrough`N=N- ` BON `-N-=N` Solution :Information

Discussion

9.	The backbone of a nucleotide strand contains the following sequence of arrangement
Answer» BASE-Sugar Sugar-Phosphate Base-Phosphate `"Base"_(1)-"Base"_(2)` ANSWER :B

Discussion

10.	The back bone for different segments in a protein is in the following form.
Answer» `ALPHA-`HELIX `alpha `-pleated Coil 1 or 3 ANSWER :D

Discussion

11.	The avrage osmotic presure of human of blood is 7.8 bar at 27^(@)C. The concentration of an aqueous NaCI solution that could be used in the blood stream is :
Answer» 0.16 mol/litre 0.32 mol/litre 0.60 mol/litre 0.45 mol/litre Solution :According to Van't Hoff equation `pi=iCRT or C=pi/(iRt)` For NaCI,i=2,pi=7.8 bar=7.8 ATM. R=.0821 L-atm `K^(-1)mol^(-1),T=27^(@)C=300` K `C=((7.8atm))/(2(0.0821 L-atm K^(-1)mol^9-1)xx300K)=0.16 mol L^(-1)`

Discussion

12.	The azeotrope of nitric acid and water has a composition by mass :
Answer» 50% nitric ACID and 50% WATER 25% nitric acid and 75% water. 32% water and 68% water. 68% nitric acid and 32% water. Solution :68% nitric acid and 32% water.

Discussion

13.	The axles are made by heating rods of iron embedded in charcoal powder. This process is known as:
Answer» Tempring Annealing Nitriding Case hardening Answer :D

Discussion

14.	The average weight of an Indian male is 150 pounds. In SI units it is equal to
Answer» 68.1 KG 75.0 kg 45.4 kg 72.0 kg Solution :1 POUND = `545 xx 10^(-3) kg ` UNIT factor ` = (545 xx 10^(-3)kg)/(1 "pound")` ` therefore 150 pound = 150 poundxx (454 xx 10^(-3) kg)/(1 pound)` ` = 150 xx 454 xx 10^(-3) kg = 68.1 kg `

Discussion

15.	The average velocity of the molecules of a gas is 400 m/s. Calculate its rms velocity at the same temperature.
Answer» SOLUTION :434.26 m/s

Discussion

16.	The average velocity of gas molecules is 400 ms^(-1) . Its r.m.s. velocity at the same temperature is
Answer» `2.62 ms^(-1)` `1.68 ms^(-1)` `5.86 ms^(-1)` `4.34 ms^(-1)` SOLUTION :Average velocity, `u_(av) = sqrt(( 8RT)/( PI M))` Root MEAN square velocity, `u _("rms") = sqrt(( 2RT)/( M))` `(u_(rms))/( u_(av)) = (sqrt((3RT)/( M)))/(sqrt((8RT)/(piM)))` `= sqrt(( 3pi )/( 8)) = sqrt( ( 3 xx 3.143)/( 8)) = 1.085` `:. u_("r.m.s.") = 1.805 xx u_(av)` `= 1.085 xx 400` `= 4.34 ms^(-1)`

Discussion

17.	The average velocity of gas molecules is 400 m/sec. Calculate its rms velocity at the same temperature.
Answer» Solution :`C_("rms") =SQRT((3RT)/(M)), C_("av") =sqrt((8RT)/(PI M))` `(C_("rms"))/(C_("av"))=sqrt((3RT)/(M))xx sqrt((pi M)/(8RT))=sqrt((3PI)/(8))=1.085` `C_("rms") =1.085 xx C_("av")` `=1.085xx400` `=434 MS^(-1)`

Discussion

18.	The average velocity of CO_(2) at T_(1) K and most probable velocity at T_(2)K is 9 × 10^(4) cm//s. Calculate the values of temperatures, T_(1) and T_(2)K.
Answer» Solution :Average VELOCITY of `CO_(2)` at `T_(1) K = sqrt((8RT_(1))/(π M))` `implies 9 xx 10^(2) m//sec= sqrt((8RT_(1))/(π (44 xx 10^(-3))))= sqrt(( 8 xx 8.314 xx T_(1))/(3.14(44 xx 10^(-3))))implies T_(1)=1682.5K` SIMILARLY, Most probable velocity of `CO_(2)` at `T_(2)K = sqrt((2RT_(2))/(M))` `implies 9 xx 10^(2) m//s= sqrt((2 xx 8.314 xx T_(2))/(44 xx 10^(-3)))implies T_(2)= 2143.37K`

Discussion

19.	The average speed of nitrogen molecule is v. if the temperature is doubled and nitrogen molecule dissociate into nitrogen atoms completely then new average speed becoems.
Answer» v `sqrt(2)v` `2v` 4v Solution :`v_(mg)=sqrt((8)/(pi)(RT)/(M))` `N_(2)(g)rarr2N(g)` `((v_(AVG))N)/((v_(avg))N_(2))sqrt((28xx2T)/(14xxT))=sqrt(4)=2` `(v_(avg))_(N)=2v`

Discussion

20.	The average velocity of an ideal gas at 27^(@)C is 0.3m sec^(-1). The average velocity at 927^(@)C will be
Answer» 0.6 m `SEC^(-1)` `0.3 m sec^(-1)` `0.9 m sec^(-1)` 1.2 m `sec^(-1)` Solution :Average velocity , `mu prop sqrt( T )` `( mu_(1))/( mu _(2)) = sqrt(( T_(1))/( T_(2)))` `mu_(1) = 0.3 m sec ^(-1) `, `T_(1) = 273 + 27 = 300K` `T_(2) = 927+ 273 = 1200k` `( 0.3)/( mu_(2)) = sqrt(( 300)/( 1200))` `mu = 0.3 XX 0.6 m sec^(-1)`

Discussion

21.	The average speed ofO_2at 273 K is equal to that of H_2 at :
Answer» Same T Higher T Lower T None of these Answer :C

Discussion

22.	The average speed of gas molecules is 400 m/sec. Calculate its rms speed at the same temperature.
Answer» Solution :`U_(AV) =SQRT((8RT)/(PI M))"….(i)"` `U_("rms") =sqrt((3RT)/(M))" ….(ii)` `therefore` By EQS. (i) and (ii) `U_("rms") = U_(AV) xx sqrt((3PI)/(8))` `because U_(AV) =400" m sec"^(-1)` `therefore U_("rms") =400xxsqrt((3xx3.14)/(8) =434" m sec"^(-1)`

Discussion

23.	The average speed of gas molecules is equal to :
Answer» `[(2RT)/(M)]^(1/2)` `[(3RT)/(M)]^(1/2)` `[(8RT)/(PIM)]^(1/2)` `[(4RT)/(piM)]^(1/2)` ANSWER :C

Discussion

24.	The average speed at T_(1)Kand the most probable speed at T_(2) K of CO_(2) gas is 9 xx 10^(4) cm s^(-1). Calculatethe value of T_(1) and T_(2).
Answer» Solution :We know, for 1 MOLE of an IDEAL gas, rms speed `= SQRT((3RT)/(M))` and rms speed : average speed : most probable speed `=1 : 0.9211 : 0.8165`. `therefore` average speed at `T_(1)K = 0.9211 xx sqrt((3RT_(1))/(M)) = 9 xx 10^(4)` …..(1) and most probable speed at `T_(2)K = 0.8165 xx sqrt((3RT_(2))/(M)) = 9 xx 10^(4)`. Substituting `R = 8.314 xx 10^(7)` ergsK/mole and M = 44 in (1) and (2), we get, `T_(1) = 1684` and `T_(2) = 2143 K`.

Discussion

25.	The average speed of an ideal gas molecule at 27^@C is 0.3 sec^(-1). The average speed at 927^@C will be …m sec^(-1).
Answer» 0.6 0.3 0.9 3 Answer :A

Discussion

26.	The average oxidation number of sulphur in Na_2 S_4 O_6
Answer» `1.2` `2.5` `3` 2 Solution :`OVERSET( +4)Na_2overset(x )(S_4) overset(-2) O_6 ` ` therefore2+ 4x - 12=0or 4x =10orx=2.5`

Discussion

27.	The average oxidation state of sulphur atom in S_4O_6^(2-)ion is
Answer» `+2` `+5` 0 `+2.5` ANSWER :D

Discussion

28.	the average osmotic pressure of human blood is 7.8 bar at 37^(@)C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream?
Answer» `"0.15 mol/L"` `"0.30 mol/L"` `"0.60 mol'L"` `"0.45 mol/L"` Solution :NACL sol. USED should be isotonic with blood STREAM. For NaCl, `i=2. PI=i` CRT `C=(pi)/(iRT)=("7.8 bar")/(2xx"0.83 bar L K"^(-1)"mol"^(-1)xx310K)` `="0.15 mol L"^(-1)`.

Discussion

29.	The average osmotic pressure of human blood is 7.7 atm at 40^(@)C. (a) What would be the total concentration of the various solutes in the blood ? (b) Assuming the concentration to be essentially the same as the molality, find the freezing point of blood (K_(f)" for water = "1.86^(@)C).
Answer» Solution :(a) We are given that `"P = 7.7 atm, T"=40^(@)=C=40+273=313K` `"R = 0.0821 litre atm/degree/mole"` `"According to van't Hoff equation, "pi=CRT therefore C=(pi)/(RT)=("7.7 atm")/("0.0821 L atm K"^(-1)"mol"^(-1)xx313K)="0.30 mole/litre"` (b) Taking the molar concentration as equal to molality (Given), we have `"m = 0.30,"K_(f)=1.86^(@)C"(Given)"` `therefore""DeltaT_(f)=K_(f)xxm=1.86xx0.30=0.558^(@)C` `therefore"Freezing POINT of blood "=0^(@)C-0.558^(@)=-0.558^(@)C`

Discussion

30.	The average osmotic pressure of human blood is 7.8 bar at 37^(@)C. The concentration of aqueous NaCl solution that could be used in the blood stream is
Answer» 7.8 mol `L^(-1)` 1.5 mol `L^(-1)` 0.075 mol `L^(-1)` 0.15 mol `L^(-1)` SOLUTION :NaCl SOL. used should be isotonic with blood STREAM. For NaCl, `i=2, pi=iCRT` or `C=(pi)/(iRT)=(7.8"bar")/(2xx0.083"bar LK"^(-1)"mol"^(-1)xx310K)` `=0.15"mol L"^(-1)`

Discussion

31.	The average O-H bond energy in H_(2)O with the help of following data. (1) H_(2)O(l)rarrH_(2)O(g), Delta H= +40.6 kJ mol^(-1) (2) 2H(g)rarrH_(2)(g), DeltaH= -435.kJ mol^(-1) (3) O_(2)(g)rarr2O(g), Delta H= +489.6 kJ mol^(-1) (4) 2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l), Delta H= -571.6 kJ mol^(-1)
Answer» `584.9 KJ MOL^(-1)` `279.8 kJ mol^(-1)` `462.5 kJ mol^(-1)` `925 kJ mol^(-1)` Solution :Assuming density of solution is `1g//c c` and specific heat is `4.2 J//gc` `=msDelta T=100xx4.2xx3` millimoles of acid neutralized `=5` `DeltaH= - 100xx4.2xx3xx(1000)/(5)= -2.52xx10^(2)KJ//"mole"`.

Discussion

32.	The average osmotic pressure of benzoic acid is 7.8 bar at 37^(@)C. What is the concerntration of aqueous KCI solution that could be used in blood stream ?
Answer» 0.16 mol `L^(-1)` 0.32 mol`L^(-1)` 0.60 mol`L^(-1)` 0.45 mol`L^(-1)` SOLUTION :`pi=CRT or C=pi/(RT) C=((7.7 atm))/((0.0821 L atm K^(-1)mol^(-1))xx310K)` `0.31 mol L^(-1)`.

Discussion

33.	The average osmotic pressure of human blood is 7.7 atm at 40^(@)C. (a) What should be the total concentration of various solutes in the blood ? (b) Assuming this concentration to be essentially the same as the molality, find the freezing point of blood. [K_(f)(H_(2)O)=1.86]
Answer» ANSWER :`0.29` mole/lite, `-0.539^(@)C`

Discussion

34.	The average molecular speed is greatest in case of a gas sample of :
Answer» 2.0 MOLE of He at 140 K 0.05 mole of NE at 500 K 0.40 mole of `O_2` at 400 K 1.0 mole of `N_2` at 560 K Answer :A

Discussion

35.	The average molecular mass of colloidal particles can be accurately determined by:
Answer» MEASUREMENT of OSMOTIC PRESSURE TYNDALL effect Brownian movement Flocculation value. Answer :A

Discussion

36.	The average molar mass of the vapour above solid NH_4Clis nearly 26.75 g "mole"^(-1)What is the composition (by wt.) of this vapour?
Answer» SOLUTION :`NH_331.8% , HCL 68.2%`

Discussion

37.	The average molar mass of the vapour above solid NH_4Clis nearly 26.5 g "mole"^(-1) . Find the composition of the vapour.
Answer» SOLUTION :0.5, 0.5

Discussion

38.	The average (mean) life of a radio nuclide which decays by parallel path is Aoverset(lambda_(1))(rarr)B : lambda_(1)=1.8xx10^(-2)sec^(-1) 2Aoverset(lambda_(2))(rarr)B , lambda_(2)=10^(-3)sec^(-1)
Answer» `52.63 sec` `500 sec` `50 sec` NONE of these Answer :C

Discussion

39.	The average (mean) life at a radio nuclide which decays by parallel path is : A overset(lambda_(1))rarr B, lambda_(1)=1.8 xx10^(-2)"sec"^(-1) 2A overset(lambda_(2))rarrC, lambda_(2)=10^(-3)"sec"^(-1)
Answer» 52.63 sec 500 sec 50 sec none of these Answer :C

Discussion

40.	The average molar mass of a mixture of methane (CH_(4)) and ethene (C_(2)H_(4)) present in the ratio of a:b is found to be 20.0g mol^(-1). If the ratio were reversed, what would be the molar mass of the mixture ?
Answer» Solution :Molar MASS of `CH_(4)="16 g mol"^(-1)` `"Molar mass of "C_(2)H_(4)="28 g mol"^(-1)` When they are PRESENT in the raio `a : b`, their AVERAGE molar mass`=(axx16+bxx28)/(a+b)="20 g mol"^(-1)` (Given) i.e., `16a+28b=20(a+b) or 4a+7b=5(a+b) or a=2b or (a)/(b)=(2)/(1)=2:1` If the ratio is REVERSED, now the ratio `a:b=1:2` `therefore"Average molar mass"=(1xx16+2xx28)/(1+2)=(16+56)/(3)=(72)/(3)="24 g mol"^(-1)`

Discussion

41.	The average life of a W gm sample of .^(200)RaE is T seconds and average energy of the beta-particles emitted is E MeV. At what rate in watts does the sample emits energy?
Answer» `(8WN_(0) E)/(T) xx 10^(-16)` `(8 (1N 2) W N_(0) E)/(T) xx 10^(-13)` `(8 WN_(0) E)/(T) xx 10^(-13)` None is correct Solution :Rate of `beta` particles emitted PER second `= LAMBDA xx (W)/(At. WT) xx N_(0) = (1)/(T) xx (W)/(200) xx N_(0)` Energy evolved per second `= (1)/(T) xx (W)/(200) xx N_(0) xx E xx 1.6 xx 10^(-19) xx 10^(6) j sec^(-1)` `= (8 WN_(0) E)/(T) xx 10^(6)` WATT per second

Discussion

42.	The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure at
Answer» `900 CAL K^(-1)`MOLECULE `^(-1)` `6.21 xx 10^(-21) JK^(-1)` molecule `^(-1)` `336.7 JK^(-1)` molecule `^(-1)` `3741.3 jk ^(-1)"molecule"^(-1)` Answer :B

Discussion

43.	The average kinetic energy of the molecules of SO_(2) at 27^(@)C is E. The average kinetic energy of CO_(2) at 27^(@)C is :
Answer» `16E //11` `11 E // 16` 2E E ANSWER :C

Discussion

44.	The average kinetic energy of an ideal gas per molecule in SI units at 25^@C will be :
Answer» ` 6.17 XX 10^(-21) KJ` `6.17 xx 10^(-21) J` `6.17 xx 10^(-21) J` `7.16 xx10^(-20) J` ANSWER :B

Discussion

45.	The average K.E. of an ideal gas in calories per mole is approximately equal to
Answer» Three times the absolute temperature Absolutetemperature Two times the absolute temperature 1.5 times the absolute temperature Solution :`K.E. = (3)/(2). RT = (3)/(2).2.T "":'R ~=2 CALK^(-1)mol^(-1)`

Discussion

46.	The average energy level of the perturbed d-orbitals in the crystal field for which the energy is taken as zero is called ….. .
Answer» SOLUTION :BARI CENTRE

Discussion

47.	The average concentration of SO_(2) in the atmosphere over a city on a cetrain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO_(2) in water at 298 K is 1.3653 mol litre^(-1) and the pK_(a) of H_(2)SO_(3) is 1.92, estimate the pH of rain on that day.
Answer» Solution :`[SO_3] = [H_2SO_3] = 1.3653 M , H_2SO_3 IFF H^(+) + HSO_3^(-) , K_a = 1.2 XX 10^(-2)` 0.913

Discussion

48.	The average atomic mass of a mixture containing 79 mol % of ""^(24)Mg and remaining 21 mole % of ""^(25)Mg and ""^(26)Mg, is 24.31,% mole of (26)Mg is
Answer» 5 20 10 15 Answer :C

Discussion

49.	The VA group element which exhibits wide range of oxidation states is
Answer» P As Bi N Answer :D

Discussion

50.	The autoreduction process is not used in the metallurgy of
Answer» Hg Cu Pb Fe Solution :`2HgS+3Oto2HgO+2SO` `2HgO+3O_(2)to 2Hg+SO_(2)` `2Cu_(2)+HgSto 2Cu_(2)O+2SO_(2)` `2PbS+3O_(2) to PbO+2SO_(2)` `2PbO+PbS to 3PbS to 3Pb+SO_(2)` This it has STRONG affinitty with OXYGEN than CARBON

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The Baeyer's angle strain is expected to be maximum in

The bakelite is made from phenol and fromaldehyde.The intial reaction between the two compounds is an example of:

The bacterial growth follows the rate law,(dN)/(dt)=kN, where k is a constant and N is the number of bacterial cell at any time. If the population of bacteria (no. of cell) is doubled in 5 minutes then the time by which the population will be eight times of the initial one.

The bad smelling substance formed by the action of alcoholic caustic potash on chloroform and aniline is :

The bad smelling substance formed by the action of alcoholic caustic potash on chloroform and aniline is

The bacteriostatic antibiotic among the following is :

The BaCl_(2) ionises to an extent of 80% in aqueous solution, the value of van't Hoff factor is -

the azocompoundsare colouredas theyabsorb thevisibleregionof electromagneticspectrumbecauseof presenceof

The backbone of a nucleotide strand contains the following sequence of arrangement

The back bone for different segments in a protein is in the following form.

The avrage osmotic presure of human of blood is 7.8 bar at 27^(@)C. The concentration of an aqueous NaCI solution that could be used in the blood stream is :

The azeotrope of nitric acid and water has a composition by mass :

The axles are made by heating rods of iron embedded in charcoal powder. This process is known as:

The average weight of an Indian male is 150 pounds. In SI units it is equal to

The average velocity of the molecules of a gas is 400 m/s. Calculate its rms velocity at the same temperature.

The average velocity of gas molecules is 400 ms^(-1) . Its r.m.s. velocity at the same temperature is

The average velocity of gas molecules is 400 m/sec. Calculate its rms velocity at the same temperature.

The average velocity of CO_(2) at T_(1) K and most probable velocity at T_(2)K is 9 × 10^(4) cm//s. Calculate the values of temperatures, T_(1) and T_(2)K.

The average speed of nitrogen molecule is v. if the temperature is doubled and nitrogen molecule dissociate into nitrogen atoms completely then new average speed becoems.

The average velocity of an ideal gas at 27^(@)C is 0.3m sec^(-1). The average velocity at 927^(@)C will be

The average speed ofO_2at 273 K is equal to that of H_2 at :

The average speed of gas molecules is 400 m/sec. Calculate its rms speed at the same temperature.

The average speed of gas molecules is equal to :

The average speed at T_(1)Kand the most probable speed at T_(2) K of CO_(2) gas is 9 xx 10^(4) cm s^(-1). Calculatethe value of T_(1) and T_(2).

The average speed of an ideal gas molecule at 27^@C is 0.3 sec^(-1). The average speed at 927^@C will be …m sec^(-1).

The average oxidation number of sulphur in Na_2 S_4 O_6

The average oxidation state of sulphur atom in S_4O_6^(2-)ion is

the average osmotic pressure of human blood is 7.8 bar at 37^(@)C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream?

The average osmotic pressure of human blood is 7.7 atm at 40^(@)C. (a) What would be the total concentration of the various solutes in the blood ? (b) Assuming the concentration to be essentially the same as the molality, find the freezing point of blood (K_(f)" for water = "1.86^(@)C).

The average osmotic pressure of human blood is 7.8 bar at 37^(@)C. The concentration of aqueous NaCl solution that could be used in the blood stream is

The average O-H bond energy in H_(2)O with the help of following data. (1) H_(2)O(l)rarrH_(2)O(g), Delta H= +40.6 kJ mol^(-1) (2) 2H(g)rarrH_(2)(g), DeltaH= -435.kJ mol^(-1) (3) O_(2)(g)rarr2O(g), Delta H= +489.6 kJ mol^(-1) (4) 2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l), Delta H= -571.6 kJ mol^(-1)

The average osmotic pressure of benzoic acid is 7.8 bar at 37^(@)C. What is the concerntration of aqueous KCI solution that could be used in blood stream ?

The average osmotic pressure of human blood is 7.7 atm at 40^(@)C. (a) What should be the total concentration of various solutes in the blood ? (b) Assuming this concentration to be essentially the same as the molality, find the freezing point of blood. [K_(f)(H_(2)O)=1.86]

The average molecular speed is greatest in case of a gas sample of :

The average molecular mass of colloidal particles can be accurately determined by:

The average molar mass of the vapour above solid NH_4Clis nearly 26.75 g "mole"^(-1)What is the composition (by wt.) of this vapour?

The average molar mass of the vapour above solid NH_4Clis nearly 26.5 g "mole"^(-1) . Find the composition of the vapour.

The average (mean) life of a radio nuclide which decays by parallel path is Aoverset(lambda_(1))(rarr)B : lambda_(1)=1.8xx10^(-2)sec^(-1) 2Aoverset(lambda_(2))(rarr)B , lambda_(2)=10^(-3)sec^(-1)

The average (mean) life at a radio nuclide which decays by parallel path is : A overset(lambda_(1))rarr B, lambda_(1)=1.8 xx10^(-2)"sec"^(-1) 2A overset(lambda_(2))rarrC, lambda_(2)=10^(-3)"sec"^(-1)

The average molar mass of a mixture of methane (CH_(4)) and ethene (C_(2)H_(4)) present in the ratio of a:b is found to be 20.0g mol^(-1). If the ratio were reversed, what would be the molar mass of the mixture ?

The average life of a W gm sample of .^(200)RaE is T seconds and average energy of the beta-particles emitted is E MeV. At what rate in watts does the sample emits energy?

The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure at

The average kinetic energy of the molecules of SO_(2) at 27^(@)C is E. The average kinetic energy of CO_(2) at 27^(@)C is :

The average kinetic energy of an ideal gas per molecule in SI units at 25^@C will be :

The average K.E. of an ideal gas in calories per mole is approximately equal to

The average energy level of the perturbed d-orbitals in the crystal field for which the energy is taken as zero is called ….. .

The average concentration of SO_(2) in the atmosphere over a city on a cetrain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO_(2) in water at 298 K is 1.3653 mol litre^(-1) and the pK_(a) of H_(2)SO_(3) is 1.92, estimate the pH of rain on that day.

The average atomic mass of a mixture containing 79 mol % of ""^(24)Mg and remaining 21 mole % of ""^(25)Mg and ""^(26)Mg, is 24.31,% mole of (26)Mg is

The VA group element which exhibits wide range of oxidation states is

The autoreduction process is not used in the metallurgy of