94420 + Interview Questions in Chemistry in Class 12 Page 1759 InterviewSolution

87901.	A radioactive isotope has a half-life of 27 days. Starting with 4g of the isotope, what will be mass remaining after 75 days
Answer» 100g 50g 0.58 g 1.58 g Solution :`k = (0.693)/(t_(1//2))` `t = (2.303)/(k) "LOG" (N_(0))/(N)` `t_(1//2) = 27` days `k = (0.693)/(27) = 0.0257 "DAY"^(-1)` `t = (2.303)/(0.0257) "log" (N_(0))/(N)` `N_(0) = 4g, N =? , t = 75` days `75 = (2.303)/(0.0257) "log"(4)/(N)` `N = 0.58g`

Discussion

87902.	A radioactive isotope having a half life of 2.3 days was received after 9.2 days. It was found that there were 300mg of the isotope in the container. The initial amount of the isotope was
Answer» 1200mg 3600mg 4800 MG 2400mg Solution :`t_(1//2) = (0.693)/(K) RARR k = (0.693)/(2.3) = 0.3013` `t= (2.303)/(k) "log" (N_(0))/(N_(t)) rArr "log" (N_(0))/(N_(t)) = (kt)/(2.303)` `rArr log N_(0) - log N_(t) = (kt)/(2.303)` `log N_(0) = (0.3013 xx 9.2)/(2.303) + 2.4771 , N_(0) = 4800 mg`.

Discussion

87903.	A radioactive isotope has a half-life of 20 days. If 100 g of the substance is taken, the weight of the isotope remaining after 40 days is
Answer» 25 g 2.5 g 60 g 40 g Solution :`N = (40)/(20) = 2` `:.` Amount LEFT `(N_(0))/(2^(n)) = (100)/(2^(2)) = 25g`

Discussion

87904.	Aradioactive isotope has t_(1//2) of 10 days. If today 125 g of it is left, what was its weight 40 days earlier ?
Answer» 2g 600g 1g `1.5g` Solution :`N=N_(0)((1)/(2))^(n)n=(40)/(10)=4` `(125)/(1000)=N_(0)((1)/(2))^(4),N_(0)=(125)/(1000)xx2xx2xx2xx2=2g`

Discussion

87905.	A radioactive isotope has a half-life of 10 days. If today 125 mg is left over, what was its original weight 40 days earlier
Answer» 2g 600 mg 1 g 1.5 g Solution :`N = N_(0) ((1)/(2))^(n) .n = (40)/(10) = 4` `(125)/(1000) = N_(0) ((1)/(2))^(4), N_(0) = (125)/(1000) xx 2 xx 2 xx 2 xx 2 = 2g`

Discussion

87906.	A radioactive isotope decays at such a rate that after 96 minutes only (1)/(8)th of the original amount remains. The half-life of this nuclide in minutes is
Answer» 12 24 32 48 Solution :`lamda = (2.303)/(t) "LOG" ([N_(0)])/([N]) = (2.303)/(96) "log" (1)/(1//8)` `:. T_(1//2) = (0.693)/(lamda) = (0.693)/(0.0216) = 32.0` MIN

Discussion

87907.	A radioactive isotope decays at such a rate that after 192 minutes only 1/16 of the original amount remains. The half-life of the radioactive isotope is
Answer» 32 MIN 48 min 12 min 24 min Solution :`K=(2.303)/(192)log16` `t_(1//2)-(0.693)/(K)=(0.693)/((2.303)/(192)xx4log2)=48min`

Discussion

87908.	A radioactive isotope decays as: ._(Z)^(M)A to ._(Z - 2)^(M - 4)B to ._(Z - 1)^(M - 4)C The half lives of A and B are 6 and 10 months respectively. Assuming that initially only A was precent, will it be possible to achieve the radioactive equilibrium for B ? If So, what would be the ratio of A and B at equilibrium? What would happen if the half lives of A and B were 10 and 6 months respectively?
Answer» Solution :At equilibrium RATIO of amounts of A and B will be `(N_(A))/(N_(B)) = (t_(1//2)A)/(t_(1//2)B) = (6)/(10) = 0.6` If the half live of A and B are 10 and 6 MONTS RESPECTIVELY, then B will decay faster than 'A' , hence equilibrium will not be ACHIEVED.

Discussion

87909.	A radioactive isotope .^(40)K with a half-life of 1.26 xx10^(9) years, decay to .^(40)Ar . A sample of rock from the moon was found to contain both element K and Ar and they are in the rartio 1:7.What is the age of the rock?(Neglect decay of .^(40)K to.^(40)Ca)
Answer» `2.52 XX 10^(9)` year `(1.26 xx 10^(9))/("In"2) xx "In" ((7)/(8))` year `(1.26 xx 10^(9))/("In" 2) xx "In" (7) ` year `3.78 xx 10^(9)` year Answer :d

Discussion

87910.	A radioactive element X decays to give two inert gases in air is:
Answer» `238_92U` `226_88RA` `90^TH` `89^Ac` ANSWER :B

Discussion

87911.	A radioactive element X (atomic weight = 200) shows activity 6.93xx10^(15)dps then find weight of X (in mg). (T_(1//2)=100 "minutes"., N_(A) = 6xx10^(23))
Answer» ANSWER :20

Discussion

87912.	A radioactive element with half-life 6.5 hr has 48 xx 10^(19) atoms. Number of atoms left after 26 hr
Answer» `24 xx 10^(19)` `12 xx 10^(19)` `3 xx 10^(19)` `6 xx 10^(19)` SOLUTION :`N_(t) = N_(0) ((1)/(2))^(n) = 48xx10^(19) ((1)/(2))^(26//65)` `= 48 xx 10^(19) ((1)/(2))^(4) = 3xx 10^(19)`

Discussion

87913.	A radioactive element undergoing decay is left 20% of its initial weight after certain period of time t. How many such periods should elapse from the start for the 50% of the element to be left over?
Answer» 3 4 5 none of these Answer :D

Discussion

87914.	A radioactive element resembling iodine in properties is:
Answer» Astatine Lead Radium Thorium Answer :A

Discussion

87915.	A radioactive element resembling iodine in properties is
Answer» ASTATINE Lead Radium Thorium Solution :Astatine (At) RESEMBLES in PROPERTIES with IODINE

Discussion

87916.	A radioactive element is:
Answer» Sulphur Polonium Tellurium Selenium Answer :B

Discussion

87917.	A radioactive element is preset in VIII group of the periodic table. If it emits one alphaparticle, the new position of the nuclide will be
Answer» VI B VIII VIIB I B Answer :a,b,C

Discussion

87918.	A radioactive element has half-life of one day. After three days, the amount of the element left will be
Answer» 1.2 of the original amount 1.4 of the original amount 1.8 of the original amount 1/16 of the original amount Solution :`n = (3)/(1) = 3, N = (N_(0))/(2^(3)) = (1)/(8)`

Discussion

87919.	A radioactive element has a half-life period of 140 days . How much of it will remain after 1120 days
Answer» `(1)/(32)` `(1)/(250)` `(1)/(512)` `(1)/(128)` Answer :A

Discussion

87920.	A radioactive element has a half life of one day. After three days, the amount of element left will be:
Answer» 1/2 of the ORIGINAL amount 1/4 of the original amount 1/8 of the original amount 1/16 of the original amount. Answer :C

Discussion

87921.	A radioactive element has a half-life of 20 minutes. How much time should elapse before the element is reduced to (1)/(8)th of the original mass
Answer» 40 minutes 60 minutes 80 minutes 160 minutes Solution :`t_(1//2) = 20` minute, `N = (1)/(9)N_(0)` USE, `t = (2.303)/(0.693) XX t_(1//2) "log"(N_(0))/(N)`

Discussion

87922.	A radioactive element has a half life of 20 minutes, How much time should elapse before the element is reduced to 1//8 of its original value
Answer» A) 160 minutes B) 60 minutes C) 30 minutes D) 40 minutes Answer :B

Discussion

87923.	A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room
Answer» 1000 days 300 days 10 days 100 days Solution :Activity `(-(dN)/(dt)) prop N` `N = N_(0) ((1)/(2))^(n) RARR (N)/(N_(0)) = ((1)/(2))^(n)` `(1)/(10) = ((1)/(2))^(n) rArr 10 = 2^(n)` `LOG 10 = n log 2 rArr n = (1)/(0.301) = 3.32` `t = n XX t_(1//2) = 3.32 xx 30 = 99.6` days

Discussion

87924.	A radioactive element decays by following three different parallel paths : A overset(lambda_(1))(to) B, lambda_(1) = 5 xx 10^(-2) sec^(-1) 2A overset(lambda_(2))(to) C,lambda_(2) = 3xx10^(-2) sec^(-1) 3A overset(lambda_(3))(to) D, lambda_(3) = 5xx10^(-3) sec^(-1) Calculate average life- time of element (in sec.) [lambda_(1), lambda_(2), lambda_(3) "are decay constans forrespective reactions"]
Answer» ANSWER :8

Discussion

87925.	A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room ?
Answer» 300 days 10 days 100 days 1000 days. Solution :LET the PERMISSIBLE activity limit be x, then INITIAL activity would be 10x `lamda=(0.693)/(30"days")` `t=(2.303)/(lamda)log"([A]_(0))/([A])` `=(2.303xx30)/(0.693)"log"(10x)/(x)` `=99.7"days"~~100` days.

Discussion

87926.	A radioactive element belongs to the group 14 of the periodic table, it undergoes beta-emission, the product obtained belongs to the following group of the periodic table
Answer» Group 12 Group 13 Group 15 Group 16 Solution :`._(6)^(14)C overset(beta " decay")RARR ._(7)^(14)N + underset(-1)overset(0)beta`

Discussion

87927.	A radioactive element decays as X to_(t_(1//2)=30min)^(alpha decay) Y to _(t_(1//2)=2 days)^((-2beta)beta decay)Z. Which of the following statement about this decay process is incorrect?
Answer» after two hours, less than 10% of the initial X is left MAXIMUM amount of Y present at any TIME before 30 min is less than 50% of the initial amount of X atomic number of X and Z are same the MASS number of Y is greater than X Answer :D

Discussion

87928.	A radioactive element (atomic mass = 90 a.m.u.) has half life of 28 years. The number of disintegration per second is :
Answer» `5.24xx10^(8)` `5.24xx10^(10)` `5.24xx10^(-10)` `5.24xx10^(12)` Solution :Decay RATE `=lamdaxxN` `lamda=(0.693)/(t_(1)//2)=(0.693)/(25xx365xx24xx36000)` `=7.83xx10^(-10)s^(-1)` No. of atoms in 1 G of element (N) `=(1)/(90)xx6.023xx10^(23)` `=6.69xx10^(19)` atoms Decay rate `=7.83xx10^(-10)xx6.96xx10^(19)` `=5.24xx10^(10)` disintegrations per sec.

Discussion

87929.	(A) Radioactive disintegration process can never go to completion (R ) Radio disintegration process follows first order kinetics
Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false Answer :A

Discussion

87930.	A radio isotope will not emit
Answer» GAMMA and alpha rays SIMULTANEOUSLY gamma rays only alpha and beta rays simultaneously beta and gamma rays simultaneously Solution :In radioactive decay `alpha and beta` particles are not emitted simultaneously, either `alpha or beta` PARTICLE is emitted, one at a TIME.

Discussion

87931.	A radiator of motor vehicle was filled with 8.2 lt of water to which 2L of methanol (d=0.8 gm//ml) were added, the lowest temp. at which the vehicle can be parked without a danger of getting water in the radiator to freeze is (K_(f) for water =1.86 Km^(-1))
Answer» `-11.34^(@)C` `-5.6^(@)C` `-6.6^(@)C` `-4.3^(@)C` ANSWER :A

Discussion

87932.	A racemic mixture of carboxylic acid having one chiral centre is treated with enantiomerically pure amine. the products formed are :
Answer» disatereomers (both OPTICALLY active) diasteremoers (only one is optically active) RACEMIC mixture two optically inactive AMMONIUM salts of acid. Answer :A

Discussion

87933.	A racemic mixture is a mixture of:
Answer» MESO and its isomers d- and its l-isomers in EQUAL proportions d- and its l-isomers in DIFFERENT proportions Meso and d-isomers Answer :B

Discussion

87934.	A racemic mixture has a net rotation
Answer» to RIGHT of ORIGINAL PLANE to LEFT of original plane to right or left of original plane zero Answer :D

Discussion

87935.	A racemic mixture consistsof
Answer» equal amounts of enantiomers DIFFERENT amounts of enantiomers UNKNOWN amounts of enentiomers unknown amounts of unknown COMPOUNDS. Solution :Optical rotation of one enantiomer is EXACTLY balanced by the optical rotation of the other enantiomer TAKEN in exactly equal amounts.

Discussion

87936.	(A) R-overset(O)overset(\|\|)(C)-Cl fails to give ketone when treated with R-Mg-X, but with R_(2)Cd, ketones is obtained easily. (R) R-Mg-X reacts with acid chloride to form ketone, but R-Mg-X further reacts with ketone readily and form tertiary alcohol.
Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) is false Answer :A

Discussion

87937.	Explain elimination reaction. Give example.
Answer» SOLUTION :ROTATION by an ENANTIOMER CANCELLED by the other.

Discussion

87938.	A qulalitative analysis of papaverine, an opium alkaloid shown carbon,hydrogen and nitrogen.A quantative analysis gave 70.8% carbon,6.2% hydrogen and 4.1% nitrogen.The empirical formula of papaverrine is :
Answer» `C_(20)H_(20)N_(2)` `C_(20)H_(21)O_(4)N` `C_(10)H_(11)O_(3)N` `C_(21)H_(20)N` SOLUTION :N//A

Discussion

87939.	A quick supply of chlorine gas may be made by reacting crystals of KMnO_4 with a concentrated solution of:
Answer» POTASSIUM chlorate Sodium chloride Bleaching powder Hydrochloric acid Answer :D

Discussion

87940.	A quantity of PCl_(5) ws heated in a 10 litre vessel at 250^(@)C: PCl_(5_(g))hArrPCl_(3(g))+Cl_(3(g)). At equilibrium the vessel contains 0.1 moles of PCl_(5)0.20 mole of PCl_(3)and 0.2 mole of Cl_(2). The equilibrium constant of the reaction is
Answer» `0.02` `0.05` `0.04` `0.025` SOLUTION :`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((0.2)/(10)XX(0.2)/(10))/([0.1//10])=0.04.`

Discussion

87941.	A quantity of PCl_5 was heated in a 10 litre vessel at 250^@ C to show PCL_5(g) hArrPCl_3 + Cl_2 AT equilibrium the vessel contains 0.1 mole of PCl_5 0.20 mole of PCl_3 and 0.20 mole od cl_2 The equilibrium constant of the reaction is :
Answer» 0.02 0.05 0.04 0.025 Answer :C

Discussion

87942.	A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C : PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)) At equilibrium, the vessel contains 0.1 mole of PCl_(5) and 0.2 mole of Cl_(2). The equilibrium constant of the reactionis
Answer» `0.04` `0.025` `0.02` `0.05` Solution :`{:(,PCl_(5),IFF,PCl_(3),+,Cl_(2)),("MOLES at EQM.","0.1 mole",,"0.2 mole",,"0.2 mole"),("Molar concs.","0.1/10",,"0.2/10",,"0.2/10"):}` `K=(0.02xx0.02)/(0.01)=0.04`

Discussion

87943.	A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C PCl_(5(g)) hArr PCl_(3 (g)) + Cl_(2(g)). At equilibrium the vessel contains 0.1 mole of PCl_(5), 0.2 moles of PCl_(3) and 0.2 moles of Cl_2. The equilibrium constant of the reaction is
Answer» 0.025 0.04 0.05 0.02 Solution :Equilibrium conc. Of `PCl_(5) = (0.1)/(10) = 0.01M` Equilibrium conc. Of `Cl_(2) = PCl_(3) = (0.2)/(10) = 0.02M` `RARR K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]) = (0.02 XX 0.02)/(0.01) = (0.0004)/(0.01)` `rArr K_(C) = 0.04`

Discussion

87944.	A quantity of ethyl acetate is mixed with an excess of sodium hydroxide at 25^(@)C. 100 c.c. of the mixture is immediately titrated against 0.05N Hydrochloric acid of which 75 c.c were required for neutralisation. After 30 minutes 50 c.c of the mixture required, similarly 25 c.c of the acid When the original reaction of ester was complete 25 c.c of the mixture required 6.25 c.c of the second order velocity constant ("in mol"//1//"min")(at time =0) of the reaction using concentration in moles per litre and time in minutes. Reactions is first order each w.r.t. NaOH and ester. Indicator chosen for above titrationis such that it gives end point when only hydrochloric acid reacts with NaOH.(log 2= 0.30, log 3= 0.48, In 10 = 2.3)(Given your Answer after multiplying with a factor of 10 and excluding the decimal places)
Answer» ANSWER :7

Discussion

87945.	A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al^(3+) at the cathode will also produce the following volume at STP of H_(2)(g) from H^(+) at the cathode:
Answer» `44.8 L` `22.4 L` `11.2 L` `2.6 L` ANSWER :D

Discussion

87946.	A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al^(3+) at the cathod will also produce the following volume (STP) of H_2(g) from H^(+) at the cathode
Answer» 44.8 L 22.4 L 11.2 L 5.6 L Solution :`AL to Al^(+3)+ 3e^(-) , 27 GM to 3F , 4.5 gm to ? = (4.5 XX 3)/(27)=0.5F` `2H^(+) + 2e^(-) to H_2 , 2F to 22.4l 0.5 F to ?` Volume of `H_2 ` from ` H^+` at cathode `= (0.5)/(2) xx 22.4 = 5.6l`

Discussion

87947.	A quantity of aluminium has a mass of 54g . What is the mass of same number of magnesium atoms?
Answer» 12.1g 24.3g 48.6g 97.2g Answer :C

Discussion

87948.	A quantity of 1.92 of methanol was burnt in a constant pressure calorimeter. The temperature of water increased by 4.2^(@)C. If the quantity of water surrounding the inner vessel was 2000 ml and the heat capacity of the inner vessel was 2.02 kJ//^(@)C. Calculate the heat of combustion of methanol. [Specific heat of capacity of H_(2)O = 4.18 J//g^(@) C]
Answer» SOLUTION :`-726.6 ` kJ/mol

Discussion

87949.	A quantity of 0.25 M NaOH is added to a solution containing 0.15 mole of aceticacid. The final volume of the solution is 375 mL and the pH of the solution is 4.45. (a) What is the molar concentration of sodium acetate? (b) How many mL of NaOH were added to the original solution? (c) What was the original concentration of the acetic acid?
Answer» SOLUTION :`(a) 0.048 M (B) 1.9 XX 10^2 ML (C) 0.81 M`

Discussion

87950.	(A): Pyrolusite is an ore of manganese. (R) : Formula of pyrolusite is Mn_(3)O_(4).
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false Solution : (A) is true but (R) is false

Discussion

Explore topic-wise InterviewSolutions in .

A radioactive isotope has a half-life of 27 days. Starting with 4g of the isotope, what will be mass remaining after 75 days

A radioactive isotope having a half life of 2.3 days was received after 9.2 days. It was found that there were 300mg of the isotope in the container. The initial amount of the isotope was

A radioactive isotope has a half-life of 20 days. If 100 g of the substance is taken, the weight of the isotope remaining after 40 days is

Aradioactive isotope has t_(1//2) of 10 days. If today 125 g of it is left, what was its weight 40 days earlier ?

A radioactive isotope has a half-life of 10 days. If today 125 mg is left over, what was its original weight 40 days earlier

A radioactive isotope decays at such a rate that after 96 minutes only (1)/(8)th of the original amount remains. The half-life of this nuclide in minutes is

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A radioactive isotope .^(40)K with a half-life of 1.26 xx10^(9) years, decay to .^(40)Ar . A sample of rock from the moon was found to contain both element K and Ar and they are in the rartio 1:7.What is the age of the rock?(Neglect decay of .^(40)K to.^(40)Ca)

A radioactive element X decays to give two inert gases in air is:

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A radioactive element with half-life 6.5 hr has 48 xx 10^(19) atoms. Number of atoms left after 26 hr

A radioactive element undergoing decay is left 20% of its initial weight after certain period of time t. How many such periods should elapse from the start for the 50% of the element to be left over?

A radioactive element resembling iodine in properties is:

A radioactive element resembling iodine in properties is

A radioactive element is:

A radioactive element is preset in VIII group of the periodic table. If it emits one alphaparticle, the new position of the nuclide will be

A radioactive element has half-life of one day. After three days, the amount of the element left will be

A radioactive element has a half-life period of 140 days . How much of it will remain after 1120 days

A radioactive element has a half life of one day. After three days, the amount of element left will be:

A radioactive element has a half-life of 20 minutes. How much time should elapse before the element is reduced to (1)/(8)th of the original mass

A radioactive element has a half life of 20 minutes, How much time should elapse before the element is reduced to 1//8 of its original value

A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room

A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room ?

A radioactive element belongs to the group 14 of the periodic table, it undergoes beta-emission, the product obtained belongs to the following group of the periodic table

A radioactive element decays as X to_(t_(1//2)=30min)^(alpha decay) Y to _(t_(1//2)=2 days)^((-2beta)beta decay)Z. Which of the following statement about this decay process is incorrect?

A radioactive element (atomic mass = 90 a.m.u.) has half life of 28 years. The number of disintegration per second is :

(A) Radioactive disintegration process can never go to completion (R ) Radio disintegration process follows first order kinetics

A radio isotope will not emit

A radiator of motor vehicle was filled with 8.2 lt of water to which 2L of methanol (d=0.8 gm//ml) were added, the lowest temp. at which the vehicle can be parked without a danger of getting water in the radiator to freeze is (K_(f) for water =1.86 Km^(-1))

A racemic mixture of carboxylic acid having one chiral centre is treated with enantiomerically pure amine. the products formed are :

A racemic mixture is a mixture of:

A racemic mixture has a net rotation

A racemic mixture consistsof

(A) R-overset(O)overset(||)(C)-Cl fails to give ketone when treated with R-Mg-X, but with R_(2)Cd, ketones is obtained easily. (R) R-Mg-X reacts with acid chloride to form ketone, but R-Mg-X further reacts with ketone readily and form tertiary alcohol.

Explain elimination reaction. Give example.

A qulalitative analysis of papaverine, an opium alkaloid shown carbon,hydrogen and nitrogen.A quantative analysis gave 70.8% carbon,6.2% hydrogen and 4.1% nitrogen.The empirical formula of papaverrine is :

A quick supply of chlorine gas may be made by reacting crystals of KMnO_4 with a concentrated solution of:

A quantity of PCl_(5) ws heated in a 10 litre vessel at 250^(@)C: PCl_(5_(g))hArrPCl_(3(g))+Cl_(3(g)). At equilibrium the vessel contains 0.1 moles of PCl_(5)0.20 mole of PCl_(3)and 0.2 mole of Cl_(2). The equilibrium constant of the reaction is

A quantity of PCl_5 was heated in a 10 litre vessel at 250^@ C to show PCL_5(g) hArrPCl_3 + Cl_2 AT equilibrium the vessel contains 0.1 mole of PCl_5 0.20 mole of PCl_3 and 0.20 mole od cl_2 The equilibrium constant of the reaction is :

A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C : PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)) At equilibrium, the vessel contains 0.1 mole of PCl_(5) and 0.2 mole of Cl_(2). The equilibrium constant of the reactionis

A quantity of PCl_(5) was heated in a 10 dm^(3) vessel at 250^(@)C PCl_(5(g)) hArr PCl_(3 (g)) + Cl_(2(g)). At equilibrium the vessel contains 0.1 mole of PCl_(5), 0.2 moles of PCl_(3) and 0.2 moles of Cl_2. The equilibrium constant of the reaction is

A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al^(3+) at the cathode will also produce the following volume at STP of H_(2)(g) from H^(+) at the cathode:

A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al^(3+) at the cathod will also produce the following volume (STP) of H_2(g) from H^(+) at the cathode

A quantity of aluminium has a mass of 54g . What is the mass of same number of magnesium atoms?

(A): Pyrolusite is an ore of manganese. (R) : Formula of pyrolusite is Mn_(3)O_(4).