94420 + Interview Questions in Chemistry in Class 12 Page 608 InterviewSolution

30351.	The half life of the homogeneousgaseous reaction SO_2Cl_2rarrSO_2+Cl_2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO_2Cl_2 to the reduced to 1% of the initial value ?
Answer» Solution :We KNOW that , k = `0.693//t_(1//2)` k = 0.693/8.0 `=0.087 "MINUTES"^(-1)` For a first order REACTION `t=(2.303)/klog.([A_0])/([A])` `t = (2.303)/(0.087"min"^(-1))LOG.(100/1)` t = 52.93 min

Discussion

30352.	The half-life of Tc^(99) is 6.0 hr. The delivery of a sample of Tc^(99) from the reactor to the nuclear medicine lab of a certain hospital takes 3.0 hr. What is the minimum amount of Tc^(99) that must be shipped in order for the lab to receive 10.0 mg?
Answer» 20.0 mg 15.0 mg 14.1 mg 12.5 mg Answer :B

Discussion

30353.	The half-life of radium (226) is 1620 years. The time taken to convert 10 grams of radium to 1.25 grams is
Answer» 810 years 1620 years 3240 years 4860 years Solution :`t = (2.303 XX t_(1//2))/(0.693) "LOG" (N_(0))/(N)`

Discussion

30354.	The half-life of Rn is 46 days, what amount will be left from 2 mole of it after 138 days?
Answer» `0.25` mole `1.25` mole `0.40` mole `0.62` mole Solution :Number of half lives gone = `(138)/(46) = 3` `N_0 = 2` mole, `N_t = ?` As, `N_t = N_0 xx (1/2)^(n) implies N_t= 2 xx (1/2)^(3) = 2/8 = 0.25` mole. HENCE 0.25 mole of .Rn. will be LEFT from 2 mole of it after 138 days.

Discussion

30355.	The half-life of radioactive sodium is 15.0 hours. How many hours would it taken for 64 gms of sodium to decay one-eighth of its original value
Answer» 3 15 30 45 Solution :`:' 2.303 "log" (No)/(N) = LAMDA t` and `lamda = (0.693)/(t_(1//2)) = (0.693)/(15)` or `lamda = 0.0462` `:.` using equation (i) we GET `2.303 "log" (64)/(8) = 0.0462 xx t` or `t = 45`hr

Discussion

30356.	The half life of radioactive sodium is 15 hours its disintegration constant is
Answer» `0.0462 h^(-1)` `0.0642 h^(-1)` `0.0462 s^(-1)` `0.0462 "min"^(-1)` SOLUTION :`K = (0.693)/(t_(1//2)) = (0.693)/(15) = 0.0462 hr^(-1)`.

Discussion

30357.	The half life of radioactive sodium is 15 hours. How many hours would it take for 64 g of sodium to decay to one-eight of its original value?
Answer» 3 HOURS 15 hours 30 hours 45 hours SOLUTION :d) `t=2.303/k log (No)/(NT)` `t=(2.303 xx(15 "hours"))/(0.693)log64/8` `=(2.303 xx 15 xx 0.9030)/(0.693) = 45` hours

Discussion

30358.	The half life of radioactive element depends upon the
Answer» NATURE of element Temperature Pressure Amount of the element Answer :A

Discussion

30359.	The half life of paracetamol with in the body is ....
Answer» 2 HOURS 2.5 hours 6 hours 10 hours Answer :B

Discussion

30360.	The half-life of cobalt - 60 is 5.26 years. Calculate the % activity remaining after 4 years.
Answer» SOLUTION :`t^(1//2)=5.26 years` t=4 years here to findthe % of activity (i.e) to find`(N)/(N_(0))` `log(N_(0))/(N)=(lambdaxxt)/(2.303)` `=(0.693)/(5.26)xx(4)/(2.303)` =0.2288 `(N_(0))/(N)` = ANTILOG (0.2288) =1.693 % of activity = `0.59 XX100` =59%

Discussion

30361.	The half-life of C^(14 is 5760 years.for a 200 mg sample ofC^(14) , the time taken to change to 25 mg is
Answer» 11520 years 23040 years 5760 years 17280 years. Solution :`R_(0) = 200 mg , R= 25 mg , R = 25 mg , t_(1//2) = 5720` years SUPPOSE REQUIRED years are x. ALSO, suppose no. of half lives = n `:. (R/(R_0))= (1/2)^n IMPLIES (25/100) = (1/2)^(n)` `implies (1/8) = (1/2)^(n) implies (1/2)^(3) = (1/2)^(n) implies n = 3` `:. n = x/(5760) implies 3 = x/(5760) implies x = 5760 xx 3 = 17280` years.

Discussion

30362.	The half-life of Co^(60) is 7 years. If one g of it decays, the amount of the substance remaining after 28 years is
Answer» 0.25 G 0.125 g 0.0625 g 0.50 g Solution :`n = (28)/(7) = 4, N = (N_(0))/(2^(n)), N = (1)/(2^(4)) = (1)/(16) = 0.0625 g`

Discussion

30363.	The half life of a substance in a certain enzyme-catalysed reaction is 138 s . The time required for the concentration of the substance to fall from 1.28 mg L^(-1) to 0.04 mg L^(-1) is
Answer» 690s 276 s 414 s 552s Solution :Enzyme catalysed reactions are initially FOLLOW FIRST ORDER kinetics when concentration decreases `1.28mg l^(-1)` to 0.04 mg `l^(-1)` . Then five half-life completed No. of half-lives =5 So , TIMES required = `5 XX 138 = 690 `s

Discussion

30364.	The half-life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28" mg L"^(-1)" to "0.04" mg L"^(-1), is
Answer» 414 s 552 s 690 s 276 s Solution :AMOUNT LEFT after n HALF lives `=([A]_(0))/(2^(n))` `0.04 = (1.28)/(2^(n)) or 2^(n) = 32 = 2^(5) therefore n = 5` Time taken ` = 5 ` half lives`= 5 xx 138 s = 690 s.`

Discussion

30365.	The half life of a reaction is inversely proportionaly to the square of the initial concentration of the reactant . Then the order of the reaction is :
Answer» 0 1 2 3 Solution :(D) `t_(1//2) prop [A]_(0)^(1-N)` where n = ORDER of reaction SINCE `t_(1//2)prop (1)/([A_(0)]^(2))` `implies t_(1//2)prop[A_(0)]^(-2)` `1-n =-2 ` n=3.

Discussion

30366.	The half-life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is
Answer» `0.5` 1 2 0 Solution :`t_(1//2) PROP (1)/([A_(0)]^(n-1)), ((t_(1//2))1)/((t_(1//2))2) = ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1)) = {([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `(t)/(t//2) = (2a)/(a)^(n-1) or 2 = (2)^(n-1)` or ` n -1 = 1 or n =2 `

Discussion

30367.	The half-life of a radionuclide is 69.3 minutes. What is its average life (in minutes)
Answer» 100 `10^(-2)` `(69.3)^(-1)` `0.693 xx 69.3` Solution :Average life `(TAU) 1.44 t_(1//2) = 1.44 xx 69.3 = 99.7 ~~ 100` MINUTES

Discussion

30368.	The half-life of a radioisotope is four hour. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is
Answer» 3.125 G 2.084 g 1.042 g 4.167 g Solution :`(N)/(N_(0)) = ((1)/(2))^((T)/(t_(1//2))), (N)/(200)= ((1)/(2))^((24)/(4)), (N)/(200) = ((1)/(2))^(6)` `N = (200)/(64) = 3.125 g`

Discussion

30369.	The half-life of a radioisotope is 20 years. If the sample has an initial activity of 640 dps,than what will be its activity after 80 years?
Answer» 20 dps 30 dps 40 dps 50 dps Solution :`t_(1//2) = 20 "years" , t = 80` years, `N_0 = 640 dps, N_t = ?` Number of half lives gone `(n) = (80/20)= 4` Now, `N_t = N_0 XX (1/2)^(n) = 640 xx (1/2)^(4) = (640)/(16) = 40 dps` The ACTIVITY of the radioisotope will be 40 dps after 80 years.

Discussion

30370.	The half life of a radioisotope is 4 hr. If the initial mass of the isotope was 200 g, the mass remaining after 24 hr undecayed is:
Answer» 3.125 g 2.084 g 1.042 g 4.167 g Solution :No. of half life periods =6 Mass left `=([A]_(0))/(2^(6))=(200)/(2^(6))` `=(200)/(64)=3.125g`

Discussion

30371.	The half-life of a radioactive nucleide 'X is 24 hours, the time required for 12.5% of the original X to remain is :
Answer» 1 days 2 days 3 days 4 days Answer :C

Discussion

30372.	The half - lifeof a radioactive isotope is three hours. If the initial mass of the isotope was 300 g. the masswhichremainsundecayed in 18 hourswould be
Answer» 2.34 g 1.17 g 9.36 g 4.68 g SOLUTION :NUMBER of half -lives(N) =` ("Total time")/( t_(1//2)) = 18/3 = 6` ` N = N_(0)/(2^(n)) = 300/2^(6)4.68 g`

Discussion

30373.	The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256g, the mass of it remaining undecayed after 18 hours would be
Answer» `4.0 G` `8.0 g` `12.0 g` `16.0 g` Solution :`N_(t) = N_(0) ((1)/(2))^(N) , N_(t) = 256 ((1)/(2))^(18//3) = 256 ((1)/(2))^(6) = 4`

Discussion

30374.	The half-life of a radioactive isotope is 3 hours. Value of its disintegration constant is
Answer» 0.231 PER HR 2.31 per hr 0.2079 per hr 2.079 per hr Solution :`K = (0.693)/(t_(1//2)) = (0693)/(3HR) = 0.231` per hr.

Discussion

30375.	The half life of a radioactive isotope is three hours. If the initial mass of the isotope was 300 g, the mass which remained undecayed in 18 hours would be :
Answer» 2.34g 1.17 g 9.36 g 4.68 g Solution :AMOUNT LEFT after N HALF lives `=(N_(0))/(2^(n))=(300)/(2^(6))` `=(300)/(64)=4.68g`

Discussion

30376.	The half life of a radioactive isotope is 2.5 hour . The mass of it that remains undecayed after 10 hour is (If the mass of the isotope was 16 g) :
Answer» 32 g 16 g 4 g 1 g SOLUTION :

Discussion

30377.	The half-life of a radioactive isotope is 1.5 hours. The mass of it that remains undecayed after 6hours is (if the initial mass of the isotope was 32g)
Answer» 32g 16g 4g 2g Answer :D

Discussion

30378.	The half-life of a radioactive element is 6 months. The time taken to reduce its original concentration to its 1.16 value is
Answer» 1 years 16 years 2 years 8 years Solution :`t = (2.303 XX t_(1//2))/(0.693) "log" (N_(0))/(N), N = (1)/(16)`

Discussion

30379.	The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in 1 g atom sample ?
Answer» `45.6 XX 10^(23)` atoms `4.56 xx 10^(23)` atoms `4.56 xx 10^(24)` atoms `4.56 xx 10^(25)` atoms SOLUTION :`N_t = N_0 (1/2)^(n)` `n = 4//10` `:. N_t = 6.022 xx 10^(23) xx (1/2)^(0.4) = 4.56 xx 10^(23)`

Discussion

30380.	The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in 1 g atom sample
Answer» `45.6 XX 10^(23)` ATOMS `4.56 xx 10^(23)` atoms `4.56 xx 10^(24)` atoms `4.56 xx 10^(25)` atoms Solution :`N_(t) = N_(0) ((1)/(2))^(n) n = 4//10` `:. N_(t) = 6.022 xx 10^(23) xx ((1)/(2))^(-.4) = 4.56 xx 10^(23)`

Discussion

30381.	The half-life of a radioactive element is 100 minutes . The time interval between the stages 50% and 87.5 % decay will be x xx 40 min . The value ofx is _____
Answer» SOLUTION :`100 overset(100)(to) 50 overset(100)(to) 25 overset(100)(to) 12.5 , t = 2 xx 100 = 200` MM

Discussion

30382.	The half life of a first order reaction x rarr products is 6.932 xx10^4 s at 500 K . What percentage of x would be decomposed on heating at 500 K for 100 min . (e^(0.06)=1.06)
Answer» Solution :Given `t_(1//2) = 0.6932xx10^4s` To solve : when t = 100 min , `([A_0]-[A])/([A_0])xx100= ? ` We known that a first order reaction , `t_(1//2)=(0.6932)/K` `k=(0.6932)/(6.932xx10^4)impliesk=10^(-5)s^(-1)` `k=(1/t)LN(([A_0])/([A]))` `10^(-5)s^(-1)xx100xx60s=ln([A_0])/([A])` `0.06=ln(([A_0])/([A]))` `([A_0])/([A])=E^(0.06)=1.06` `:. ([A_0]-[A])/([A_0])xx100%=(1-([A])/([A_0]))xx100%=(1-1/(1.06))xx100%=5.6%`

Discussion

30383.	The half -life of a first order reaction [k=(2.303)/(t)"log"((a)/(a-x))] is
Answer» DIRECTLY proportional to a inversely proportional to 'a' independent proportional to (a - x) SOLUTION :It is a FACT .

Discussion

30384.	The half- lifeof a firstorderreactionis 900min at820 k Estimateits half- lifeat 720 kif theenergyof activationof thereactionis 250kJ "mol"^(-1) (1.464 xx10^(5)min)
Answer» Solution :HINT `: (k_(2))/(k_(1))= ((t_(1//2))_(1))/(t_(1//2))_(2)` Half- LIFE period`-1.46xx 10^(5) min`

Discussion

30385.	The half-life of a first order reaction [k=2.303/t log (a/(a-x))] is
Answer» directly proportional to 'a' inversely proportional to 'a' independent of 'a' proportional to (a - x) SOLUTION :For a first order reaction `k = (2.303)/(t) "LOG" (a)/(a-x) ` , when `t = t_(1//2) , x= (a)/(2)` `therefore t_(1//2)= (2.303)/(k) "log" (a)/(a- (a//2))` or `t_(1//2) = (2.303)/(k) "log" 2 = (2.303 xx 0.3010)/(k) = (0.693)/(k)` i.e., Half LIFE of a1 order reaction is independent of INITIAL concentration 'a' of reactant.

Discussion

30386.	The half life of a first order reaction is 69.35 sec . The value of the rate constant of the reaction is
Answer» `1. 0 s^(-1)` `0.1 s^(-1)` `0.01 s^(-1)` `0.001 s^(-1)` SOLUTION :`K = (0.693)/(t_(1//2)) = (0.693)/(69.35) = 9.99 xx 10^(-3) = 0.01 s^(-1)` .

Discussion

30387.	The half- life of afirstorderreactionis 1.7hours. Howlongwill it takefor 20%of thereactanttodisappear ?
Answer» SOLUTION :Timerequiredfor 20%reaction=32 .86 MIN

Discussion

30388.	The half-life of a first order reaction is 30 min and the initial concentration of the reactant is 0.1 M. If the initial concentration of reactant is doubled, then the half-life of the reaction will be
Answer» 1800 s 60 min 15 min 900 s Answer :A

Discussion

30389.	The half life of a first order reaction is
Answer» INDEPENDENT of the initial CONCENTRATION of the reactant directly PROPORTIONAL to the initial concentration of the reactant inversely proportional to the initial concentration of the reactant directly proportional to the square of the initial concentration of the reactant. ANSWER :A

Discussion

30390.	Thehalf life of a firstorderreactionis 30 minand theinitialconcentrationof thereactant is 0.1M . Ifthe initialconcentrationofreactantisdoubled,thenthehalflife ofthe reactionwill be
Answer» 1800S 60 min 15 min 900s Answer :A

Discussion

30391.	The half life of a first order reaction is 100 seconds. What is the time required for 90% completion of the reaction?
Answer» 100 sec. 200 sec. 333 sec. 500 sec. Answer :C

Discussion

30392.	The half life of a first order reaction is 100 seconds at 280 K. If the temperature coefficient is 3.0 its rate constant at 290 K in s^(-1) is
Answer» `2.08xx10^(-3)` `2.08xx10^(-2)` `6.93xx10^(-3)` `6.93xx10^(-2)` ANSWER :B

Discussion

30393.	The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mole/litre and concentration at some instant ‘t’ is 0.01 mol/litre, then the value of ‘t’ is :
Answer» 10 MINUTES 30 minutes 20 minutes 40 minutes Solution :0.08 MOL `l^(-1)` to `0.01` mol `l^(-1)` involves 3 half-life . So the t is 30 minutes .

Discussion

30394.	The half-life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/L and concentration at some instant is 0.01 mol/L, then t is
Answer» 10 minutes 30 minutes 20 minutes 40 minutes. Solution :`K=(0.693)/(t_(1//2))=(0.693)/(10)` `t=(2.303)/(k)"LOG"(0.08)/(0.01)=(2.303xx10)/(0.693)log` 8=30 minutes

Discussion

30395.	The half-life of a first order reactionhaving rate constant K = 1.7 xx 10^(-5) s^(-1) is
Answer» 12.1 H 9.7 h 11.3 h 1.8 h Solution :k = `1.7 xx 10^(-5) s^(-1)` `t_(1//2) = (0.693)/(k) = (0.693)/(1.7) xx 10^(5) = 11.32` h.

Discussion

30396.	The half life of a first order reaction AtoB+C is 10 minutes. The concentration of 'A' would be reduced to 10% of the original concentration in
Answer» 10 minutes 33 minutes 90 minutes 70 minutes Answer :B

Discussion

30397.	The half life of a 1^(st) order reaction is 1 min 40 seconds. Calculate its rate constant (k)
Answer» `k=6.9xx10^(-3) "MIN" ^(-1)` `k=6.9xx10^(-3) s^(-1)` `k=6.9xx10^(-3) s` `k=100 s` ANSWER :B

Discussion

30398.	The half-life of ._(92)U^(238) " is " 4.5 xx 10^(9) years. After how many years, the amount of ._(92)U^(238) will be reduced to half of its present amount
Answer» `9.0 XX 10^(9)` years `13.5 xx 10^(9)` years `.45 xx 10^(9)`years `4.5 xx 10^(4.5)` years Solution :Because `t_(1//2) = 4.5 xx 10^(9)` years, so after `4.5xx10^(9)` years the amount of `._(92)U^(238)` will be half decayed

Discussion

30399.	The half life of ._(92)^(238)U is 4.5 xx 10^(9) years. Uranium emits an alphaparticle to give thorium. Calculate the time required to get the product which contains equal masses of thorium and uranium.
Answer» SOLUTION :`N_(0) = (1)/(238) + (1)/(234)`,`N = (1)/(238)` Use, `(0.693)/(t_(1//2)) = (2.303)/(t) log_(10) ((N_(0))/(N))`

Discussion

30400.	The half-life of ._(6)^(14)C is 5730 year. What fraction of its original C^(14) would left after 22920 year of storage?
Answer» 0.5 0.25 0.125 0.0625 Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

The half life of the homogeneousgaseous reaction SO_2Cl_2rarrSO_2+Cl_2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO_2Cl_2 to the reduced to 1% of the initial value ?

The half-life of Tc^(99) is 6.0 hr. The delivery of a sample of Tc^(99) from the reactor to the nuclear medicine lab of a certain hospital takes 3.0 hr. What is the minimum amount of Tc^(99) that must be shipped in order for the lab to receive 10.0 mg?

The half-life of radium (226) is 1620 years. The time taken to convert 10 grams of radium to 1.25 grams is

The half-life of Rn is 46 days, what amount will be left from 2 mole of it after 138 days?

The half-life of radioactive sodium is 15.0 hours. How many hours would it taken for 64 gms of sodium to decay one-eighth of its original value

The half life of radioactive sodium is 15 hours its disintegration constant is

The half life of radioactive sodium is 15 hours. How many hours would it take for 64 g of sodium to decay to one-eight of its original value?

The half life of radioactive element depends upon the

The half life of paracetamol with in the body is ....

The half-life of cobalt - 60 is 5.26 years. Calculate the % activity remaining after 4 years.

The half-life of C^(14 is 5760 years.for a 200 mg sample ofC^(14) , the time taken to change to 25 mg is

The half-life of Co^(60) is 7 years. If one g of it decays, the amount of the substance remaining after 28 years is

The half life of a substance in a certain enzyme-catalysed reaction is 138 s . The time required for the concentration of the substance to fall from 1.28 mg L^(-1) to 0.04 mg L^(-1) is

The half-life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28" mg L"^(-1)" to "0.04" mg L"^(-1), is

The half life of a reaction is inversely proportionaly to the square of the initial concentration of the reactant . Then the order of the reaction is :

The half-life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is

The half-life of a radionuclide is 69.3 minutes. What is its average life (in minutes)

The half-life of a radioisotope is four hour. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is

The half-life of a radioisotope is 20 years. If the sample has an initial activity of 640 dps,than what will be its activity after 80 years?

The half life of a radioisotope is 4 hr. If the initial mass of the isotope was 200 g, the mass remaining after 24 hr undecayed is:

The half-life of a radioactive nucleide 'X is 24 hours, the time required for 12.5% of the original X to remain is :

The half - lifeof a radioactive isotope is three hours. If the initial mass of the isotope was 300 g. the masswhichremainsundecayed in 18 hourswould be

The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256g, the mass of it remaining undecayed after 18 hours would be

The half-life of a radioactive isotope is 3 hours. Value of its disintegration constant is

The half life of a radioactive isotope is three hours. If the initial mass of the isotope was 300 g, the mass which remained undecayed in 18 hours would be :

The half life of a radioactive isotope is 2.5 hour . The mass of it that remains undecayed after 10 hour is (If the mass of the isotope was 16 g) :

The half-life of a radioactive isotope is 1.5 hours. The mass of it that remains undecayed after 6hours is (if the initial mass of the isotope was 32g)

The half-life of a radioactive element is 6 months. The time taken to reduce its original concentration to its 1.16 value is

The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in 1 g atom sample ?

The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in 1 g atom sample

The half-life of a radioactive element is 100 minutes . The time interval between the stages 50% and 87.5 % decay will be x xx 40 min . The value ofx is _____

The half life of a first order reaction x rarr products is 6.932 xx10^4 s at 500 K . What percentage of x would be decomposed on heating at 500 K for 100 min . (e^(0.06)=1.06)

The half -life of a first order reaction [k=(2.303)/(t)"log"((a)/(a-x))] is

The half- lifeof a firstorderreactionis 900min at820 k Estimateits half- lifeat 720 kif theenergyof activationof thereactionis 250kJ "mol"^(-1) (1.464 xx10^(5)min)

The half-life of a first order reaction [k=2.303/t log (a/(a-x))] is

The half life of a first order reaction is 69.35 sec . The value of the rate constant of the reaction is

The half- life of afirstorderreactionis 1.7hours. Howlongwill it takefor 20%of thereactanttodisappear ?

The half-life of a first order reaction is 30 min and the initial concentration of the reactant is 0.1 M. If the initial concentration of reactant is doubled, then the half-life of the reaction will be

The half life of a first order reaction is

Thehalf life of a firstorderreactionis 30 minand theinitialconcentrationof thereactant is 0.1M . Ifthe initialconcentrationofreactantisdoubled,thenthehalflife ofthe reactionwill be

The half life of a first order reaction is 100 seconds. What is the time required for 90% completion of the reaction?

The half life of a first order reaction is 100 seconds at 280 K. If the temperature coefficient is 3.0 its rate constant at 290 K in s^(-1) is

The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mole/litre and concentration at some instant ‘t’ is 0.01 mol/litre, then the value of ‘t’ is :

The half-life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/L and concentration at some instant is 0.01 mol/L, then t is

The half-life of a first order reactionhaving rate constant K = 1.7 xx 10^(-5) s^(-1) is

The half life of a first order reaction AtoB+C is 10 minutes. The concentration of 'A' would be reduced to 10% of the original concentration in

The half life of a 1^(st) order reaction is 1 min 40 seconds. Calculate its rate constant (k)

The half-life of ._(92)U^(238) " is " 4.5 xx 10^(9) years. After how many years, the amount of ._(92)U^(238) will be reduced to half of its present amount

The half life of ._(92)^(238)U is 4.5 xx 10^(9) years. Uranium emits an alphaparticle to give thorium. Calculate the time required to get the product which contains equal masses of thorium and uranium.

The half-life of ._(6)^(14)C is 5730 year. What fraction of its original C^(14) would left after 22920 year of storage?