94420 + Interview Questions in Chemistry in Class 12 Page 1751 InterviewSolution

87501.	A second order reaction in which both the reactants have same concentration , is 20% completed in 500 seconds. How much it will take for 60% completion ?
Answer» Solution :The second order equation when both the REACTANTS have same CONCENTRATION is `k=(1)/(t)[(1)/((a-x))-(1)/(a)]` If `a=100`, `x=20`, `t=500sec` `k=(1)/(500)xx(20)/(100xx(100-20))` When `a=100`, `x=60`, `t=?` `t=(1)/(k)[(1)/((100-60))-(1)/(100)]=(1)/(k)*(60)/(100xx40)` Substituting the value of `k` `t=(500xx100xx80)/(20)xx(60)/(100xx40)` `=3000` seconds.

Discussion

87502.	A second order reaction in which both the reactants have same concentration , is 20% completed in 500 seconds . How much time it will take for 60% completion ?
Answer» SOLUTION :The second order EQUATION when both the REACTANTS have same CONCENTRATION is `k = 1/t .x/(a(a-x))` If a = 100, x20 , t = 500 seconds Then ` k =1/500xx(20)/(100xx(100-20))` When `a = 100,x = 60, t = ? ` `t = 1/ k . (60)/(100xx40)` Substituting the value of k , `t = (500xx100xx80)/(20)xx60/(100xx40)"or "t = 3000 ` seconds

Discussion

87503.	A sealed tube which can withstand a pressure of 3 atm sphere is filled with air at 27^(@)C and 760 mm pressure. The temperature above which the tube will burst will be:
Answer» `900^(@)C` `627^(@)C` `727^(@)C` `1173^(@)C` ANSWER :B

Discussion

87504.	A second order reaction can be altered to a first order reaction by taking one of the reactant inlarge excess, it is called... ......
Answer» SOLUTION :PSEUDO FIRST ORDER REACTION

Discussion

87505.	A second order reaction between A and B is elementary reaction: A+B to Product rate law expression of this reaction will be:
Answer» Rate = K[A][B] Rate = `K[A]^0[B]^2 ` Rate = `K[A]^2[B]^0 ` Rate = `K[A]^(3//2)[B]^(1//2)` ANSWER :A

Discussion

87506.	(A) Second ionization enthalpies of Cr and Cn are very low (R) Both Cr and Cu are stable in their +2 oxidation state
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 :Both (A) and (R) are false

Discussion

87507.	(A) Seais the greatest source of some halogens (R ) Halogens are highly reactive
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 :B

Discussion

87508.	Assertion : Sea water boils at higher temperature than distilled water Reason :Addition of non volatile solute to a solvent lowers the vapour pressure
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 (A) is false but (R) is true ANSWER :A

Discussion

87509.	A schematic plot of In k_(eq) versus inverse of temperature for a reaction is shown below : The reaction must be :
Answer» EXOTHERMIC endothermic one with negligible ENTHALPY change highly spontaneous at ordinary temperature . high spontaneous at ordinary temperature . Solution :(A) `In ""(k_(2))/(k_(1))= (DeltaH)/( R ) [(1)/(T_(1))-(1)/(T_(2))]` In `(6)/(2) = (DeltaH)/(R ) [1.5xx10^(-3) - 2 xx10^(-3) ]` `DELTA` H of the reaction comes out to be - ve Therefore the reaction is exothermic .

Discussion

87510.	(A) Schottky and Frenkel defects are also called thermodynamic defects (R) Both Schottky and Frenkel defects increases with increase in temperature.
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 (A) is false but (R ) is true Answer :A

Discussion

87511.	(A) Schottky type defect is shown by crystals with high coordination number. (R) In crystals with Schottky defect cations always occupy interstitial positions
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 :C

Discussion

87512.	A scarlet compound A is treated with conc. HNO_(3) to give a chocolate brown precipitate B. The precipitate is filtered and the filtrate is neutralised with NaOH. Addition of KI to the resulting solution gives a yellow ppt C. The brown ppt B on warming with conc. HNO_(3) in the presence of Mn(NO_(3))_(2) produces a pink coloured solution due to the formation of D. Identify A, B, C and D. Write the reaction sequence.
Answer»

Discussion

87513.	A scarlet compound (A) is treated with conc. HNO_(3) to give a chocolate brown precipitate (B) and a colourless solution of C. On adding Kl to this solution, a yellow ppt. (D) is obtained. The precipitate (B) on warming with conc. HNO_(3) in the prese
Answer» `PbCrO_(4)` `Pbl_(2)` `MnO_(2)` `HMnO_(4)` ANSWER :D

Discussion

87514.	A scarlet red precipitate is obtained on treating mercuric chloride solution with:
Answer» `H_2S` KI NaOH `NH_4OH` ANSWER :B

Discussion

87515.	A scarlet compound (A) is treated with conc. HNO_(3) to give a chocolate brown precipitate (B) and a colourless solution of C. On adding Kl to this solution, a yellow ppt. (D) is obtained. The precipitate (B) on warming with conc. HNO_(3) in the presence of Mn(NO_(3))_(2) produces a pink coloured solution due to the formation of (E). What is (E)?
Answer» `PbCrO_(4)` `Pbl_(2)` `MnO_(2)` `HMnO_(4)` Answer :D

Discussion

87516.	A scarlet compound (A) is treated with conc. HNO_3 to give a chocolate brown precipitate (B) and a colourless solution of C, On adding Kl to this solution, a yellow ppt. (D) is obtained. The precipitate (B) on warming with conc. HNO_3 in the presence of Mn(NO_(3))_2 prdouces a pink coloured solution due to the formation of (E). What is (E)?
Answer» `PbCrO_4` `Pbl_2` `MnO_2` `HMnO_4` ANSWER :D

Discussion

87517.	A scaletcompound A is treated with concenbtrated HNO_(3)to gave achocolate brownprecipitate B. The precipitate is filtered and the filtrateisneurralised with NaOH Addition of KI to theresultingsolutiongives a yellowprecipitate C the brownprecipitate B on warmingwith concentrated HNO_(3) in the presence of Mn(NO_(3))_(2) produces a pinkcolouredsolutiondue to theformation of D identify A, B,C, and D write the reaction sequence.
Answer» Solution :A : `Pb_(3)O_(4) B : PbO_(2) C : PbI_(2)O_(4) B : PB(MnO_(4))_(2)` Thepink colouredcompound is PRECIPITATE saltwhich is FORMED by the oxidationof maganese saltlead dioxide .Theyellow precipitate to be red lead . `underset((A))(Pb_(3)O_(4)+)4HNO_(3)rarrunderset((B))(PbO_(2)darr)+underset("Filtrate")(2Pb(NO_(3))_(2))+2H_(2)O` `Pb (NO_(3))_(2) +2KI rarr underset("Yellow"(C))(PbI_(2)darr)+2KNO_(3)` `5PbO_(2) +2MN(NO_(3))_(2) +4HNO_(3) rarr` `Pb(MnO_(4))_(2)+underset((D))(4Pb(NO_(3))_(2))+2H_(2)O`

Discussion

87518.	(A) Scandium is the lightest transition metal, (R) Atomic radius of scandium is largest among all the transition elements.
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

87519.	(A) Sc exhibit variable oxidation states. (R) Sc is not a transitional element.
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 :Both (A) and (R) are false

Discussion

87520.	(A): Sc^(+3) ion in aqueous solutions is colorless (R) : Ions with do configuration are colorless
Answer» Both (A) and (R) are TRUE and (R) is thecorrect 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 (A) is false but (R) is true Solution :Both (A) and (R) are true and (R) is thecorrect explanation of (A)

Discussion

87521.	A saturated solution, prepared by dissolving CaF_2(s) in water, has [Ca^(2+)]=3.3 times 10^-4 M. What is the K_(sp) of CaF_2?
Answer» Solution :`CaF)_2(s)=Ca^(2+) (aq)+2F^(-) (aq)` `[F^-]=2[Ca^(2+)]=2 times3.3 times10^-4M` `=6.6 times10^-4M` `K_(sp)=[Ca^(2+)][F^-]^2` `=(3.3 times10^-4)(6.6 times10^-4)^2` `=1.44 TIMES 10^-10`

Discussion

87522.	A saturated solution of Mg(OH)_2 in water at 25^oC contains 0.11 g Mg(OH)_2 per litre of solution. Thesolubility product of Mg(OH)_2 is :
Answer» `(0.11)^2` `(0.11)^3` `4XX(0.11)^3` `4xx(0.11)^3/(58)^3` ANSWER :D

Discussion

87523.	A saturated solution of calcium fluoride contains 2xx10^(-4) mole of the salt per litre of the solution its K_sp is :
Answer» `8xx10^(-18)` `3.2xx10^(-11)` `4XX10^(-6)` `1.43xx10^(-9)` ANSWER :B

Discussion

87524.	A saturated solution of Ag_2SO_4 is 2.5 xx 10^-2 M. The value of its solubility product is:
Answer» `62.5xx10^(-6)` `6.25xx10^(-4)` `15.625xx10^(-6)` `3.125xx10^(-6)` ANSWER :A

Discussion

87525.	A saturated solution of Ag_(2)SO_(4) is 2.5 xx 10^(-2) M, The value of its solubility product is
Answer» `62.5 xx 10^(-6)` `6.25 xx 10^(-4)` `15.625 xx 10^(-6)` `3.125 xx 10^(-6)` Solution :`{:("For",AgSO_(4)rarr,2Ag^(+)+,SO_(4)^(-)),(,,2X,x):}` `K_(sp) = (2x)^(2).x , K_(sp) = 4X^(3) , K_(sp) = 4 xx (2.5 xx 10^(-2))^(3)` `K_(sp) = 62.5 xx 10^(-6)`.

Discussion

87526.	A saturated hydrocarbon 'X' having molecular formula (C_6H_12) will form only one monochloro product on reaction with Cl_2 in presence of ultravoilet light.Calculate the number of dichloroproducts (structural only ) of the same compound 'X'.
Answer» SOLUTION :

Discussion

87527.	A saturated hydrocarbon C_(6)H_(14) gives two monochloro compounds on chlorination. Indentify the hydrocarbon
Answer» SOLUTION :Since the HYDROCARBON gives TWO monochloro derivatives, it indicates that it contains two EQUIVALENT HYDROGENS. The hydrocarbon is 2, 3-dimethylbutane

Discussion

87528.	Protein found in hair and nail is:
Answer» palmiticacid LINOLENIC acid Oleic acid Linoleic acid Answer :A

Discussion

87529.	A saturated fatty acid found in oil and fats is:
Answer» palmiticacid LINOLENIC acid Oleic acid Linoleic acid Answer :A

Discussion

87530.	A saturated compound C_(2)H_(4)Cl_(2) permits the existence of,
Answer» OPTICAL ISOMERISM Cis-trans isomerism position isomerism FUNCTIONAL isomerism Answer :C

Discussion

87531.	A saturated calomel electrode is coupled through a salt bridge with a quinhydrone electrode dipping in 0.1 M NH_(4)Cl. The observed EMF at 25^(@)C is 0.152V. Find the dissociated constant of NH_(4)OH. The oxidation potential of saturated calomel electrode=-0.699V at 25^(@)C.
Answer» Solution :For the cell`:` Saturated CALOMEL `\|\|0.1 MNH_(4)Cl\|H_(2)Q,Q\|Pt` `E_(cell)=E_(qui nhydron e el etrode)-E_(saturated calomel)` `E_(cell)=0.152,,,,[{:(E_(saturated calomel(o x i d)),=,-0.242V),(E_(saturated calomel (red)),=,-0.242V):}]` `(E_(qui nhydron e)=0.6994-0.059pH)` `( :. E^(C-)._(o x i de)=-0.6994,so E^(c-)._(Red)=0.6994)` `0.152=0.6994-0.059pH-0.242` `:. PH=(0.457-0.152)/(0.059)=5.17` For `0.1 M NH_(4)Cl,` salt of `S_(A)//W_(B)` which on hydrolysis GIVES `NH_(4)OH` `pH=(1)/(2)(pK_(W)=pK_(b)-logc)` `5.17=(1)/(2)(14-pK_(b)-log0.1) Solving , we get `K_(b)=2.18xx10^(-5)`

Discussion

87532.	A sample_(92)^(238)U when kept at 298 K in a closed cassel shows alpha-decay and beta-decay to given _(82)^(206)Pb. After t hour the volume of the gas collected in the vessel at 1 atm, 273K is found to be 89.6 ml. Then which of the the following is/are correct for the nucler process and the products of the process ?
Answer» Mass of Pb OBTAINED in time t hour =`103`mg. If the temperatures is increased by `10^(@)C` then the time taken for the same amount of gas to be produced will be `(t)/(2)` hour. `alpha` PARTICALES show more IONISATION power rhan `BETA` particles. `lmbda` RADIATIONS undergoes no deviation while traveling through electric or magnetic field. Answer :A::C::D

Discussion

87533.	A sample was weighted using two different balances. The result's were (i) 3.929 g (ii) 4.0 g. how would the weight of the sample be reported. If it has to be reported in 3 significant no
Answer» 3.929 g 3 g 3.9 g 3.93 g Solution :Round off the DIGIT at `2^(nd)` POSITION of decimal `3.929=3.93`.

Discussion

87534.	A sample of water has a hardness expressed as 80 ppm of Ca^(2+).This sample is passed through an ion exchange column and the Ca^(2+) is replaced by H^+ What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]
Answer» 3 2.7 5.4 2.4 Answer :d

Discussion

87535.	A sample of water has a hardness expressed as 80 ppm of Ca^(2+) This sample is passed through an ion exchange column and the Ca^(2+) is replaced by H^(+).What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]
Answer» 3 2.7 5.4 2.4 Solution :`10^6` ml water CONTAINS 80 GM of `CA^(2+)=80/40` moles =2 moles of `Ca^(2+)=2xx2` moles of `H^+` ions so `10^3` ml of `H_2O` will have =`4xx10^(-3)` moles of `H^+` ions so pH=3-log 4=3-0.6=2.4

Discussion

87536.	A sample of water contains 12% w//w MgSO_(4) and 9.5% w//w MgCl_(2). If the sulphate dissociates to 8 extent and chloride to 60%extent then calculate the boliling point of the solution. Given : K_(b)[H_(2)O]=0.785 K-kg//"mole"^(-1) [Express your answer in Kelvin scale]
Answer» ANSWER :377

Discussion

87537.	A sample of water contains 0.012 g of MgSO_(4) per litre. The hardness of the water sample in ppm is
Answer» 20 PPM 10 ppm 25 ppm 15 ppm Solution :Equivalent mass of `MgSO_(4)=60` `rArr"60 g of "MgSO_(4)-="50 g of "CaCO_(3)` `therefore"0.012 g of "MhSO_(4)-=(50xx0.012)/(60)="0.01 g of "CaCP_(3)` `rArr"1 LITRE of water contains 0.01 parts by weight of "CaCO_(3)` `"i.e. 1000 parts of water contains 0.01 parts by weight of "CaCO_(3)` `rArr ""10^(6)` parts by weight of water contains 10 parts by weight of `CaCO_(3)`. `therefore` The hardness of the water sample is 10 ppm.

Discussion

87538.	A sample of water containing KCL does not behave as hard water, but a sample of water containing CaCl_(2) behaves as hard water-why?
Answer» Solution :The POTASSIUM salt of SOAP is soluble in water and forms lather while calcium or MAGNESIUM salt of soap is insoluble in water and does not FORM lather.

Discussion

87539.	A sample of uranium mineral was found to contain .^(206)Pb and ^(238)U in the ratio of 0.008 : 1 . Estimate the age of the mineral (half-life of .^(238)U is 4.51xx10^(9) years).
Answer» Solution :`t=(2.303)/(LAMBDA)log[1+(.^(206)Pb)/(.^(238)U)]` `t=(2.303)/(0.693)xxt_(1//2)xxlog[1+(.^(206)Pb)/(.^(238)U)]` `implies` Ratio by mass `.^(206)Pb : .^(238)U=0.008 : 1` Ratio by moles `.^(206)Pb : .^(238)U=(0.008)/(206) : (1)/(238)` `=0.0092` `:. t=(2.303xx4.51xx10^(9))/(0.693)log(1+0.0092)` `=(2.303xx4.51xx10^(9))/(0.693)xx0.00397` `=(0.0412)/(0.693)xx10^(9)=5.945xx10^(7)` years

Discussion

87540.	A sample of uranite a uranium confainig mineral was found on analysis to contain 0.214gm of .^(206Pb for evert gm of .^(238)U. If all the lead came from the radioactive disintegration of uranium and assume that all isotopes of uranium other than .^(238)U can be neglected, estimate the date when the mineral was formed in the earth crust. (The half-life of .^(238)U is 4.5xx10^(9) years).
Answer» SOLUTION :`1.5xx10^(9)` YEARS

Discussion

87541.	A sample of Uraninite, a Uranium containing mineral was found on analysis to contain 0.214 gm of Pb^(206) for every gram of Uranium. Assuming that the lead all resulted from the radioactive disintegration of the Uranium since the geological formation of the Uraninite and all isotopes of Uranium other than""^( 238)U can be neglected. Estimate the day when the mineral was formed in the Earth's crust.[t_(1//2) " of" ^(238)U = 4.5 xx 10^(9)" years " ]
Answer» ANSWER :`1.4 XX 10^(9)` YRS

Discussion

87542.	A sample of U - 238 (half-life = 4.5 xx 10^(9)yr) ore is found to contain 23.8 g of U-238 and 20.6 g of Pb - 206. What will be the age of theore?
Answer» `5.5 XX 10^(12)` years `6.0 xx 10^(8)` years `4.5 xx 10^(9)` years `5.5 xx 10^(14)` years Solution :Disintegration occurs as : `""_(92)U^(238) to ""_(82)Pb^(206) + 8 ""_(2)He^(4) + e ""_(-1)e^(0)` ATOMS of Pb present = `(20.6)/(206) = 0.1 g ` atom = U DECAYED Atoms of U present = `(23.8)/(238) = 0.1 g` atom `:. N = 0.1 g` atom `N_0` = U present + U decayed = `0.1 + 0.1 = 0.2 g` atom `t = (2.303 xx 4.5 xx 10^(9))/(0.693) "LOG"_(10)(0.2)/(0.1)` `:. t = 4.5 xx 10^(9) ` years.

Discussion

87543.	A sample of rock from moon contains equal number of atoms of uranium and lead (t_(1//2)" for "U=4.5xx10^(9) years). The age of the rock would be :
Answer» `4.5xx10^(9)` YEARS `9.0xx10^(9)` years `13.5xx10^(9)` years `2.25xx10^(9)` years Answer :A

Discussion

87544.	A sample of rock from moon contains equal number of atoms of uranium and lead (t_(1//1) " for " U = 4.5 xx 10^(9) " years"). The age of the rock would be
Answer» `9.0 xx 10^(9)` years `4.5 xx 10^(9)` years `13.5 xx 10^(9)` years `2.25 xx 10^(9)`years Solution :`N = (N_(0))/(2^(n)), " use " t = (2.303 xx t_(1//2))/(0.693) " LOG" (N_(0))/(N)`

Discussion

87545.	A sample of radon emitted initially 7 xx 10^(4) alpha-particles per second. After some time, the emission rate became 2.1 xx 10^(4). If t_((1)/(2)) for radon is 3.8 days, find the age of the sample
Answer» ANSWER :6.6 DAYS

Discussion

87546.	A sample of rock from moon contains equal number of atoms of uranium and lead (t_((1)/(2)) for U= 4.5 xx 10^(9) years). The age of the rock would be
Answer» `4.5 XX 10^(9)` YEARS `9 xx 10^(9)` years `13.5 xx 10^(9)` years `2.25 xx 10^(9)` years Answer :A

Discussion

87547.	A sample of radioactive substance shows an intensity of 2.3 millicurie at a time t and an intensity of 1.62 millicurie after 600 s. The half-life period of the radioactive metal is
Answer» Solution :Here `(N^(0))/(N)= (2.30)/(1.62)` We have, `lamda= (2.303)/(600) "LOG" (2.30)/(1.62)= 0.000584` Now, `t_((1)/(2))= (0.6932)/(lamda)= (0.6932)/(0.000584)=1187` SECONDS

Discussion

87548.	A sample of radioactive ""^(133)I gave with a Geiger counter 3150 counts per minute at a certain time and 3055 counts per minute exactly one hour later. Calculate the half-life period of ""^(133)I
Answer» SOLUTION :`N^(0)= 3150, N= 3055, t=1` hours We have, `lamda= (0.6932)/(t_(1//2))` Substituting these VALUES in `lamda= (2.303)/(t) "log" (N^(0))/(N)` `(0.6932)/(t_((1)/(2)))= (2.303)/(1) "log" (3150)/(3055)` `t_((1)/(2))= 22.63` hours.

Discussion

87549.	A sample of pyrolusite weighing 0.5 g is distilled with conc. HCI. The evolved CI_2 when passed through a solution of KI liberates sufficient I_2 to react with 125 ml of N//12.5 hypo solution (Na_2 S_2 O_3. 5 H_2 O). The percentage of MnO_2 in the sample of pyrolusite is:
Answer» 0.87 `43.5%` `72.8%` `22.1%` ANSWER :A

Discussion

87550.	A sample of pure Cu (4.00 g) heated in a stream of oxygen for some time, gains in weight with the formation of black oxide of copper (CuO) . The final mass is 4.90 g. What percent of copper remains unoxidized (Cu=64)
Answer» `90%` `10 %` `20 %` `80 %` ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in .

A second order reaction in which both the reactants have same concentration , is 20% completed in 500 seconds. How much it will take for 60% completion ?

A second order reaction in which both the reactants have same concentration , is 20% completed in 500 seconds . How much time it will take for 60% completion ?

A sealed tube which can withstand a pressure of 3 atm sphere is filled with air at 27^(@)C and 760 mm pressure. The temperature above which the tube will burst will be:

A second order reaction can be altered to a first order reaction by taking one of the reactant inlarge excess, it is called... ......

A second order reaction between A and B is elementary reaction: A+B to Product rate law expression of this reaction will be:

(A) Second ionization enthalpies of Cr and Cn are very low (R) Both Cr and Cu are stable in their +2 oxidation state

(A) Seais the greatest source of some halogens (R ) Halogens are highly reactive

Assertion : Sea water boils at higher temperature than distilled water Reason :Addition of non volatile solute to a solvent lowers the vapour pressure

A schematic plot of In k_(eq) versus inverse of temperature for a reaction is shown below : The reaction must be :

(A) Schottky and Frenkel defects are also called thermodynamic defects (R) Both Schottky and Frenkel defects increases with increase in temperature.

(A) Schottky type defect is shown by crystals with high coordination number. (R) In crystals with Schottky defect cations always occupy interstitial positions

A scarlet compound (A) is treated with conc. HNO_(3) to give a chocolate brown precipitate (B) and a colourless solution of C. On adding Kl to this solution, a yellow ppt. (D) is obtained. The precipitate (B) on warming with conc. HNO_(3) in the prese

A scarlet red precipitate is obtained on treating mercuric chloride solution with:

(A) Scandium is the lightest transition metal, (R) Atomic radius of scandium is largest among all the transition elements.

(A) Sc exhibit variable oxidation states. (R) Sc is not a transitional element.

(A): Sc^(+3) ion in aqueous solutions is colorless (R) : Ions with do configuration are colorless

A saturated solution, prepared by dissolving CaF_2(s) in water, has [Ca^(2+)]=3.3 times 10^-4 M. What is the K_(sp) of CaF_2?

A saturated solution of Mg(OH)_2 in water at 25^oC contains 0.11 g Mg(OH)_2 per litre of solution. Thesolubility product of Mg(OH)_2 is :

A saturated solution of calcium fluoride contains 2xx10^(-4) mole of the salt per litre of the solution its K_sp is :

A saturated solution of Ag_2SO_4 is 2.5 xx 10^-2 M. The value of its solubility product is:

A saturated solution of Ag_(2)SO_(4) is 2.5 xx 10^(-2) M, The value of its solubility product is

A saturated hydrocarbon 'X' having molecular formula (C_6H_12) will form only one monochloro product on reaction with Cl_2 in presence of ultravoilet light.Calculate the number of dichloroproducts (structural only ) of the same compound 'X'.

A saturated hydrocarbon C_(6)H_(14) gives two monochloro compounds on chlorination. Indentify the hydrocarbon

Protein found in hair and nail is:

A saturated fatty acid found in oil and fats is:

A saturated compound C_(2)H_(4)Cl_(2) permits the existence of,

A saturated calomel electrode is coupled through a salt bridge with a quinhydrone electrode dipping in 0.1 M NH_(4)Cl. The observed EMF at 25^(@)C is 0.152V. Find the dissociated constant of NH_(4)OH. The oxidation potential of saturated calomel electrode=-0.699V at 25^(@)C.

A sample was weighted using two different balances. The result's were (i) 3.929 g (ii) 4.0 g. how would the weight of the sample be reported. If it has to be reported in 3 significant no

A sample of water has a hardness expressed as 80 ppm of Ca^(2+).This sample is passed through an ion exchange column and the Ca^(2+) is replaced by H^+ What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]

A sample of water has a hardness expressed as 80 ppm of Ca^(2+) This sample is passed through an ion exchange column and the Ca^(2+) is replaced by H^(+).What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]

A sample of water contains 12% w//w MgSO_(4) and 9.5% w//w MgCl_(2). If the sulphate dissociates to 8 extent and chloride to 60%extent then calculate the boliling point of the solution. Given : K_(b)[H_(2)O]=0.785 K-kg//"mole"^(-1) [Express your answer in Kelvin scale]

A sample of water contains 0.012 g of MgSO_(4) per litre. The hardness of the water sample in ppm is

A sample of water containing KCL does not behave as hard water, but a sample of water containing CaCl_(2) behaves as hard water-why?

A sample of uranium mineral was found to contain .^(206)Pb and ^(238)U in the ratio of 0.008 : 1 . Estimate the age of the mineral (half-life of .^(238)U is 4.51xx10^(9) years).

A sample of U - 238 (half-life = 4.5 xx 10^(9)yr) ore is found to contain 23.8 g of U-238 and 20.6 g of Pb - 206. What will be the age of theore?

A sample of rock from moon contains equal number of atoms of uranium and lead (t_(1//2)" for "U=4.5xx10^(9) years). The age of the rock would be :

A sample of rock from moon contains equal number of atoms of uranium and lead (t_(1//1) " for " U = 4.5 xx 10^(9) " years"). The age of the rock would be

A sample of radon emitted initially 7 xx 10^(4) alpha-particles per second. After some time, the emission rate became 2.1 xx 10^(4). If t_((1)/(2)) for radon is 3.8 days, find the age of the sample

A sample of rock from moon contains equal number of atoms of uranium and lead (t_((1)/(2)) for U= 4.5 xx 10^(9) years). The age of the rock would be

A sample of radioactive substance shows an intensity of 2.3 millicurie at a time t and an intensity of 1.62 millicurie after 600 s. The half-life period of the radioactive metal is

A sample of radioactive ""^(133)I gave with a Geiger counter 3150 counts per minute at a certain time and 3055 counts per minute exactly one hour later. Calculate the half-life period of ""^(133)I

A sample of pyrolusite weighing 0.5 g is distilled with conc. HCI. The evolved CI_2 when passed through a solution of KI liberates sufficient I_2 to react with 125 ml of N//12.5 hypo solution (Na_2 S_2 O_3. 5 H_2 O). The percentage of MnO_2 in the sample of pyrolusite is:

A sample of pure Cu (4.00 g) heated in a stream of oxygen for some time, gains in weight with the formation of black oxide of copper (CuO) . The final mass is 4.90 g. What percent of copper remains unoxidized (Cu=64)