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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 29501. |
The initial concentration of N_(2)O_(5) in the following first order reaction N_(2)O_(5(g))to2NO_(2(g))+(1)/(2)O_(2(g)) was 1.24xx10^(-2) Mol L^(-1) at 318 K.The concentration of N_(2)O_(5) after 60 minutes was 0.20xx10^(-2) mol L^(-1). calculate the rate constant of the reaction at 318 K. |
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Answer» Solution :Given reaction is first ORDER in it k=(?) INITIAL t=0 time ,the concentration of `N_(2)O_(5)` is `1.24xx10^(-2) mol L^(-1)` So.`t_(1)=0` and `[R]_(1)=1.24xx10^(-4)` mol `L^(-1)` After 60 MINUTES the concentration of `N_(2)O_(5)=0.2xx10^(-2) mol L^(1)` So, `t_(2)=60` minutes and `[R]_(2)=0.2xx10^(-2)` mol `L^(-1)` For a first order reaction , `log([R]_(1))/([R]_(2))=(k(t_(2)-t_(2)))/(2.303)` `therefore log (1.24xx10^(-2))/(0.2xx10^(-1))=k((60-0)"MINUTE")/(2.303)` `therefore log 6.2=k(60 "minute")/(2.303)` `therefore 0.7924=k(60)/(2.303)` `therefore k=(0.7924xx2.303)/(60 "minute") =0.0304 "minute"^(-1)` |
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| 29502. |
The IUPAC name ofbeta -methyl valerladehyde |
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Answer» 2-METHYL pentanal |
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| 29503. |
The initial concentration of N_(2)O_(5) in the following first order reaction N_(2)O_(5)(g) to 2NO_(2)(g) + (1)/(2) O_(2)(g) was 1.24 xx 10^(-2) mol L^(-1) at 318K. The concentration of N_(2)O_(5) after 60 mintues was 0.2 xx10^(-2) mol L^(-1). Calculate the rate constant of the reaction. |
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Answer» Solution :DATA : `[R]_(0) = 1.24 xx 10^(-2) molL^(-1)` `[R] = 0.2 xx 10^(-2) mol L^(-1)` t = 60 MIN Formula `K = (2.303)/(t) log ""([R]_(0))/([R])` ` k = (2.303)/(60) log (1.24 xx 10^(-2))/(0.2 xx 10^(2))` `k = (2.303)/(60) xx log 6.2,` `k = 0.03 min^(-1)` |
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| 29504. |
The IUPAC name of 'B' is |
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Answer» 3- methylbutan - 2 - OL 
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| 29505. |
The initial concentration of the first order reaction,N_2O_(5(g))to2NO_(2(g))+1/2O_(2(g))was 1.24xx10^(-2)molL^(-1).The concentration of N2O5 after ‘1’ hour was 0.20 x 10-2 mol L-1 Calculate the rate constant of the reaction at 300 K. |
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Answer» SOLUTION :For a first ORDER REACTION,(n=1)K=2.303/tlog` ([R_0])/([R])` `=2.303/60`log`(1.24xx10^(-2))/(0.2xx10^(-2))` |
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| 29506. |
The initial concentration of N_2O_5 in the first order reaction N_2O_5 to 2NO_2 + 1//2O_2 was 1.24 xx10^(-2) "mol L"^(-1) at 318 K. The concentration of N_2O_5 after 60 minutes was 0.20 xx 10^(-2) "mol L"^(-1). Calculate the rate constant of the reaction at 318 K. |
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Answer» SOLUTION :`K=2.303/t "LOG"([A]_0)/([A])=2.303/t "log"[N_2O_5]_0/[N_2O_5]_t=2.303/60 "log" (1.24xx10^(-2))/(0.2xx10^(-2))` `=2.303/60"log"6.2 =2.303/60xx0.7924 "min"^(-1)` `=0.0304 "min"^(-1)` |
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| 29507. |
The IUPAC name of an unsymmetrical ether with the molecular formula, C_(4)H_(10)O is |
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Answer» ETHOXYPROPANE |
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| 29508. |
The initial concentration of N_(2)O_(5) in the first order reactionN_(2)O_(5)to2NO_(2)+1/2O_(2) was 1.24xx10^(-2) mole /lit at 318 k. The concentration of N_(2)O_(5) after 60 min is 0.20xx10^(-2) mole/lt. Calculate the rate constant of reaction at 318 K |
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Answer» `0.0304"MIN"^(-1)` `K=2.303/50xxlog(6.2)=2.303//60xx0.7924=0.0304` |
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| 29509. |
The IUPAC name of alpha-methyl valerladehyde is |
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Answer» 2-methyl pentanal |
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| 29510. |
The IUPAC name of Acrolein (or) CH_2 = CH- CHO is…………………… . |
| Answer» SOLUTION :prop-2-enal | |
| 29511. |
The IUPAC name of Acrolein is ............... |
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Answer» Prop - 2 - ENAL |
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| 29512. |
The IUPAC name of Acetone is : |
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Answer» Propanal |
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| 29513. |
The IUPAC name of Acetaldol is.......... |
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Answer» 3 – HYDROXY BUTANAL |
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| 29514. |
Theinitial concentrationof ethyl acetaneis0.85 mol L^(-1) . Followingthe acid catalysed hydrolysis the , concentrations of ester after 30 min and 60 minof the reactionare respectively0.8 and 0.754 mol L^(-1) .Calculatethe rateconstantand pseudo rateconstant. |
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Answer» Solution :Acid catalysed ester follows PSEUDO first ORDER kinetics. The rate constant k is given as `k=(2.303)/(t)log.(a)/(a-x)` `k=(2.303)/(30)log.(0.85)/(0.05)=2.020xx10^(-3)"min"^(-1)("or")` `k=(2.303)/(30)log.(0,80)/(0.046)=1.997xx10^(-3)"min"^(-1)` The rate constant (k) is the product of pseudo first order rate constant (k.). Concentration of water, TAKEN as constant as `55.5` "MOL"L^(-1)` `k=k.[H_(2)O]` Substituting the values, `1.997xx10^(-3)=k.[55.5]` Pseudo rate constant=k. `3.6xx10^(-5)L "mol"^(-1)"min"^(-1)` |
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| 29515. |
The IUPAC name of |
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Answer» 1-chloro-1-oxe-2,3-dimenthylpentane |
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| 29516. |
The IUPAC name of |
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Answer» but-3-enoicacid 
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| 29517. |
The IUPAC name of |
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Answer» 1-chloro-1-oxo-2,3dimenthyl PENTANE |
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| 29518. |
The initial concentration of cane sugar is presence of an acid was reduced from 0.20 to 0.10 M in 5 hours and to 0.05 M in 10 hours, value of K ? (in hr^(-1)) |
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Answer» `0.693` |
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| 29519. |
The IUPAC name for,(CH_(3))_(2)NC(CH_(3))_(3) |
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Answer» N,N-dimethyl -2 -methylpropen -1-AMINE |
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| 29520. |
The initial concentration of A in the reactionA rarrB is 0.01 (M) . If the initial concentration of A is changed to 0.02(M) , then - |
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Answer» the values of both REACTION -rate and rate CONSTANT will INCREASE |
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| 29521. |
The IUPAC name for the coordination compound Ba[BF_(4)]_(2) is : |
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Answer» BARIUM tetraflouroborate (III) |
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| 29522. |
The IUPAC name for the compound : [Co(en)_(2)Cl(ONO)]Cl is : |
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Answer» Chloridonitrito-N-bis (ETHYLENEDIAMINE) COBALT (III) CHLORIDE. |
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| 29523. |
The initial cell potential of Daniel cell is __________ |
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Answer» 0.5 V |
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| 29524. |
The IUPAC name for the compound CH_(3)CH=underset(""C-=CH)underset(|"")(CHCHCH_(3)) is : |
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Answer» 4-Ethyl PENT -2-ene |
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| 29525. |
The IUPAC name for the compound : CH_(3)-underset(O)underset(||)(C)-CH_(2)-CH_(2)OH is |
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Answer» 3-Keto-1-butanol |
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| 29526. |
The inhibitors : |
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Answer» RETARD the rate of a CHEMICAL REACTION |
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| 29527. |
The infrated frequency of the CX vibration for CH_(3)X depends on which of the following? (P) Mass of X (Q) Strength of the CX bond (R) Type of CX vibration (stretch or bend) |
| Answer» Answer :D | |
| 29528. |
The IUPAC name for the compound : CH_(3)-underset(C_(2)H_(5))underset(|"")(CH)-CH_(2)-underset(OH" ")underset(|"")(CH)-CH_(3) is |
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Answer» 4-Ethylpentan-2-ol |
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| 29529. |
The influence of temperature on the rate of reaction is given by : |
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Answer» GIBBS HELMHOLTZ EQUATION |
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| 29530. |
The IUPAC name for the complex [Co(NO_(2))(NH_(3))_(5)]Cl_(2) is : |
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Answer» Nitro-N-pentaamminecobalt(III) chloride |
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| 29531. |
The IUPAC name for tertiary butyl iodide is |
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Answer» 1- IODO -3- methylpropane |
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| 29532. |
The inertness of nitrogen is due to its |
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Answer» HIGH ELECTRONEGATIVITY.. |
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| 29533. |
The IUPAC namefor diglyme i.e. CH_(3)OCH_(2)CH_(2)OCH_(2)CH_(2)OCH_(3) is/are: |
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Answer» 2,5,8-trioxanonane |
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| 29534. |
The inert gas used as substituent for nitrogen in oxygen used by deep sea divers for breathing is: |
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Answer» Xe |
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| 29535. |
The IUPAC name for [PtCl(NH_(2)CH_(3))(NH_(3))_(2)]Cl is: |
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Answer» diamminechloro(methylamine)PLATINUM(II)chloride |
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| 29536. |
The inert gases present in the atmosphere are: |
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Answer» He and NE |
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| 29537. |
The IUPAC name for, CH_3CHOHCH_2-C(CH_3)_2-OH is: |
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Answer» 1,1-dimethyl butane-1,2-diol |
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| 29538. |
The IUPAC name for : CH_(3)CH=CHCH_(2)underset(NH_(2)"")underset(|"")(CHCH_(2))COOH is |
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Answer» 5-Amino-2-heptenoic acid |
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| 29539. |
The IUPAC name for, (CH_3)_2CH-CH_2-CH_2-Cl is: |
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Answer» 1-chloropentane |
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| 29540. |
The inert gas obtained from monazite sand is: |
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Answer» He |
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| 29541. |
The IUPAC name for [Be_(4)O(CH_(3)COO)_(6)] is |
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Answer» Basic beryllium acetate (II) |
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| 29542. |
The inert gas obtained from monazite send is : |
| Answer» Solution :He' was believed to be the decay product of RADIO elements like Thorim, Uranium and radium. Minerals like PITCH blende and `""_(88)Ra^226 to ""_(86)Ra^222 + ""_2He^(4)` monozite sand | |
| 29543. |
The IUPAC name for |
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Answer» 1-Chloro-2-nitro-4-methylbenzene |
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| 29544. |
The IUPAC naem(s) of the following compound is (are): |
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Answer» 4-methylchlorobenzene |
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| 29545. |
The inert form of carbon is : |
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Answer» Diamond |
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| 29546. |
The IUPAC definition of a transition element is that it is an element that has an incomplete d-subshell in either the neutral atom or its ion. Thus the group 12 elements are member of the d-block but are not transition elements. Chemically solft members of the d-block occurs as sulphide minerals and are partially oxidised to obtain the metal, the more electropositive 'hard' metals occurs as oxides and are extracted by reduction. Opposite to p-block elements, the higher oxidation states are favoured by the heavier elements of d-block Metals on the right of the d-block tend to exist in low oxidation states and form complexes with the ligands. Square-planar complexes are common for the platinum metals and gold in oxidation states that yield d^8 electronic configuration, which include RH(I),Ir(I),Pd(II),Pt(II) and Au(III). The most distinctive features/properties of transition metal complex is their wide range of colours.The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.It is important to note that (a) in absence of ligand, crystal field spilling does not occur and hence the substances is colourless, (b) the type of ligand also influences the colour of the complexes. Select the correct statement. |
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Answer» `MnO_4^(-) lt Cr_2O_7^(2-) lt VO_2^(+)` respresents the CORRECT increasing order of their oxidising power (B)According to crystal field theory, the colour of the complex is due to d-d transition of the electrons . In the absence of ligand crystal field splitting does not occur and hence `TiCl_3` becomes colourless. (C )Copper pyrites is `CuFeS_2` It has iron SULPHIDE as impurity which is removed as slag. `FeSiO_3` is blast furnace during smelting . `2CuFeS_2 + 2O_2 to Cu_2S+2FeS+SO_2` `2FeS+3O_2 to 3FeO+2SO_2` `FeO+SiO_2 to FeSiO_3` (slag) (D)In both complexes, for the square planar shape (because of `5d^8` configuration ) the hybridisation is `dsp^2`.Hence the unpaired electron in 5d-orbital pair up to make one d-orbital empty for `dsp^2` hybridisation.Thus there is no unpaired electron in `[PtCl_4]^(2-)` and `[Pt(CN)_4]^(2-)` complexes. |
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| 29547. |
The IUPAC definition of a transition element is that it is an element that has an incomplete d-subshell in either the neutral atom or its ion. Thus the group 12 elements are member of the d-block but are not transition elements. Chemically solft members of the d-block occurs as sulphide minerals and are partially oxidised to obtain the metal, the more electropositive 'hard' metals occurs as oxides and are extracted by reduction. Opposite to p-block elements, the higher oxidation states are favoured by the heavier elements of d-block Metals on the right of the d-block tend to exist in low oxidation states and form complexes with the ligands. Square-planar complexes are common for the platinum metals and gold in oxidation states that yield d^8 electronic configuration, which include RH(I),Ir(I),Pd(II),Pt(II) and Au(III). The most distinctive features/properties of transition metal complex is their wide range of colours.The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.It is important to note that (a) in absence of ligand, crystal field spilling does not occur and hence the substances is colourless, (b) the type of ligand also influences the colour of the complexes. Which of the following has dsp^2 hybridisation and is diamagnetic in nature ? (i)Na_4[Cr(CO)_4] , (ii)[Ni(DMGH)_2] , (iii)[PtHBr(PEt_3)_2] (iv)[As(SCN)_4]^(3-) , (v)[AuBr_4]^(-) |
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Answer» I,II,IV only  CO is strong field ligand so complex is diamagnetic and tetrahedral . (II)`._28Ni(+II)-3d^8 , [Ni(DMGH)_2]`  DMGH is a chelating ligand so complex is diamagnetic and square planar. (III)`._78Pt(+II)-5d^8, [PtHBr(PEt_3)_2]`  `5d^8` configuration has greater CFSE, so complex is diamagnetic and square planar. (IV)`._47Ag(+I)-4d^10 , [Ag(SCN)_4]^(3-)`  No empty d-orbital is available for `dsp^2` hybridisation and `SCN^-` is weak field ligand so complex is tetrahedral and diamagnetic . (V)`._79Au(+III)-5d^8 , [AuBr_4]^-`  `5d^8` configuration has greater CFSE, so complex is diamagnetic and square planar. |
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| 29548. |
The inert electrode is ________. |
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Answer» Cu |
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| 29549. |
The IUPAC definition of a transition element is that it is an element that has an incomplete d-subshell in either the neutral atom or its ion. Thus the group 12 elements are member of the d-block but are not transition elements. Chemically solft members of the d-block occurs as sulphide minerals and are partially oxidised to obtain the metal, the more electropositive 'hard' metals occurs as oxides and are extracted by reduction. Opposite to p-block elements, the higher oxidation states are favoured by the heavier elements of d-block Metals on the right of the d-block tend to exist in low oxidation states and form complexes with the ligands. Square-planar complexes are common for the platinum metals and gold in oxidation states that yield d^8 electronic configuration, which include RH(I),Ir(I),Pd(II),Pt(II) and Au(III). The most distinctive features/properties of transition metal complex is their wide range of colours.The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.It is important to note that (a) in absence of ligand, crystal field spilling does not occur and hence the substances is colourless, (b) the type of ligand also influences the colour of the complexes. Which of the following statements is incorrect ? |
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Answer» In d-block elements , the d-orbitals of the penultimate energy level of their atoms RECEIVE ELECTRONS and thus give rise to three complete and one incomplete rows of the transition metals. (B)`Cu_2S` is chalcocyte or copper glance and `Ag_2S` is argentite. (C )Siderite is `FeCO_3` i.e. carbonate ore. `3Fe_2O_3+COto2Fe_3O_4+CO_2` 500 - 800 K `Fe_3O_4+COto3FeO+CO_2` 500 - 800 K `FeO+COtoFe+CO_2` 900 - 1500 K `FeO+CtoFe+CO` above1500 K (D)As the involvement of greater number of (n-1)d electrons in addition to ns electrons RESULTS into strong interatomic metallic bonding. |
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| 29550. |
The IUPAC na me of CH_(3)-Chunderset(CH_(2)CH_(2)CH_(3)^(-))underset(|)(C)- CH_(2)CH_(3) is: |
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Answer» 3-propylhex-2-ene |
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