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25601.

The presence of carbon in an organic compound can be shown by :

Answer»

Heating with COPPER which goes black
Burning it to PRODUCE which goes black
Heating it with copper oxide to convert it into `CO_2`
None of the above

Answer :C
Discussion
25602.

The presence of a base is needed in the ammonolysis of alkyl halides. Explain.

Answer»

Solution :During ammonolysis of alkyl halides, the acid liberated during the REACTION combines with the AMINE formed to FORM the amine salt. To liberate FREE amine from the amine salt, a base is NEEDED. The base used is excess of alcoholic `NH_(3)` which serves both as a nucleophile as well as a base. for example.
.
Discussion
25603.

The presence of a catalyst

Answer»

INCREASES the HEAT of reaction
decreases the heat of reaction
has no effect on the heat of reaction
has no effect on the ENERGY of activation

Solution :Presence of CATALYST has no effect on the heat of reaction.
Discussion
25604.

The prescibed upper limit concentration of lead in drinking water is about :

Answer»

30 ppb
70 ppb
50 ppb
90 ppb

Answer :C
Discussion
25605.

The preparation of salicylaldehyde phenol is known as ....

Answer»

DOW's REACTION
REIMER- Tiemann reaction
Kolbe's reaction
Cumene PROCESS

SOLUTION :Reimer- Tiemann reaction
Discussion
25606.

The preparation of ethyl acetoacetate involves:

Answer»

Witting REACTION
Claisen condensation
Reformatsky reaction
Cannizzaro.s reaction

Answer :B
Discussion
25607.

The prefixes erythro and threo are used for distinguishing a pair of enantriomers containing two chiral carbon atoms when two of the atoms or groups on each chiral carbon atom are the two identical groups are on the same side of the Fischer projection formula, is called the erythro isomer while the isomer, in which the two identical groups are on the opposite sides of the fischer projection formula, is called the Threo isomer. Which among the following is a threo isomer?

Answer»




ANSWER :B
Discussion
25608.

The prefixes erythro and threo are used for distinguishing a pair of enantriomers containing two chiral carbon atoms when two of the atoms or groups on each chiral carbon atom are the two identical groups are on the same side of the Fischer projection formula, is called the erythro isomer while the isomer, in which the two identical groups are on the opposite sides of the fischer projection formula, is called the Threo isomer. Erythro isomer among the following is:

Answer»




ANSWER :D
Discussion
25609.

The preparation of alkyl fluoride from alkyl choloride, in presence of metallic fluorides is known as,

Answer»

WILLIAMSON's REACTION
Finkelstein reaction
SWARTS reaction
Wurtz reaction

ANSWER :C
Discussion
25610.

The prefixes erythro and threo are used for distinguishing a pair of enantriomers containing two chiral carbon atoms when two of the atoms or groups on each chiral carbon atom are the two identical groups are on the same side of the Fischer projection formula, is called the erythro isomer while the isomer, in which the two identical groups are on the opposite sides of the fischer projection formula, is called the Threo isomer. Select thecorrect representation among the following:

Answer»




ANSWER :B::C::D
Discussion
25611.

The preparation of ethyl acetoacetate involves

Answer»

Witting REACTION
Claisen CONDENSATION
CANNIZARO's reaction
Reformatsky reaction

Answer :B
Discussion
25612.

The prefix 10^(18) is:

Answer»

giga
exa
kilo
nano

Answer :B
Discussion
25613.

The prefered conformation of trans-1, 2- dibromocyclohexane is:

Answer»

diaxial
diequatorial
axial/equatorial
neither A, B nor C

Answer :B
Discussion
25614.

The predominat form of histamine present in human blood is (pK_(a) , Histimine = 6.0)

Answer»




SOLUTION :At PH 7.4 MAJOR form of histamine is protonated PRIMARY amine.
Discussion
25615.

The predominant product formed, when 3-methyl-2-pentene reacts with HOCl, is

Answer»

`CH_3-CH_2-undersetunderset(CH_3)|oversetoverset(Cl)|C-oversetoverset(OH)|CH-CH_3`
`CH_3-CH_2-undersetunderset(CH_3)|oversetoverset(Cl)|C-oversetoverset(Cl)|CH-CH_3`
`CH_3-CH_2-undersetunderset(CH_3)|overset(OH)overset|C-undersetunderset(Cl)|CH-CH_3`
`CH_3-undersetunderset(CH_3)|oversetoverset(CH_3)|C-oversetoverset(OH)|CH-CH_3`

SOLUTION :`CH_3-CH_2-undersetunderset(CH_3)(|)C=CH-CH_3+Hoverset-Ooverset+Cl to CH_3-CH_2-underset(CH_3)underset|overset(OH)overset(|)C-overset(Cl)overset(|)CH-CH_3`
Addition takes place ACCORDING to Markownikoff's rule in which `Cl^+` goes to that carbon ATOM which is more HYDROGENATED .
Discussion
25616.

The predominant form of histamine present in human blood is (pk_(a) Histidine = 6.0)-

Answer»




ANSWER :C
Discussion
25617.

The predominant form of histamine present in human blood is (pK_(a) Histidine = 6.0)

Answer»




Solution :The pH of BLOOD is approx 7.0, THEREFORE the ACIDS with `pK_(a)` with less than 7 will LOOSE `H^(+)`.
Discussion
25618.

The predominant form of histamine present in human blood is (pK_(a), Histidine = 6.0) -

Answer»




SOLUTION :
Discussion
25619.

The predominant form of histamine present in human blood is (pK_(a), "Histidine" = 6.0)

Answer»




ANSWER :D
Discussion
25620.

The precipitation power of ions are in the order Al^(3+) gt Ba^(3+) gtNa^(+). Similarly [Fe(CN)_6]^(4-) gt SO_4^(2-) gt Cl^(-). Give the reason behind this .

Answer»

Solution :(i) A negative ion causes the precipitation of positively changed sol and vice versa.
(ii) When the VALENCY of ion is high, the precipitation POWER increased. So in cations the precipitation power ORDER is `Al^(3+) GT Ba^(2+) gt Na^(+)`and in anions the precipitation power order is `[Fe(CN)_6]^(4-) gt SO_4^(2-) gt Cl^(-)`.
(iii) The precipitation power of electrolyte is determined by FINDING the minimum concentration required to cause precipitation of sol in 2hrs. This value is called flocculation value. The smaller the flocculation value, greater will be precipitation.
Discussion
25621.

The precipitation of protein is called ______.

Answer»

PEPTISATION
DENATURATION
RENATURATION
NONE of these

SOLUTION :denaturation
Discussion
25622.

The precipitation is noticed when an aqueous solution of :

Answer»

`NaNO_2`
`BA(NO_3)_2`
`ZNSO_4`
`HgNO_3`

ANSWER :D
Discussion
25623.

The precipitateofsecond group sulphides qualitatative analysis is carried oud withhydrogen in the presence of hydrochloricacidbutnot with nitric acid .Explain.

Answer»

SOLUTION :The precipitate of second GROUP sulphidesin qualitative anlysis is CARRIED out with hydrogensulphide in THEPRESENCE of hydrochloride acid but not in mitricacid.
Nitric acid is a strongoxidisingagent and oxides `H_(2)S` into S, So it cannot be used in thereactioninvolvinghydrogensulphide
Discussion
25624.

The precipitate of second group sulphide in qualitative analysis is carried out with hydrogen sulphide in the presence of hydrochloric acid but not in nitric acid. Explain.

Answer»
Discussion
25625.

The precipitate of CaF_(2)(K_(sp) = 1.7 xx 10^(-10)) is obtained when equal volumes of the following are mixed

Answer»

`10^(-4) M Ca^(2+) + 10^(-4) M F^(-)`
`10^(-2) M Ca^(2+) + 10^(-3) M F^(-)`
`10^(-5) M Ca^(2+) + 10^(-3) M F^(-)`
`10^(-3) M Ca^(2+) + 10^(-5) M F^(-)`

Solution :CALCULATE the ionic product `[Ca^(2+)] [ F^(-)]^(2)` in each CASE and see that ionic product MUST exceed `K_(sp)` for precipitation. Concentration of each ion is halved due to dilution effect (Volume DOUBLED).
Discussion
25626.

The precipitate of calcium fluoride(K_(sp)=1.7xx10^(-10)) is obtained when equal volumes of the following are mixed :

Answer»

`10^(-4)M Ca^(2+)`and`10^(-4)M F^(-)`
`10^(-2)M Ca^(2+)`and`10^(-3)M F^(-)`
`10^(-5)M Ca^(2+)`and`10^(-3)M F^(-)`
`10^(-3)M Ca^(2+)`and`10^(-5)M F^(-)`

ANSWER :B
Discussion
25627.

The precipitate of CaF_2 (K_s = 1.7 xx 10^-10) is obtained when equal volmes of the following are mixed:

Answer»

`10^(-4)` M `CA^(2+) + 10^(-4)M F^-`
`10^(-2)` M `Ca^(2+) + 10^(-3)M F^-`
`10^(-5)` M `Ca^(2+) + 10^(-3)M F^-`
`10^(-3)` M `Ca^(2+) + 10^(-5)M F^-`

Answer :B
Discussion
25628.

The precipitate of AgCl is soluble in :

Answer»

`NH_4OH`
`HNO_3`
`H_2O`
HCl

Answer :A
Discussion
25629.

The precipitate obtained by adding lead nitrate solution to a sodium carbonate solution is

Answer»

`PB(OH)_(2)`
`PbCO_(3)`
`PbCO_(3).Pb(OH)_(2)`
`PbO_(2)`

Solution :Hydroxy carbonate is precipitated. To get PURE `PbCO_(3),NaHCO_(3)` should be USED.
Discussion
25630.

The precious stone aquamarine is :

Answer»

Mg-Al SILICATE
Be-Al silicate
Na-Al silicate
Fluoro silicate of Al

Answer :C
Discussion
25631.

The precious Ruby stone is :

Answer»

ALUMINIUM silicate
Alumina
Sodium aluminium silicate
Sodium silicate

Answer :B
Discussion
25632.

The prdouct/s formed when diborane is hydrolysed is/are

Answer»

`B_(2)O_(3)` and `H_(3)BO_3`
`B_(2)O_(3)` only
`H_(3)BO_(3)` and `H_(2)`
`H_(3)BO_(3)` only

Solution :When diborane is hydrolysed one can get both ORTHOBORIC acid and `H_(2)`.
`B_(2)H_(6) +6H_(2)O to 2H_(3)BO_(3) +6H_(2)O`
Discussion
25633.

The P.P. of ethane over a saturated solution containing 6.56 xx 10^(-2) g of ethane is 1 bar. If the solution contains 5 xx 10^(-2) g of ethanol, then what shall be its P.P? (P.P = Partial pressure).

Answer»


SOLUTION :`S propto P`
`S_(1)/S_(2) =P_(1)/P_(2)`
`P_(2) =("1 bar" xx 5 xx 10^(-2))/(5.56 xx 10^(-2)) = 0.762`
Discussion
25634.

The p.pt. of alcohols are _______ than corresponding thiols :

Answer»

More
Less
Same
Either of these

Answer :A
Discussion
25635.

The power output for a certain laswer transition was found to be 2.79 watt per square meter. Givne lambda = 520nm, calculate the number of quanta emitted per square meter per second.

Answer»

Solution :Energy corresponding to the photon `= ( 6.626 xx 10^(-34) xx 3 xx 10^(8))/( 520 xx 10^(-9))`
`= 3.822 xx 10^(-19) J `
the power output `= 2.79 ` Watt `//m^(2)`
`= 2.79 J // `SEC `//m^(2)`
`:.`no. of photons EMITTED per sq. METER sec`= ( 2.79 )/( 3.822 xx 10^(-18)) = 7.20 xx 10^(18)`
Discussion
25636.

The potential of the cell for the reaction M(s)+2H^(+)(1M)toH_(2)(g)(1atm)+M^(2+)(0.1M) is 1.500 V. the standard reduction potential for M^(2+)//M(s) couple is

Answer»

`0.1470V`
`1.470V`
`1.47V`
none of these

Solution :The cell is `M|M^(2+)(0.1M)||H^(+)(1M)|H_(2)(1atm)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"[M^(2+)]`
`therefore1.50=E_(cell)^(@)-(0.0591)/(2)xxlog(0.1)`
`=E_(cell)^(@)+0.02955`
or `E_(cell)^(@)=1.50-0.02955=1.47045V`
But `E_(cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(M^(2+)//M)^(@)`
i.e., `1.47=0-E_(M^(2+)//M)^(@)`
or `E_(MG^(2+)//M)^(@)=-1.47M`
Discussion
25637.

The powdered pyrolusite ore when fused with KOH in presence of air forms"…...................".

Answer»


ANSWER :`K_(2)MnO_(4)`\
Discussion
25638.

The potential of the cell for the reaction M(s)+2H^(+)(1M)rarrH_(2)(g)(1"atm")+M^(2+)(0.1M)is 1.500V the standard reduction potential for M^(2+)//M couple is

Answer»

0.1470V
1.470V
`-1.47V`
NONE of these

Solution :The cell is
`M|M^(2+)(0.1M)||H^(+)(M)|H^(+)(M)|H _(2)(1"atm")`
`E_("cell")=E_("cell")^(@)-(0.0591)/(2)"log"[M^(2+)]`
`1.5=E_("cell")^(@)-(0.0591)/(2)"log"(0.1)`
`1.5=E_("cell")^(@)+0.02955`
or `E_("cell")^(@)=E_(2H^(+)//H_(2))^(@)-E_(M^(2+)//M)^(@)`
`therefore 1.47=0-E_(M^(2+)//M)^(@)`
or `E_(M^(2+)//M)^(@)=-1.47V`.
Discussion
25639.

The potential of the cell for the reaction M_((s))+2H^(+)(1M)toH_(2(g)),(1atm)+M^(2+)(0.1M) is 1.500 V. the standard reduction potential for M^(2+)//M_((s)) couple is:-

Answer»

0.1470V
1.470V
14.70V
None of these

Solution :`E_(cell)=E_(cell)^(o)=-(0.0591)/(2)LOG((["products"])/(["REACTANTS"]))`
`1.5=E_(cell)^(@)-0.0296log[(0.1)/(1)]`
`E_(Cell)^(o)=1.470V`
`E_(Cell)^(o)=E_(H^(+)//H_(2))^(o)-E_(M//M^(2+))^(o)`
`E_(M//M^(2+))^(o)=-1.470`
So, `E_(M^(2+)//M)^(o)=1.470V`.
Discussion
25640.

The potential of the cellCu, Cu^(2+) (0.1M)"//"HCl (xM), Cl_(2), Ptis 1.07 V. If the standard potential of copper and chlorine electrodes are 10.34 Vand 1.36 V,calculate the concentration of HCI.

Answer»


ANSWER :0.01 M
Discussion
25641.

The potential of standard hydrogen electrode is zero. This implies that

Answer»

`DeltaG_(f)^(o)(H^(+),aq)=0`
`DeltaH_(f)^(o)(H^(+),aq)=0`
`DeltaG_(f)^(o)(H^(+),aq)LT0`
`DeltaG_(f)^(o)(H^(+),aq)gt0`

Solution :As `DeltaG_(f)^(o)=-nFE^(o)`
if `E^(o)=0`
then `DeltaG_(f)^(o)=0`.
Discussion
25642.

The potential of the cell containing two hydrogen electrodes as represented belowPt, H_2(g) |H^(+) (10^(-6) M)|| H^(+) (10^(-4) M)|H_2(g) , Pt at 298 K is

Answer»

`-0.118V`
`-0.0591V`
`0.118V`
`0.0591 V`

ANSWER :C
Discussion
25643.

The potential of single electrode depends upon

Answer»

the name of the ELECTRODE
temperature
concentration of the iron with RESPECT to which it is REVERSIBLE
all the above
Discussion
25644.

The potential of hydrogen electrode is -118 mV . The H^(+) concentration of the solution is

Answer»

`0.01 M `
` 2M`
`10^(-4) M`
1 M

ANSWER :A
Discussion
25645.

The potential of hydrogen electrode at pH=10 and 25^(@)Cis

Answer»

0.59V
zero VOLTS
`-0.59V`
`-0.059V`

ANSWER :C
Discussion
25646.

The potential of calomel electrode depends upon the concentration of

Answer»

KCL solution
`HgCl_(2)` solution
`Hg_(2)Cl_(2)`
HG only

Solution :PRINCIPLE of calomel electrode.
Discussion
25647.

The potential of an electrode when each species involved in it exists in the standard state is called its standard potential. The standard reduction potential of a couple is the measure of its tendency to get reduced. A series obtained by arranging the various couples in order of their decreasing standard potensital is called electrochemical series. Any of the two couples of this series joined together gives an electrochemical cell in which reduction occurs at the electrode which occupiesthe higher position. The standard potential of hydrogen electrode is taken to be zero by convention. Given below are the sequence of half-reactions (acidic media) with relevant E^(@) values in volt at 298 K. MnO_(4)^(-)overset(0.50)rarr MnO_(4)^(2-)overset(2.26)rarr MnO_(2)overset(0.95)rarr Mn^(3+)overset(1.50)rarr Mn^(2+)overset(-1.18)rarr Mn Also given some more data are : Fe^(2+)overset(-0.44)rarr Fe_((s))overset(+0.036)rarr Fe^(3+) For the cell Fe_((s))|Fe^(2+)(0.1M)||100 mL 0.3M HA (pKa=5.2) mixed with 50 mL 0.4M NaOH|H_(2)(1 atm), Pt the E_("cell") at 298(2.303(RT)/(F) and "log"_(10)20=1.3) is equal to

Answer»

`-0.15 V`
`+0.185 V`
`-0.44 V`
`+0.145 V`

ANSWER :D
Discussion
25648.

The potential of an electrode when each species involved in it exists in the standard state is called its standard potential. The standard reduction potential of a couple is the measure of its tendency to get reduced. A series obtained by arranging the various couples in order of their decreasing standard potensital is called electrochemical series. Any of the two couples of this series joined together gives an electrochemical cell in which reduction occurs at the electrode which occupiesthe higher position. The standard potential of hydrogen electrode is taken to be zero by convention. Given below are the sequence of half-reactions (acidic media) with relevant E^(@) values in volt at 298 K. MnO_(4)^(-)overset(0.50)rarr MnO_(4)^(2-)overset(2.26)rarr MnO_(2)overset(0.95)rarr Mn^(3+)overset(1.50)rarr Mn^(2+)overset(-1.18)rarr Mn Also given some more data are : Fe^(2+)overset(-0.44)rarr Fe_((s))overset(+0.036)rarr Fe^(3+) The value of E^(@) for the reaction is equal to MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarr Mn^(2+)+5Fe^(3+)4H_(2)O,

Answer»

`0.734 V`
`0.282 V`
`4.498 V`
Can't be CALCULATED as CONCENTRATIONS of species are not given.

Answer :A
Discussion
25649.

The potential of an electrode when each species involved in it exists in the standard state is called its standard potential. The standard reduction potential of a couple is the measure of its tendency to get reduced. A series obtained by arranging the various couples in order of their decreasing standard potensital is called electrochemical series. Any of the two couples of this series joined together gives an electrochemical cell in which reduction occurs at the electrode which occupiesthe higher position. The standard potential of hydrogen electrode is taken to be zero by convention. Given below are the sequence of half-reactions (acidic media) with relevant E^(@) values in volt at 298 K. MnO_(4)^(-)overset(0.50)rarr MnO_(4)^(2-)overset(2.26)rarr MnO_(2)overset(0.95)rarr Mn^(3+)overset(1.50)rarr Mn^(2+)overset(-1.18)rarr Mn Also given some more data are : Fe^(2+)overset(-0.44)rarr Fe_((s))overset(+0.036)rarr Fe^(3+) The E^(@) for MnO_(4)^(-)rarr Mn^(2+) is equal to

Answer»

`1.054 V`
`1.506 V`
`5.27 V`
`7.53 V`

ANSWER :B
Discussion
25650.

The potential of a hydrogen electrode at pH=10 is :

Answer»

`+0.59" V "`
`0.00" V"`
`-0.59" V "`
`-0.059" V "`

Solution :(C ) `H^(+)+e^(-) to 1//2 H_(2)`
`E=E^(@)-(0.059)/(N)"log"(1)/([H^(+)])=-(0.059)/(1)PH`
`=(0-0.059)/(1)xx10=0-0.59=-0.59" V"`
Discussion