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29351.

The ion that cannot be precipitated by both HCl and H_(2)S is

Answer»

`PB^(2+)`
`CU^(+)`
`Ag^(+)`
`Sn^(2+)`

Solution :`Cu^(+)` DISPROPORTIONATES in the aqueous solution to a very large EXTENT.
`2Cu^(+)hArrCu^(2+)+Cu`
`K=([Cu^(2+)])/([Cu^(+)]^(2))=1.6xx10^(6)`
THUS, the concentration of `Cu^(+)` ion is very small and hence it cannot be precipitated.
Discussion
29352.

The ion that cannot be precipitated by both HCl and H_2 S is

Answer»

`Pb^(2+)`
`Cu^(2+)`
`Ag^(+)`
`Sn^(2+)`

SOLUTION :All except `Cu^(2+)` are precipitated both with HCl with `H_(2)S`. (`Cu^(2+)` is not precipitated with HCl, it is precipitated only with `H_(2)S`)
Discussion
29353.

The ion that cannot be precipitate by both HCI and H_(2)S is

Answer»

`Pb^(2+)`
`Cu^(o+)`
`AG^(o+)`
`SN^(2+)`

Solution :`Cu^(o+)` disproportio nates in the aqueous solution to a very large extent
`2Cu^(o+) hArr Cu^(2+) +Cu`
`K = ([Cu^(2+)])/([Cu^(o+)]^(2)) = 1.6 xx 10^(6)`
THUS , the concentration of `Cu^(o+)` ion is verysmall , and HENCE it cannotbeprecipitate
Discussion
29354.

The ion that can be precipitated by HCl as well as H_(2)S is

Answer»

`PB^(2+)`
`FE^(3+)`
`Zn^(2+)`
`Cu^(2+)`

Solution :`Pb^(2+)` as it precipitated as CHLORIDE and sulphide in `I^(st)` and `II^(nd)` group respectively.
Discussion
29355.

Theionthatbeprecipitate by both HCI and H_(2)S is

Answer»

`Pb^(2+)`
`Cu^(o+)`
`Ag^(o+)`
`Sn^(2+)`

SOLUTION :`Sn^(2+)` cannot be precipitate by HCI but is precipitate by `H_(2)S` as a chocolate -coloured precipitate of `SNS`, SOLUBLEIN yellow ammonium SULPHIDE.
Discussion
29356.

The ion most difficult to removes as precipitate is

Answer»

`Ag^(+)`
`NH_(4)^(+)`
`Fe^(+2)`
`CO^(+2)`

Answer :B
Discussion
29357.

The ion not precipitated by H_(2)S in presence of HCl is :

Answer»

`Cu^(+)`
`Ag^(+)`
`PB^(2+)`
`AL^(3+)`

Answer :D
Discussion
29358.

The ion having maximum value of hydration energy is :

Answer»

`LI^+`
`NA^+`
`K^+`
`CS^+`

ANSWER :A
Discussion
29359.

The iodo lactisation of CO_(2)H with I_(2) gives the compound

Answer»




SOLUTION :
Discussion
29360.

The (i)_____of a syste can be changed by transfer of heat from the surrounding to the system of cive-verse without expenditure of (ii)_____. Identify (i) and (ii) in order to complete the above statement.

Answer»

(i)-ENTHALPY, (ii)-WORK
(i)-internal energy, (ii) heat
(i)-enthalpy, (ii)- heat
(i)-internal energy, (ii)-work

Solution :The internal energy of a system can be changed by transfer of heat from the SURROUNDINGS to the system or vice-versa WIHTOUT expenditure of work.
Discussion
29361.

An iodine molecule dissociates into atom after absorbinglight of wavelength 4500Å. If quantum of radiation isabsorbedby eachmolecule calculate thekineticenergyof iodine (Bond energy of I_(2) is 240 kJ mol^(-1))

Answer»

Solution :Bond ENERGY PER molecule of `I_(2)= (240 xx 1000)/(6.022 xx 10^(23))J= 3.984 xx 10^(-19)J`
Energy absorbed `=(hc)/(lamda)`
`=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(4500 xx 10^(10))= 4.417 xx 10^(-19)J`
`THEREFORE` KE of ONE `I_(2)` molecule `=4.417 xx 10^(-19)- 3.984 xx 10^(-19)J = 4.33 xx 10^(-20)J`
KE of one I atom `=(4.33 xx 10^(-20))/(2)= 2.165 xx 10^(-20)J`
Discussion
29362.

The 'iodine value' of oil indicates

Answer»

Its boiling point
Inflammability
Unsaturation present in acid contains
Solubility of salt in oils

Solution :Iodine number is the number of GMS of `I_(2)` which combine with 100 GM of oil or fat. It SHOWS the degree of unsaturation of acdis in fat or oil.
Discussion
29363.

The iodide is quantitatively converted to chloride when it is heated in a stream of chlorine. XI_(3) + 3Cl_(2) rarr XCl_(3) + I_(2) It 1 gm of XI_(3) is converted into (1)/(3) gm of XCl_(3) then atomic mass of X is :

Answer»

`30.75 gm`
`20.5 gm`
`61.5 gm`
`10.25 gm`

Solution :`{:(2XI_(3),+,3Cl_(2),rarr,2XCl_(3),+,3I_(2),),(1 gm,,,,(1)/(3) gm,,,):}`
Let atomic mass of X be M
`n_(XI_(3)) = n_(XCl_(3))`
`(1)/(M + 381) = ((1)/(3))/(M + 106.5)`
`M + 106.5 = (M)/(3) + 127`
`(2M)/(3) = 20.5`
`M = 30.75 gm`
Discussion
29364.

The iodinated derivative of amino acid tyrosine is known as ........................

Answer»

EPINEPHRINE
THYROXINE
Testosterone
Norepinephrine

Solution :Thyroxine
Discussion
29365.

Compounds with antiseptic properties are

Answer»

`CH_(3)I`
`CHI_(3)`
`CH_(2)I_(2)`
KI

Answer :B
Discussion
29366.

The inversion temperature (T_i) for a gas is given by :

Answer»

a /Rb
2a/Rb
Rb/a
`(2Rb)/a`

ANSWER :B
Discussion
29367.

The inversion temperature for a van der Waals' gas is:

Answer»

`T_i=2a/(RB)`
`T_i=a/(Rb)`
`T_i=a/2(Rb)`
`T_i` =0.5T Boyle

Answer :A
Discussion
29368.

The inversion of sucrose (C_(12)H_(22)O_(11)) is a first order reaction and is studied by measuring angle of rotation at different time intervals. C_(12)H_(22)O_(11) + H_(2)O overset(H^(+)) to underset("glucose")(C_(6)H_(12)O_(6)) + underset("fructose")(C_(6)H_(12)O_(6)). if (r_(infty)-r_(0)) and (r_(infty)-r_(t))(a-x), then 50% invertion will be completed when:

Answer»

`r_(0) = 2r_(t)-r_(infty)`
`r_(0) = r_(t) - r_(infty)`
`r_(0) = r_(t) - 2r_(infty)`
`r_(0) = r_(t) + r_(infty)`

SOLUTION :`k=(2.303)/t log (r_(infty)-r_(0))/((r_(infty)-r_(t))`…………..(i)
`(2.303/t) log (a/(a//2))` …………..(ii)
`=(2.303)/t LOG2` ………….(iii)
Discussion
29369.

The inversion of cane sugar proceeds with half-life of 600 minutes at Ph = 5 for any concentration of sugar. However if Ph = 6 , the half-life changes to 60 minute. The rate law expression for sugar inversion can be written as

Answer»

`r= k . [ "SUGAR"]^(2) [ H^(+) ] ^(0)`
`r= k . [ "sugar"]^(1) [ H^(+) ] ^(0)`
`r= k . [ "sugar"]^(1) [ H^(+) ] ^(1)`
`r= k . [ "sugar"]^(0) [ H^(+) ] ^(1)`

Answer :C
Discussion
29370.

The inversion of cane sugar was studied in 1 N HCl at 298 K. The following polarimetric readings were obtained at different intervals of time : {:("Time (minutes)",,,:,,,0.0,,,7.18,,,18.00,,,27.05,,,oo),("Reading (degree)",,,:,,,+24.09,,,+21.41,,,+17.74,,,+15.00,,,-10.74):} Show that the inversion of cane sugar is a unimolecular reaction.

Answer»

SOLUTION :In the present case, `r_(0)=+24.09,r_(oo)=-10.74:.r_(0)-r_(oo)=+24.09-(-10.74)=34.83`
The value of k at different instants may be calculated as under :
`{:("t (MIN)",,," "r_(t),,,""r_(t)-r_(oo),,,""k=(2.303)/(t)log""(r_(0)-r_(oo))/(r_(t)-r_(oo))),(7.18,,,+21.41,,,21.41-(-10.74)=32.15,,,k=(2.303)/(7.18"min")log""(34.83)/(32.15)=0.01113" min"^(-1)),(18.00,,,+17.74,,,17.74-(-10.74)=28.48,,,k=(2.303)/(18.00"min")log""(34.83)/(28.48)=0.01118" min"^(-1)),(27.05,,,+15.00,,,15.00-(-10.74)=25.74,,,k=(2.303)/(27.05"min")log""(34.83)/(25.74)=0.01118" min"^(-1)):}`
The constancy in the value of k proves that the REACTION is of the first order.
Discussion
29371.

The inversion of cane sugar proceeds with half- life of 500 minutes at pH=6 for any concentration of sugar. However, if pH=5, the half-life changes to 50 minutes. The rate law expression for the sugar inversion can be written as

Answer»

`r=k["sugar"]^(2)[H]^(6)`
`r=k["sugar"]^(1)[H^(+)]^(0)`
`r=k["sugar"]^(1)[H^(+)]^(1)`
`r=k["sugar"]^(0)[H^(+)]^(1)`

SOLUTION :`((t_(1//2))_(1))/((t_(1//2))_(2))=([H^(+)]_(2))/([H^(+)]_(1))`
`(500)/(50)={(10^(-5))/(10^(-6))}^(x)`
`1=x`
Discussion
29372.

The inversion of cane sugar proceeds with half- life of 500 minutes at pH=5 for any concentration of sugar. However, if pH=6, the half-life changes to 50 minutes. Derive the rate law for inversion of cane-sugar.

Answer»

Solution :At `pH=5`, the half-life is `500` for all CONCENTRATIONS of sugar, i.e. `t_(1//2) PROP ["sugar"]^(0)`. Thus, the REACTION is I order with respect to sugar,
Now rate `=K["sugar"]^(1)[H^(+)]^(m)`
Also, for `[H^(+)]t_(1//2) prop [H^(+)]^(1-m)`
`500 prop [10^(-5)]^(1-m)`
`500 prop [10^(-6)]^(1-m)`
`10=(10)^(1-m)`
`1-m=1`
`m=0`
Threfore, rate `-K["sugar"]^(1)[H^(+)]^(0)`
Discussion
29373.

The inversion of cane sugar proceeds with half life of 500 minute at pH=5 for any concentration of sugar. However if pH=6 the half life changes to 50 minutes. The rate law expression for sugar inversioncan be written as

Answer»

`r=K["sugar"]^(2)[H^(+)]^(0)`
`r=k["sugar"]^(1)[H^(+)]^(0)`
`r=k["sugar"]^(1)[H^(+)]^(1)`
`r=K["sugar"]^(0)[H^(+)]^(1)`

SOLUTION :`5""500,((t_(1//2))_(1))/((t_(1//2))_(2))=[([H^(+)]_(2))/([H^(+)]_(1))]^(n-1),(10)^(1)=(10^(-1))^(n-1),(10)^(1)=(10)^((-n+1)),n=0`
`6""50`
Discussion
29374.

The inversion of cane sugar proceeds with half life of 50 minutes pH = 5 for any concentration of sugar. However if pH = 6, the half life changed to 500 minutes. The law expression of sugar inversion can be written as:

Answer»

`r=K["SUGAR"]^(2)[H^(+)]^(0)`
`r=K["sugar"]^(1)[H^(+)]^(0)`
`r=K["sugar"]^(1)[H^(+)]^(1)`
`r=K["sugar"]^(0)[H^(+)]^(0)`

Answer :C
Discussion
29375.

The inversion of cane sugar proceeds with half life of 500 minute at pH = 5 for any concentration of sugar. However, if pH=6, the half life changes to 50 minute. The rate law expression for the sugar inversion can be written as

Answer»

`r=k(SUGAR)^2(H^+)^0`
`r=k(sugar)^1(H^+)^0`
`r=k(sugar)^1(H^+)^1`
`r=k(sugar)^0(H^+)^1`

ANSWER :B
Discussion
29376.

The inversion of cane sugar is represented by C_(12) H_(22) O_(11) + H_(2)O to C_(6) H_(12)O_(6) + C_(6)H_(12) O_(6) It is a reaction of

Answer»

SECOND order
Unimolecular
Pseudo unimolecular
None of the three

SOLUTION :INVERSION of cane SUGAR is a Pseudo unimolecular reaction.
Discussion
29377.

The inversion of cane sugar is first order in [sugar] and proceeds with half-life of 600 min at pH=4 for a given concentration of sugar. However , if pH=5, the half-life changes to 60 min. The rate law expression for the sugar inversion can be written as

Answer»

`rate=K[sugar]^(1)[H^+]^(2)`
`rate=k[sugar][H^(+)]^(1)`
`rate=k[sugar][H^(+)]^(4)`
`rate=k[sugar][H^(+)]^(0)`

Solution :`"Rate"=k[A]^(a)[B]^(b)`
Overall order of reaction = a+b
Inversion of cane sugar is FIRST order reaction (GIVEN).
HENCE,`"rate"=k[sugar]^(a)[H^(+)]^(b)anda+b=1`
In the given options, only (d) has value of a+b=1
Discussion
29378.

The inversion of cane sugar is represented by: C_12 H_22O_11 + H_2O to C_6H_12O_6 + C_6H_12O_6 It is a reaction of :

Answer»

SECOND order
Unimolecular
PSEUDO unimolecular
NONE of the three

Answer :C
Discussion
29379.

The inversion of cane sugar into glucose and fructose is:

Answer»

`I` ORDER
`II `order
`III` order
Zero order

Answer :A
Discussion
29380.

The introduction of neutron into the nuclear composition of an atom would lead to a change in

Answer»

Its atomic WEIGHT
the CHEMICAL nature of the atom
Its atomic number
The number of the electron

Answer :A
Discussion
29381.

The inversion of cane sugar into glucose and fructose is reaction of

Answer»

I order
III order
II order
Zero order

Answer :A
Discussion
29382.

The inversion of cane sugar into glucose and fructose is

Answer»

1ST ORDER
2ND order
3rd order
ZERO order

Answer :A
Discussion
29383.

the intial state A has the temperature T_(A)U_(A)as the internal energy of the system . By appling the mechanical work . New state B is achieved with the temperatutre T_(B) and having the interanal energy U_(B). Given that t_(B) gt T_(A). What is the correct expression for the change in internal energy (DeltaV) ?

Answer»

`U_(B)= U_(A)`
`U_(B) -U_(A)`
`U_(A) - U_(B)`
none of these

Solution :the change in the INTERNAL ENERGY is `DeltaU=U_(B)-U_(A)`
Discussion
29384.

The intial concentration of "N"_(2)"O"_(5) in the first order reaction, "N"_(2)"O"_(5)(g)to2"NO"_(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.

Answer»

Solution :`k=(2.303)/(t)log""([A]_(0))/([A])=(2.303)/(t)log(["N"_(2)"O"_(5)]_(0))/(["N"_(2)"O"_(5)]_(t))=(2.303)/(60" min")log""(1.24xx10^(-2)"mol L"^(-1))/(0.2xx10^(-2)"mol L"^(-1))`
`=(2.303)/(60)log6.2min^(-1)=(2.303)/(60)xx0.7924min^(-1)=0.0304min^(-1)`
Discussion
29385.

The introduction of a neutron into the nuclear composition of an atom would lead to a change in:

Answer»

the NUMBER of electrons
the CHEMICAL NATURE of the atom
its ATOMIC number
its atomic weight.

Answer :D
Discussion
29386.

The intestinal antiseptic Salol is correctly represented as:

Answer»




ANSWER :B
Discussion
29387.

The interplanar distance in a crystal used for X- ray diffraction is 2A^(0). The angle of incidence for first order diffraction is 9^(@), what is the wave length of X-rays ?

Answer»

SOLUTION :`0.63 A^(@)`
Discussion
29388.

The interparticle forces present in Nylon-66 are:

Answer»

Vauder WALL's FORCES
Hydrogen bonding
DIPOLE dipole INTERACTIONS
None of these

Answer :D
Discussion
29389.

The internuclear distances in 0-0 bonds for O_(2)^(+),O_(2),O_(2)^(-) and O_(2)^(2-) respectively are :

Answer»

`1.30Å,1.49Å,1.12Å,1.21Å`
`1.49Å,1.21Å,1.12Å,1.30Å`
`1.21Å,1.12Å,1.49Å,1.30Å`
`1.12Å,1.21Å,1.30Å,1.49Å`

Solution :The BOND length FOLLOWS the order:
`O_(2)^(+)O_(2) O_(2)^(-) LT O_(2)^(2-)`
According to this the possible values are `1.12Å,1.21Å,1.30Å,1.49Å`
Discussion
29390.

The interparticle forces between linear chains in Nylon-6,6are:

Answer»

H-bonds
Covalent bond
Ionic-bonds
Coordinate bonds

Answer :A
Discussion
29391.

The internuclear distance in H_2 and Cl_2 molecules are 74 and 198 pm respectively.The bond length of H-Cl may be :

Answer»

272 pm
70 pm
136 pm
248 pm

Answer :C
Discussion
29392.

The interparticle forces in solid hydrogen are :

Answer»

VAN DER Waals' forces
Covalent bonds
HYDROGEN bonds
Coordinate bonds

Solution :The interparticle forces in SOLID hydrogen are van der Waals forces.
Discussion
29393.

The internal resistance to flow in liquids which one layer offers to the other layer trying to pass over it is called :

Answer»

Fluidity
Specific resistance
Viscosity
SURFACE TENSION.

ANSWER :D
Discussion
29394.

The internal energy (U) of an ideal gas decreased by the same amount as the work done by the system.

Answer»

the process must be adiabatic
the process must be isothermal
the process must be isobaric
the temperature must DECREASE.

ANSWER :A::D
Discussion
29395.

The internal energy of one mole of a gas is:

Answer»

`3//2RT`
`KT//2`
`RT//2`
`3KT//2`

ANSWER :A
Discussion
29396.

The internal energy changes in the conversion of 1.0 mole of the calcite from of CaCO_(3) to the aragonite from is +0.21 kJ . Calculate the enthalpy change when the pressure is 0.1 bar, given that the densities of the solids are 2.71 g cm^(-3) and 2.93 g cm^(-3) respectively.

Answer»

<P>

Solution :`DELTAH = DeltaE + PDeltaV`
Given `DeltaE = +0.21 kJ mol^(-1) = 0.21 XX 10^(3) J mol^(-1)`
P = 1 bar = `1.0 xx 10^(5)` Pa
`DeltaV = V_("argonite") - V_("calcite")` (mol. Wt. of `CaCO_(3)=100`)
`=(100/2.93)- (100/2.71) CM^(3) mol^(-1)` of `CaCO_(3)`
`=-2.77 cm^(3) = -2.77 xx 10^(-6) m^(3)`
`DeltaH = 0.21 xx 10^(3) - 1 xx 10^(5) xx 2.77 xx 10^(-6)`
`=2.0972 J mol^(-1)`
`=0.20972 kJ mol^(-1)`
Discussion
29397.

The internal energy change when a system goes from the state A to B is 40kJ/mol. If the system goes from A to B by a reversible path and returns to the state A by an irreversible path, what would be the net change in internal energy?

Answer»

`gt40kJ`
`lt40kJ`
Zero
40 kJ

Answer :C
Discussion
29398.

The internal energy of a substance

Answer»

INCREASES with increase in TEMPERATURE
Decreases with increase in temperature
Can be CALCULATED by the relation `E=mc^(2)`
REMAINS unaffected with change in temperature

Solution :`DeltaE` increases with temperature. '
Discussion
29399.

The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversiblepath and return to state A by an irreversible path what would be the net change in internal energy

Answer»

40 KJ
`gt 0` kJ
`lt 40` kJ
Zero

Solution :`AOVERSET(40)RARRB`
`Aoverset(-40)rarrB`
`DeltaH=40-40=0`.
Discussion
29400.

The internal energy change when a system goes from state A to B is 40 kJ/mol. I the system goes from A to B by a reversible path and return to state A by a irreversible path, what would be the net change in internal energy?

Answer»

40 KJ
`gt 40kJ`
`lt 40kJ`
zero

Solution :In a CYCLIC PROCESS, `DeltaE=0`.
Discussion