Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Barbiturates are the drugs which act on the____________.

Answer»

CENTRAL NERVOUS system
respiratory system
heart
liver

Answer :A
Discussion
2.

Barbiturates acts as

Answer»

HYPNOTIC i.e., SLEEP producing AGENTS
non-narcotic analgesics
ACTIVATOR of neutrotransmitters
ANTIALLERGIC drugs .

Answer :A
Discussion
3.

Ba(OH)_(2) solution , Which of the following reaction is incorrect for compound 'X', 'Y' and 'Z'.

Answer»

`Ba(OH)_(2)+Na_(2)SO_(3)RARR'X'`
`'X' +H_(2)SO_(3)("EXCESS")rarr'Y'`
`Z+"DIL". HCL rarrSO_(2)("gas")`
`Y+H_(2)O_(2)RARRZ`

Solution :By interaction of U.V radiations on `O_(2)` molecule `O_(3)` is formed in stratosphere.
Discussion
4.

Ba(OH)_(3)+NaOH hArr Na[B(OH)_(4)]: To keep the above reaction on forward direction, which reagent should be used?

Answer»

CIS -1, 2 - diol
trans -1, 2 - diol
Borax
`Na_(2)HPO_(4)`.

ANSWER :A
Discussion
5.

Ba(OH)2 is used to estimate the amt. of :

Answer»

`N_2`
`CO_2`
CO
`N_2O`

ANSWER :B
Discussion
6.

Band theory predicts that magnesiums is an insulator. However, in practice it acts as a conductor due to

Answer»

presence of filled 3S - orbital
overlap of filled 2p and filled 3s - orbital
overlap of filled 3s and EMPTY 3p - orbital
presence of UNFILLED 3p - orbital

Answer :C
Discussion
7.

Balz-Schiemann'sreactionis usedto convert

Answer»

Aromaticaldehydeto ALDOL
benzenetochlorobenzeme
chorobene diazoniumto phenol
benzenediazoniumchlorideto FLUOROBENZENE

ANSWER :D
Discussion
8.

Balance the following skeleton equation : Mg_(3)N_(2)+H_(2)OrarrMg(OH)_(2)+NH_(3).

Answer»

Solution :The biggest formula is `Mg_(3)N_(2)`. Hence, the various atoms are balanced in the order : Mg, N and H.
(i) to equalise the number of Mg atoms on both sides, MULTIPLY Mg `(OH)_(2)` by 3. We get
`Mg_(3)N_(2)+H_(2)Orarr 3Mg (OH)_(2)+NH_(3)`
(ii) To balance the nitrogen atoms, multiply `NH_(3)` by 2 on the R.H.S. We get
`Mg_(3)N_(2)+H_(2)Orarr 3Mg (OH)_(2)+2NH_(3)`
(iii) to equalise the number of H and O atoms on both side of the above equation, multiply `H_(2)O` OCCURRING on L.H.S. of the equation by 6. We get
`Mg_(3)N_(2)+6H_(2)Orarr 3Mg(OH)_(2)+2NH_(3)`
This is the required balanced chemical equation.
Discussion
9.

Balance the following skeleton equation by the method of Partial Equations : P+HNO_(3)rarr H_(3)PO_(4)+H_(2)O+NO_(2).

Answer»

Solution :This reaction is involves oxidation of phosphorus by nitric acid and takes place in the following steps :
(i) Nitric acid decomposes to liberate nascent oxygen.
`2HNO_(3)rarr 2NO_(2)+H_(2)O+(O)"…(1)"`
(ii) Nascent oxygen oxidises phosphorus to phosphorus pentoxide.
`2P+5(O)rarr P_(2)O_(5)"...(2)"`
(iii) Phosphorus pentoxide, `P_(2)O_(5)`, dissolves in water to produce phosphoric acid.
`P_(2)O_(5)+3H_(2)Orarr 2H_(3)PO_(4)"...(3)"`
(iv) To cancel the intermediate products, multiply the partial EQUATION (1) by 5 and adding to eqns. (2) and (3), we have
`10HNO_(3)+2Prarr 2H_(3)PO_(4)+2H_(2)O+10NO_(2)`
Dividing by the common factor 2, we have
`5HNO_(3)+Prarr H_(3)PO_(4)+H_(2)O+5NO_(2)`
This represents the ATOMIC equation since phosphorus which is known to exist in the molecular form `(P_(4))` has been written in the atomic form.
(V) To make the above atomic equaiton molecular, multiply it throughout by 4. Thus, we have the required balanced molecular equaiton :
`20HNO_(3)+P_(4)rarr 4H_(3)PO_(4)+4H_(2)O+20NO_(2)`
Discussion
10.

Balance the following skeleton equation by the method of Partial Equations : KMnO_(4)+H_(2)SO_(4)+(COOH)_(2) rarr K_(2)SO_(4)+MnSO_(4)+CO_(2)+H_(2)O.

Answer»

Solution :The oxidation of oxalic ACID, `(COOH)_(2)`, by potassium permaganate, `KMnO_(4)`, takes place in the FOLLOWING steps :
(i) `KMnO_(4)` reacts with dil. `H_(2)SO_(4)` to produce nascent oxygen.
`KMnO_(4)+H_(2)SO_(4)rarr K_(2)SO_(4)+MnSO_(4)+H_(2)O+(O)`
By balancing this skeleton equation by Hit and Trial method, we ge
`2KMnO_(4)+3H_(2)SO_(4) rarr K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5(O)"...(1)"`
(ii) Oxalic acid is oxidised to `CO_(2)` and `H_(2)O` by the nascent oxygen produced in equation (1).
The balanced partial equation for this reaction is :
`(COOH)_(2)+(O)rarr 2CO_(2)+H_(2)O"...(2)"`
To cancel the intermediate product, i.e., nascent oxygen, MULTIPLY equaiton (2) by 5 and ADDING to (1), we have
`2KMnO_(4)+3H_(2)SO_(4)+5(COOH)_(2)rarr K_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O`
This represents the balanced chemical equation for the above reaction.
Discussion
11.

Balance the following reactions by ion electron method : KMnO_(4) + H_(2) SO_(4) + HCl rarr K_(2) SO_(4) + MnSO_(4) + Cl + H_(2) O

Answer»

SOLUTION :Ionic FORM of the given reactionis
Step I `:` Ionic form of the given reaction is CL^(-) rarr 2K^(+)+SO_(4)^(2-) + Mn^(2+) + SO_(4)^(2-) +Cl_(2) + H_(2)O`
or `MnO_(4) + H^(+) + Cl^(-) rarr Mn^(2-) + Cl_(2) + H_(2) O`
Step II `:` Oxidation `Cl^(-) rarr overset( 0 ) ( C) l_(2)`
REDUCTION `:``overset( + 7)(M) n O_(4)^(-) rarr Mn^(2+)`
Step II `:` Oxidation `Cl^(-) rarr overset( 0 ) ( C) l_(2)`
Reduction`: overset( +7)(M) nO_(4)^(-) rarr Mn^(2+)`
Step III `:` Oxidation `2Cl^(-) rarr Cl_(2)`
Reduction `: MnO_(4)^(-)rarr Mn^(2+)`
Step IV ` :`Oxidation `2Cl^(-) rarrCl_(2)`
Reduction `: MnO_(4)^(-) + 8 H^(+) rarr Mn^(2+)+ 4H_(2) O `
Step V `: ` Oxidation `2Cl^(-) rarr Cl_(2) + 2e`
Reduction `MnO_(4) ^(_) + 8 H^(+) + 5e rarr Mn^(2+) + 4H_(2) O`
Step VI` :` Oxidation `2Cl^(-) rarr Cl_(2) + 2e ]xx 5 `
Reduction `: MnO_(4)^(-) + 8 H^(+) +5e rarr Mn^(2+) + 4H_(2) O ] xx 2 ` It is balanced reaction in ionic form.
Step VII `: 2KMnO_(4) + 10 HCl + 3H_(2) SO_(4) rarr 2MnSO_(4) + 8 H_(2)O + 5Cl_(2) + K_(2) SO(4)`is the balanced reaction.
Discussion
12.

Balance the following reactions by ion electron method :Cl_(2) + OH^(-) rarr ClO_(3)+Cl + H_(2)O

Answer»

Solution :Step I `:` Oxidation`: Cl_(2) rarr ClO_(3)^(-)`
Reduction`: Cl_(2) rarr CL^(-)`
Step II `:` Oxidation `Cl_(2) rarr 2ClO_(3)^(-)`
Reduction `:` `Cl_(2) rarr2Cl^(-)`
Step III `:`
Oxidation `: Cl_(2) +12OH^(-) rarr 2ClO_(3)^(-) + 6 H_(2) O`
Reduction `:` `Cl_(2) rarr 2Cl^(-)`
Sep IV `:`
Oxidation `:` `Cl_(2) +12OH^(-) rarr 2ClO_(3) + 6 H_(2) O + 10E`
Reduction `:Cl_(2) + 2e rarr 2Cl^(-)`
Step V `:`
Oxidation`:``Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 6H_(2)O+10e `
Reduction `: Cl_(2) + 2e rarr 2Cl^(-) ] xx5`
`6Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 10 Cl ^(-) + 6H_(2) O`
or `3Cl_(2) + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) + 3 H_(2)O`is the balanced reaction.
Step III onwards may be replaced as `:`
Step III `:` Oxidation`: Cl_(2) +6H_(2)O rarr 2ClO_(3) + 12H^(+)`
Reduction `: Cl_(2) rarr 2Cl^(-)`
Step IV `:` Oxidation `: Cl_(2) +6 H_(2) O rarr 2ClO_(3) +12H^(+) + 10e`
Reduction `: Cl_(2) + 2e rarr 2Cl^(-)`
Step V `:` Oxidation`: Cl_(2) + 6 H_(2) rarr 2ClO_(3) ^(-) + 12H ^(+) + 10e`
Reduction `: Cl_(2) + 2e rarr 2eCl^(-) ] xx5`
`6 Cl_(2) + 12OH^(-) rarr 2eCl^(-) + 10Cl^(-) + 6 H_(2)O`
or , `3Cl_(2) + 3H_(2) O rarr ClO_(3)^(-) + 5Cl^(-) + 6 H^(+)`
To REMOVE `H^(+)` ion, and equal NUMBER of `OH^(-)` IONS in both sides.
`3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) +6OH^(-)`
or, `3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl+ 6 H_(2) O `
or, `3Cl_(3) + 6OH^(-) rarr ClO_(3)^(-) + 5Cl^(-) +3H_(2)O` is the balanced reaction.
Discussion
13.

Balance the following reactionaZn+NO_(3)+bH_(2)Ooverset("OH")tocZn^(2+)+NH_(4)^(+)+dOH^(-)

Answer»

a = 4 , b = 7,C= 4, d = 10
a = 2, b = 5,c= 6 , d = 8
a = 3, b = 3 , c = 1 , d = 9
a = 1 , b = 3 , c = 7 , d = 8

Answer :1
Discussion
14.

Balance the following in basic medium KOH + K_(4) Fe(CN)_(6)+Ce(NO_(3))_(4) rarr Fe(Ohl)_(3) + Ce(OH)_(3) + K_(2) CO_(3) +KNO_(3) + H_(2)O

Answer»

SOLUTION :`258KOH + K_(4) Fe(CN)_(6)+ 61Ce(NO_(3))_(4) rarr 61Ce(OH)_(3) + Fe(OH)_(3) + 36H_(2)O + 6 K_(2) CO_(3) + 250 KNO_(3)`
Discussion
15.

Balance the following in basic mediumCrI_(3) + H_(2)O_(2) + OH^(-)rarr CrO_(4)^(2-) + IO_(4)^(-) + H_(2)O

Answer»

Solution :`2CrI_(3) + 27 H_(2)O_(2) + 10 OH^(-) rarr 2CrO_(4)^(2-) + 6 IO_(4)^(-) + 32 H_(2)O`
Discussion
16.

Balance the following equations : XeF_6 + H_2O to XeO_2F_2 + HF

Answer»

SOLUTION :`XeF_6 + 2H_2O to XeO_2 F_2 + 4HF`
Discussion
17.

Balance the following equations using desired medium : SbCl_(3) + KIO_(3) + HCl rarr SbCl_(5) + ICI + H_(2) O + KCl

Answer»

SOLUTION :`2SbCl_(3) + KIO_(3) + 6HCL rarr 2SbCl_(5) +ICI+3H_(2)O + KCL`
Discussion
18.

Balance the following equations using desired medium : FeC_(2)O_(4) + KMnO_(4) + H_(2)SO_(4) rarr Fe_(2) ( SO_(4))_(3) + CO_(2) + MnSO_(4) + K_(2)SO_(4) + H_(2)O

Answer»

Solution :`10 FeC_(2) O_(4) + 6 KMnO_(4) + 24H_(2)SO_(4) RARR 5Fe_(2) ( SO_(4))_(3) + 20 CO_(2) + 6 MnSO_(4) + 3K_(2)SO_(4) + 24H_(2)O`
Discussion
19.

Balance the following equations using desired medium : Pb(N_(3))_(2) + Co(MnO_(4))_(3) rarr CoO +MnO_(2) + Pb_(3) O_(4) + NO

Answer»

SOLUTION :`30Pb(N_(3))_(2) + 44Co(MnO_(4))_(3) rarr132MnO_(2) + 44CoO + 180 NO+ 10 Pb_(3) O_(4)`
Discussion
20.

Balance the following equations using desired medium : FeCr_(2)O_(4) + K_(2) CO_(3) + KCIO_(3) rarr Fe_(2) O_(3) +K_(2) CrO_(4) + KCl + CO_(2)

Answer»

SOLUTION :`6FeCr_(2)O_(4) + 12K_(2)CO_(3) + 7 KClO_(3) RARR 3Fe_(2) O_(3) + 12K_(2) CrO_(4) + 7 KCl + 12CO_(2)`
Discussion
21.

Balance the following equations in proper mediumCr_(2)O_(7)^(2-) +C_(2)H_(4)O+H^(+) rarr C_(2)H_(4)O_(2) +Cr^(3+)

Answer»

Solution :`Cr_(2)O_(7)^(2-) + 3C_(2) H_(4) O+ 8 H^(+) rarr 3C_(2) H_(4) O_(2) + 2Cr^(3+) + 4H_(2)O`
Discussion
22.

Balance the following equations in proper medium C_(2)H_(5)OH+ MnO_(4)^(-) rarr C_(2) H_(3) O^(-) +MnO_(2)(s) + H_(2)O

Answer»

SOLUTION :`3C_(2)H_(5) OH + 2MnO_(4)^(-)+OH^(-) RARR 3C_(2) H_(3) O^(-) + 2MnO_(2)(s) + 5H_(2)O`
Discussion
23.

Balance the following equations in basic medium MnO_(4) + C_(2)O_(4)^(2-) +H^(+) rarr Mn^(2+ ) + CO_(2) + H_(2)O

Answer»

Solution :`2MnO_(4)^(-) + 5C_(2) O_(4)^(2-) + 16H^(+) rarr 2MN^(2+) +10CO_(2) + 8 H_(2)O`
Discussion
24.

Balance the following equations in acidic medium [Fe(CN)_(6)]^(4-) + MnO_(4)^(-) rarr Fe^(3+) +CO_(2) + NO_(3)^(-) +Mn^(2+)

Answer»

SOLUTION :`5[FE(CN)_(6)]^(4-) + 188 H^(+) + 61 MnO_(4)^(-) rarr 5Fe^(3+) + 30 CO_(2) + 30 NO_(3)^(-) + 61 Mn^(2+) + 94 H_(2)O`
Discussion
25.

Balance the following equations in acidic medium KClO_(3) + H_(2) SO_(4) rarr KHSO_(4) + HClO_(4) + ClO_(2) + H_(2)O

Answer»

SOLUTION :`3KClO_(3) + 3H_(2)SO_(4) rarr 3KHSO_(4) + HClO_(4) + 2ClO_(2) + H_(2)O`
Discussion
26.

Balance the following equations in acidic medium H_(2) S+Cr_(2) O_(7)^(2-) + H^(+) rarr Cr_(2) O_(3) + S_(8) +H_(2)O

Answer»

SOLUTION :`24H_(2)S+ 8Cr_(2)O_(7)^(2-) + 16 H^(+) RARR 8 Cr_(2)O_(3) + 3S_(8) + 32H_(2)O`
Discussion
27.

Balance the following equations in acidic medium Cu_(2)O +H^(+) +NO_(3) ^(-) rarr Cu^(2+) + NO + H_(2) O

Answer»

SOLUTION :`3Cu_(2) O+ 14H^(+) + 2NO_(3)^(-) RARR 6Cu^(2+) + 2NO + 7 H_(2)O`
Discussion
28.

Balance the following equations in acidic medium Br^(-) + BrO_(3)^(-) + H^(+)rarr Br_(2) + H_(2)O

Answer»

Solution :`5Br^(-) + BrO_(3)^(-) + 6 H^(+) rarr 3Br_(2) + 3H_(2)O`
Discussion
29.

Balance the following equations by Hit and Trial Method : (i) SO_(2)+H_(2)S rarr S+H_(2)O (ii) Al_(4)C_(3)+H_(2)Orarr Al(OH)_(3)+CH_(4) (iii) KMnO_(4)+HClrarrKCl+MnCl_(2)+H_(2)O+Cl_(2) (iv) KMnO_(4)+KOHrarrK_(2)MnO_(4)+MnO_(2)+O_(2) (v) FeS_(2)+O_(2) rarr Fe_(2)O_(3)+SO_(2) (vi) Zm+NaOH rarr Na_(2)ZnO_(2)+H_(2) (vii) Na_(2)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+NaI (viii) C_(2)H_(6)+O_(2) rarr CO_(2)+H_(2)O (ix) Ca_(2)P_(2)+H_(2)O rarr Ca(OH)_(2)+PH_(3)

Answer»

SOLUTION :
Discussion
30.

Balancethe followingequations (b )Cr_(2) O^(2-)+ Sn^(2+)+ H^(+) to

Answer»

Solution :`(b)Cr_(3) O_(7)^(-2) + SN^(2+)+ H^(+)to CR^(+3)+ Sn^(+4)`
Discussion
31.

Balancethe followingequations (a )MnO_(4)^(-)+ Fe^(2+)+ H^(+)to

Answer»

SOLUTION :(a) `MnO_(4) ^(-)+ Fe^(2+)H^(+) toMn^(+3) + Fe^(+3)`
Discussion
32.

Balance the following equation : XeF_(6) + H_(2)O to XeO_(2)F_(2) + 4HF

Answer»

Solution :`XeF_(6) + 2H_(2)O to XeO_(2)F_(2) + 4HF`
Discussion
33.

Balance the following equation : XeF_(6) + H_(2)O rarr XeO_(2)F_(2) + HF

Answer»

SOLUTION :`XeF_(6) + 2 H_(2)O rarr XeO_(2) F_(2) + 4 HF`
Discussion
34.

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) P_(4)(s) + OH^(-) (aq) rarr PH_(3) (g) + HPO_(2)^(-) (aq) (b) N_(2)H_(4)(I) + ClO_(3)^(-) (aq) rarr NO(g) + Cl^(-) (g) (c) Cl_(2)O_(7) (g) + H_(2)O_(2) (aq) rarr ClO_(2)^(-) (aq) + O_(2)(g) + H^(+)

Answer»

Solution :a) `P_(4)(s)+OH^(-)(aq)toPH_(3)(g)+H_(2)PO_(2)^(-)(aq)`
Ion electron method : `underset(0)(P_(4))+underset(-2,+1)(OH^(-))tounderset(-3, +1)(PH_(3))+underset(+1, +1, -2)(H_(2)PO_(2)^(-))`


Note : Here `P_(4)` acts both as oxidant and reductant.
Oxidation NUMBER method :
`underset((0))(P_(4)(s))+OH^(-)(aq)tounderset((-3))(PH_(3)(g))+underset((+1))(H_(2)PO_(2)^(-)(aq))`

In order to balance the change in oxidation number `H_(2)PO_(2)^(-)` is to be multiplied by 3
`P_(4)+OH^(-)toPH^(3)+3H_(2)PO_(2)^(-)`
Since the reaction is taking place in basic medium, `H_(2)O` is to be added on the side which has ledder H atom and `OH^(-)` are to be added on the side which has lesser O atoms.
`P_(4)+3H_(2)O+3OH^(-)toPH_(3)+3H_(2)PO_(2)^(-)`
b) `N_(2)H_(4)(l)+ClO_(3)^(-)(aq)toNO(g)+Cl^(-)(g)`

Step - III : EQUALISE the increase and decrease in ON by multiplying `N_(2)O_(4)` with 3 and `ClO_(3)^(-)` with 4.
`3N_(2)O_(4)+4ClO_(3)^(-)to6NO+4Cl^(-)`
Step - IV : Balance the atoms except H and O. Here they are balanced.
Step - V : Balance O atoms by adding `OH^(-)` ions and H atoms by adding `H_(2)O` on the sides deficient of O and H atoms respectively
`3N_(2)O_(4)+4ClO_(3)^(-)to6NO+4Cl^(-)+12OH^(-)`
c) `Cl_(2)O_(7)(g)+H_(2)O_(2)(aq)toClO_(2)^(-)(aq)+O_(2)(g)+H^(+)`
Ion electron method :

Oxidation number method :
Step - I : Skeleton equation is
Step - II : Equalise the increases/decrease in ON by multipling `H_(2)O_(2)` with 4 since in each CHLORINE of `Cl_(2)O_(7)` decrease in ON is 4. For 2 Cl atoms it is 8. In `H_(2)O_(2)` increase in ON for each 0 is 1 and for two 0 atoms it is 2.
`Cl_(2)O_(7)+4H_(2)O_(2)to2ClO_(2)^(-)+4H_(2)O+2O_(2)`
Step - III : Balance the O atoms by adding `OH^(-)` and H atoms by adding `H_(2)O` to the sides deficient of O and H atoms respectively.
`Cl_(2)O_(7)+4H_(2)O_(2)+2OH^(-)to2ClO_(2)^(-)+4H_(2)O+2O_(2)`
Discussion
35.

Balance the following equation by partial Equation method: (i) PbS+O_(3)rarr PbSO_(4)+O_(2) (ii) K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+SO_(2) rarr K_(2)SO_(4)+Cr_(2)(CO_(4))_(3)+H_(2)O (iii) KMnO_(4)+FeSO_(4)+H_(2)SO_(4) rarr K_(2)SO_(4)+MnSO_(4)+Fe_(2)(SO_(4))_(3)+Fe_(2)(SO_(4))_(3)+H_(2)O (iv) Mg+HNO_(3) rarr Mg(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O (v) Cu+HNO_(3) rarr Cu(NO_(3))_(2)+NO+H_(2)O (vi) C+H_(2)SO_(4) rarr CO_(2)+SO_(2)+H_(2)O (vii) P_(4)+HNO_(3) rarr H_(3)PO_(4)+NO_(2)+H_(2)O (viii) CuSO_(4)+KI rarr K_(2)SO_(4)+Cu_(2)I_(2)+I_(2) (ix) Fe_(2)(SO_(4))_(3)+NH_(3)+H_(2)O rarr Fe(OH)_(3)+(NH_(4))_(2)SO_(4) (x) I_(2)+HNO_(3) rarr HIO_(3)+NO_(2)+H_(2)O

Answer»

SOLUTION :
Discussion
36.

Balance the chemical reaction : (i) overset(+7)MnO_(4)^(-) + 3H_(2) overset(-2)O overset("noutral medium") to overset(+4)MnO_(2) + overset(0)O_(2) (ii)Write the balanced equation when ferrous sulphate is treated with acidified (H_(2)SO_(4)) potassium permanganate . (iii) Balance the equation MnO_(4) + C_(2)O_(4) + H^(+) toCO_(2) + Mn^(+2) + H_(2)O (iv) Balance the equation K_(2)Cr_(2)O_(7) + HCl to KCl + CrCl_(3) + H_(2)O + Cl_(2)

Answer»

Solution :(i) ` 2MnO_(4)^(-) + 3H_(2)O to 2MnO_(2) + 3O_(2) + 2OH^(-) + 2H_(2)O`
(II) `10 FeSO_(4) + 2KMnO_(4) + 8H_(2)SO_(4) to 5Fe_(2)(SO_(4))_(3) + 2MnSO_(4) + K_(2)SO_(4) + H_(2)O`
(iii) `2MnO_(4)^(-) + 5C_(2)O_(4)^(2-) + 16H^(+) to 2MN^(2+) + 10CO_(2) + 8H_(2)O`
(iv) ` K_(2)Cr_(2)O_(7) + 14HCl to 2KCL + 2CrCl_(3) + 7H_(2)O + 3Cl_(2)`
Discussion
37.

Balance the equation by lon-electron method (vi) Cl_(2) +OH^(-) to S^(2-) + S_(2)O_(3)^(2-) (In basic medium)

Answer»

SOLUTION :`Cl_(2) + 2OH^(-) to CL^(-) + CLO^(-) + H_(2)O`
Discussion
38.

Balance following equations in proper medium Na_(2)S_(2)O_(2) + KMnO_(4)+H_(2)O rarr Na_(2) S_(4) S_(6) +MnO_(2) + KOH + NaOH

Answer»

SOLUTION :`6Na_(2)S_(2)O_(3) + 2KMnO_(4) + 4H_(2)O rarr3Na_(2) S_(4) O_(6) + 2MnO_(2) + 2KOH + 6NaOH`
Discussion
39.

Balance given following half reaction for the unbalanced whole reaction : CrO_(4)^(2-)rarrCrO_(2)^(-)_OH^(-) is :

Answer»

`CrO_(4)^(-2)+2H_(2)O+3e^(-)rarrCrO_(2)^(-)+4H_(2)O^(-)`
`2CrO_(4)^(2-)+8H_(2)OrarrCrO_(2)^(-)+4H_(2)O+8OH^(-)`
`CrO_(4)^(-2)+H_(2)OrarrCrO_(2)^(-)+H_(2)O+OH^(-)`
`3CrO_(4)^(-2)+4H_(2)O+6e^(-)rarr2CrO_(2)^(-1)+8OH^(-)`

Answer :A
Discussion
40.

Balance following equations in proper medium S+OH^(-) rarr S^(2-) +S_(2)O_(3)^(2-)

Answer»

SOLUTION :`4S+ 6OH^(-) rarr 2S^(2-) + S_(2) O_(3)^(2-)`
Discussion
41.

Balance following equations in proper medium P + OH^(-) +H_(2)O rarr H_(2)PO_(4)^(-) +PH_(3)

Answer»

Solution :`8P + 3OH^(-) + 9 H_(2)O rarr 3H_(2)PO_(4)^(-) + 5PH_(3)`
Discussion
42.

Balance following equations in proper medium FeC_(2)O_(4) + KMnO_(4) + H_(2) SO_(4) rarr Fe_(2) ( SO_(4))_(3) + CO_(2) + MnSO_(4) + K_(2) SO_(4) + H_(2)O

Answer»

Solution :`10FeC_(2)O_(4) + 6 KMnO_(4) + 24H_(2)SO_(4) rarr 5Fe_(2) ( SO_(4))_(3) + 20 CO_(2) +6 MnSO_(4) + 3K_(2) SO_(4) + 24H_(2)O`
Discussion
43.

Baking soda or baking powder is :

Answer»

WASHING Soda
Caustic Soda
Soda ash
Sodium bicarbonate

Answer :D
Discussion
44.

Baking soda is :

Answer»

`Na_(2)CO_(3)*10H_(2)O`
`Na_(2)CO_(4)`
`NaHCO_(3)`
`CaCO_(3)`

Answer :C
Discussion
45.

Bakelite plastic is formed, when phenol reacts with

Answer»

`CH_3CHO`
HCHO
acetone
HCOOH

Answer :B
Discussion
46.

Bakelite is the condensation polymer of formaldehyde and .................

Answer»


ANSWER :PHENOL
Discussion
47.

Bakelite is the polymer of

Answer»

BENZALDEHYDE and phenol
Acetaldehyde and phenol
Formaldehyde and phenol
Formaldehyde and BENZYL alcohol

Answer :C
Discussion
48.

Bakelite is prepared by the reaction between

Answer»

UREA and FORMALDEHYDE
Tetramethylene glycol and hexamethylene diisocyanate
PHENOL and formaldehyde
ETHYLENE glycol and dimethyl terephhalate.

Answer :C
Discussion
49.

Bakelite is _____ polymer.

Answer»

SOLUTION :CONDENSATION
Discussion
50.

Bakelite is _____ .

Answer»

SOLUTION :CONDENSATION
Discussion