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.

Give method of preparation of ethyne (Acetelene) from vicinal dihalides.

Answer»

Solution :(a) (i) Alkenlhalide is OBTAINED by reaction of vicinal dihlaide with alcoholic potassium hdyroxide reaction and this alkynyl HALIDE on reaction with sodamide `(NaNH_(2))` gives alkyne. This reaction is called dehydrohalogenation.
(ii) This alkinyl on reaction with sodamide `(NaNH_(2))` gives alkyne.

(b) It can obtained by the following reaction too,
`underset("1,2-dibromoethane")(CH_(2)Br-CH_(2)Br)+underset("Alcoholic")(2KOH) rarr underset("Ethyne")(CH-=CH)+2KBr+2H_(2)O`
According to reaction (a), it is carried out in two or according to reaction (b), it is carried out in ONE STEP. But second step of reaction (a) is difficult and hence, it can be made easy by ADDITION of `NaNH_(2)`.
Discussion
2.

Give method to convert solid into gas.

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Solution :When SOILD is heated it will CONVERT into LIQUID. When liquid is heated it will convert into gas.
Discussion
3.

Give methods of preparation for benzene.

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Solution :Industrially, benzene is prepared from the coal tar. In LESS proportional benzene is prepared by following methods :
(i) Benzene is prepared by cyclic polymerisation of ethyne:

(ii) Benzene is formed by cyclization of n-hexane :

(III) By decarboxylation of benzoic acid : Heating of benzoic acid with soda lime gives benzene.

(iv) By REDUCTION of PHENOL gives benzene :
Discussion
4.

Give mehod of preparation of ethyne from calcium carbonate or calcium carbide.

Answer»

SOLUTION :(i) Quick lime can be OBTAINED by heating limstone as shown in the following reactions :
`underset("Limestone")(CaCO_(3(s))) overset(Delta)rarr underset("Lime")(CaO_((s))) + CO_(2(g))`
(ii) Now heating calcium oxide with COKE, calcium carbide and carbon monoxide are obtained.
`underset("Quick lime")(CaO) + 3C underset(Delta)overset("Closed furnace")rarr underset("Calcium carbide")(CaC_(2)) + CO`
(iii) By REACTION of calcium carbide with water, ETHYNE is obtained as follows.
`underset("carbide")underset("Calcium")(CaC_(2))+underset("Water")(2H_(2)O)rarr underset("hydroxide")underset("Calcium")(Ca(OH)_(2)) + underset("Ethyne")(C_(2)H_(2))`
By this way ethyne is produced in industries.
Discussion
5.

Give mechanism of chhlorination of methane.

Answer»

Solution :Halogenation is supposed to proceed via free radical chain mechanism involving three steps namely (a) Initiation, (b) Propagation and (c) Termination as given below :
(a) Initiation (Step-I) : The reaction is initiated by homolysis of chlorine molecule in the presence of light or heat. The Cl-Cl bond is weaker than the C-C and C-H bond and hence, is easiest to break.
`underset("Dichlorine")(Cl-Cl)underset("Homolysis")overset(hv)rarr underset("Chlorine free radicals")(overset(.)(Cl)+overset(.)(Cl))` ....(i)
(b) Propagation (Step-II) : Chlorine free radical attacks the methane molecule and takes the reaction in the forward direction by breaking the C-H bond to generate METHYL free radical with the formation of H-Cl.
(i) `underset("Methane")(CH_(4))+underset("radicals")underset("free")underset("Chlorine")(overset(.)(Cl))overset(hv)rarr underset("radicals")underset("free")underset("Methyl")(overset(.)(Cl))+H-Cl` ...(ii)
(ii) The methyl radical thus obtained attacks the second molecule of chlorine to form `CH_(3)-Cl` with the liberation of another chlorine free radical by homolysis of chlorine molecule.
`underset("radicals")underset("free")underset("Methyl")(overset(.)(CH_(3)))+underset("of chlorine")underset("Moelcule")(Cl-Cl)overset(hv)rarr underset("Chloromethane")(H_(3)C)- underset("radicals")underset("free")underset("Chlorine")(Cl+overset(.)(Cl))`...(ii)
The chlorine and methy free radicals generated above repeat steps (a) and (b) respectively and thereby setup a chain of reactions.
The propagation steps (a) and (b) are THSE which directly give principal products, but manyother propagation steps are possible and may occur. Two such steps given below explain how more highly halogenated products are formed.
`CH_(3)Cl + overset(.)(Cl) rarr overset(.)(C)H_(2)Cl+HCl`
`overset(.)(C)H_(2)Cl + (Cl)-Cl rarr overset(.)(C)H_(2)Cl_(2)+overset(.)(Cl)`
(c) Termination (Step -II) : The reaction stops after some TIME DUE to consumption of reactants and /or due to the following side reactions :
The possible chain terminating steps are :
(i) `overset(.)(Cl) + overset(.)(Cl)rarrCl + Cl`
(ii) `H_(3)overset(.)(C)+ overset(.)(C)H_(3) rarr H_(3)C-CH_(3)` (Ethane)
(iii) `H_(3)overset(.)(C)+ overset(.)(Cl) rarr CH_(3)-Cl`
Through in (c), `CH_(3)-Cl`, the one of the product is formed but free radicals are CONSUMED and the chain is terminated.
Chlorination of methane ethane is obtained because according to step-III (b) ethane is obtained when joining of methyl free radical.
Discussion
6.

Give line and condensed formula of following (i) CH_(3)CH_(2) CH_(2) CH_(2) CH_(2) CH_(2) CHBr CH_(2) CHO (ii) CH_(3) - underset(underset(CH_(3))(|))(C H) - CH_(2)OH (iii) NH_(2) - CH_(2) CH_(2) CH_(2) CH_(2) CH_(2) CH_(2) COOH (iv) CH_(3) CH_(2) CH_(2)OH

Answer»

SOLUTION :
Discussion
7.

Give main use of Quartz.

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Solution :Quartz is extensively used as a piezoelectric material , it has MADE possible to DEVELOP EXTREMELY accurate clocks, modern RADIO and television broadcasting and mobile radio COMMUNICATIONS.
Discussion
8.

Give main product of the following reaction. (in one step) (i) 1, 2-dibromo propane + alcoholic KOH (ii) 1, 1, 2,2-tetrabromo ethane + zinc powder in methanol (iii) 1,1,2,2-tetrabromo propane + zinc powder in methanol. (iv) Reaction occur on passing of hydrogen gas through carbon electrode at high temperature in electric arc.

Answer»

Solution :
(iii) `underset("1,1,2,2-tetrabromo PROPANE")(CH_(3)CBr_(2)CHBr_(2) + 2Zn) rarr underset("PROPYNE")(CH_(3)C)-=CH+2ZnBr_(2)`
(iv) `2C+H_(2) underset(3270K)overset("Electric arc")rarr HC -= CH`
Discussion
9.

Give main characteristics of hybridization.

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Solution :The main characteristics of hybridization are under:
(i) The number of hybrid ORBITALS is equal to the number of the atomic orbitals that get hybridised.
The hybridised orbitals are always equivalent in energy and shape.
The hybrid orbitals are more EFFECTIVE in fomtlng stable bonds them the pure atomic orbitals.
(iv) These hybrid orbitals are directed in space in some PREFERRED direction to have minimum repulsion between electron pairsand thus a stable arrangement.
THEREFORE the type of hybridization indicates the geometry of the MOLECULES.
Discussion
10.

Give magnitude of energy of force and value of interaction process with example.

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Solution :DIPOLE - Dipole force : 1 to 3 K cal `mol^(-1)`
e.g., : `SO_(2), NO, HCl`
INTERACTION energy in rotational molecule `prop (1)/(r^(6))`
Interaction energy energy in stable molecule `prop (1)/(r^(3))`
(We go to closer energy release).
If distance between molecules `lt` 500 pm. Then this forces present.
Discussion
11.

Give law of multiple proportion ?

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Solution :When TWO elements combine from more then one COMPOUND then mass of one element COMBINING with mass of other element is in simple RATION.
Discussion
12.

Give limitations of Arrhenious acid-base concept.

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SOLUTION :(i) This concepts is apply only for aqueous solution.
(ii) This is UNABLE to explain the basic character of `NH_3` LIKE substance because, the HYDROXYL group is absent in `NH_3`.
Discussion
13.

Give law of chemical equilibrium for equilibrium process.

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Solution :A general reversible reaction A + B `hArr` C + D, where A and B are the reactants, C and D are the products in the balanced chemical equation.
On the basis of experimental studies of many reversible reactions, the Norwegian chemists (Cato Maximillian Guldberg) and (PETER Waage) proposed in 1864 that the concentrations in an EQUILIBRIUM mixture are related by the following equilibrium equation,
`K_c=([C][D])/([A][B])` (Equilibrium equation) ....(Eq-i)
where , `K_c`=Equilibrium constant and in it C= Concentration MOLE / litre
(Right side expression ) =`([C][D])/([A][B])`=(Equilibrium constant expression)
In expression of `K_c` , concentrations are equilibrium concentration.
The equilibrium equation-(i) is also known as the law of .mass action..
Equilibrium law of Chemical equilibrium law : At a given temperature, the product of concentrations of the reaction products raised to the respective STOICHIOMETRIC coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium law or Law of Chemical equilibrium.
The equilibrium constant for a general reaction,
aA + bB `hArr` cC + dD ....(ii)
For the equation (7.11) the mathematical FORM of the law of chemical equilibrium is eq.-(iii).
`K_c=([C]^c [D]^d)/([A]^a[B]^b)`...(iii)
Discussion
14.

Give Kelvin statement of second law of thermodynamics.

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SOLUTION :Kelvin-Planck statement: It is impossible to take heat from a HOTTER reservoir and CONVERT it completely into work by a CYCLIC PROCESS without transferring a part of heat to a cooler reservoir.
Discussion
15.

Give IUPAC names of (CH_(3))_(3)CH(C_(2)H_(5))_(2).

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Solution :Its structure and IUPAC nameis as FOLLOWS :
`{:(""CH_(3)),("|"),(H_(3)overset(1)(C)-""^(2)C-overset(3)(C)H-overset(4)(C)_(2)H-overset(5)(C)_(3)H),("|""|"),(""CH_(3) ""CH_(2)-CH_(3)),("3-ethyl-2,2-dimethyl pentane"):}`
Discussion
16.

Give IUPAC names of (CH_(3))_(3)C CH(C_(2)H_(5))_(2). Why 3-methyl-4,4 dimethyl pentane is not true ?

Answer»

Solution :In this NAME substitution ORDER is t 3,4 and 4 so in its true name 3-ethyl-2,2 dimethyl PENTANE is possible but above this PREFERENCE is not possible.
Discussion
17.

Give IUPAC names for the following compounds (CH_(3))_(3)C-C-=C-CH(CH_(3))_(2)

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SOLUTION :`underset((2,2,5-"triemethyl-hexa"-3-"YNE"))((CH_(3))_(3)C-C-=C-CH(CH_(3))_(2))`
Discussion
18.

Give IUPAC names for the following compounds CH_(3)-CH=CH-CH=CH-C-=C-CH_(3)

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SOLUTION : `underset("OCTA"-4,6"DIENE"-3-"yene")(CH_(3)-CH=CH-CH=CH-C-=C-CH_(3))`
Discussion
19.

Give IUPAC namesfor the following comounds

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Solution :`(i) (CH_(3)) C-C =C- CH (CH_(3))_(2)`
`(II) CH_(3) -underset(CH_(3)) underset(|) overset( CH_(2) ) overset(|)(C )- underset(H ) underset(|) overset(CH_(3)) overset(|) (C ) - C= C - CH_(3)`
`(III)underset(2,2,5, " trimethyl-1-hexata -3- yne ")(CH_(3) - underset(|) overset(CH_(3)) overset( |) (C ) - overset(3)(C )- underset( CH_(3))underset( |) (CH ) - overset(3) (C H_(3)))`
(IV) Ethylisopropylacetylene
Discussion
20.

Give IUPAC name of following: (i) CH_(3)CH_(2) CH= CHCOOH (ii) CH_(2)= overset(overset(CH_(3))(|))(C )- underset(underset(CH(CH_(3))_(2))(|))(CH)-C(CH_(3))_(3)

Answer»

SOLUTION :(i) pent-2-enicacid (II) 2, 4, 4-trimethyl-3-isopropylpent-1-en
Discussion
21.

Give IUPAC name of following (i) CH_(3)COOCH_(3) (ii) CH_(3)COOCH_(2)CH_(3) (iii) CH_(3)CH_(2)COOCH_(3) (iv) HCOOCH_(3)

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Solution :(i) `CH_(3)COOCH_(3)` (Methyl ethanoate)
(ii) `CH_(3)COOCH_(2)CH_(3)` (ETHYL ethanoate)
(III) `CH_(3)CH_(2)COOCH_(3)` (Methyl propanoate)
(iv) `HCOOCH_(3)` (Methyl methanoate)
Discussion
22.

Give IUPAC name of (CH_(3))_(3)C.C-=C.C(CH_(3))_(3).

Answer»

DI(TRI METHYL)but-2-yne
3,3,4,4-tetramethylhex-3-yne
2,2,5,5-tetramethylhex-3-yne
2,2,5,5-tetramethylhex-4-yne

Answer :C
Discussion
23.

Give IUPAC name and its functional group of the following (i) CH_(3)SO_(3)H (ii) CH_(3)CH_(2)CH_(2)NO_(2) (iii) CH_(3)CH_(2)CH_(2)CH_(2)CONH_(2) (iv) CH_(3)CH_(2)CH_(2)NH_(2)CH CH_(3) (v) CH_(3)CH_(2)CH_(2)CH_(2)COCl (vi) CH_(3)CH_(2)CN (vii) CH_(3)CH_(2)CHO (viii) CH_(3)CH_(2)CH_(2)OH

Answer»

SOLUTION :
Discussion
24.

Give IUPAC name of CH_(3)-CH_(2)COOC_(2)H_(5)

Answer»

ETHYL but-1-anoate
ethylpropanoate
ethylprop-2-anoate
None of these

Answer :B
Discussion
25.

Give IUPAC name of above compound.

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SOLUTION :`underset("Hex-4-ene-1-yne")(overset(6)(CH_(3))-overset(5)(CH)-overset(4)(CH)-overset(3)(CH_(2))-overset(2)(C)-=overset(1)(CH))`
Discussion
26.

Give IUPAC name and product structure obtained from the reaction of but -1-ene with HBr in presence of peroxide.

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SOLUTION :
Discussion
27.

Give is the difference in the structure of the following pair of comounds : CO_(2) and SiO_(2) .

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Solution :`CO_(2)` a linear molecule and exists as monomer . It is a gas while `SiO_(2)` is solid at room temperature due to its THREE DIMENSIONAL network in which each Si atom is covalently BONDED to four oxygen ATOMS tetrahedrally. In `CO_(2)` is sp hybridised while in `SiO_(2)` is `sp^(3)` hybridised . `CO_(2)` is a discrete molecule whereas `SiO_(2)` is a 3- dimensional solid
Discussion
28.

Give isotopes and their names of C_(3)H_(6).

Answer»

Solution :(i) `UNDERSET("Propene")(CH_(3) CH=CH_(2))` (II)
Discussion
29.

Giveisotopes of hydrogen carbon andchlorineand writedownnumberof electronneutron andprotons.

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SOLUTION :
Discussion
30.

Deuterium Tritium are the Isotopes of -

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SOLUTION :`._(17) ^(35) CI ` and `._(17)^(37) CI`are theisotopsof clorineisotopes
Discussion
31.

Give ionic equilibrium of aqueous solution of weak base NH_2NH_2and pyridine C_6H_5N.

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SOLUTION :`NH_2NH_(2(aq)) + H_2O_((l)) hArr NH_2NH_(3(aq))^(+) + OH_((aq))^(-)`
`C_6H_5N_((aq)) + H_2O_((l)) hArr C_6H_5N^(+)H_((aq)) + OOH_((aq))^(-)`
Discussion
32.

Give interaction energy of the following : (i) London forces(ii) Dipole - Dipole forces (iii) Dipole - induce forces.

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Solution :DISTANCE between TWO molecule = r
Interaction energy of london force `PROP (1)/(r^(6))`
Energy of forces of interaction of stationary dipole - dipole `prop (1)/(r^(3))`
Interaction energy of rotational Dipole - Dipole forces `prop (1)/(r^(6))`
Interaction energy of Dipole - INDUCE Dipole forces `prop (1)/(r^(6))`
Discussion
33.

Give intermediates produced by fission of bond and write its main characteristics

Answer»

Solution :
These all three intermediate are very UNSTABLE and reactive and produce during reaction. The stability of these three INTERMEDIATES is depends on the inductive EFFECT and hyperconjugation of attached ALKYL group. There are three covalent bond with the carbon of these intermediate species.
`underset("1-electron less")(H- underset(underset(H)(|))overset(overset(H)(|))(C ^(+))) underset("1-electron more")(H-underset(underset(H)(|))overset(overset(H)(|))(C ):^(-)) underset("Neutral")(H- underset(underset(H)(|))overset(overset(H)(|))(C ).)`
Discussion
34.

Give information on uses of green chemistry in day to day life.

Answer»

Solution :Uses of green chemistry in our life is as below:
(i) Dry Cleaning of Clothes : Tetra chlroroethene `(Cl_2 C = C CI_2)` was earlier used as solvent for dry cleaning. The compound contaminates the ground water and is also a suspected carcinogen. So insted of this compound liquefied carbondioxide with a suitable detergent is used.
REPLACEMENT of halogenated solvent by liquid `CO_2` will result in LESS harm to ground water. Hydrogen peroxide `(H_2 O_2)` is used for the purpose of bleaching clothes which gives beller results and makes use of lesser amount of water.
(ii) Bleaching of Paper : Earlier chlorine gas was used for bleaching paper whereas hydrogen peroxide `(H_2 O_2)` is used nowaday.
(iii) Synthesis of Chemicals : Ethanal `(CH_3 CHO)` is now commercially PREPARED by oxidation of ETHENE in presence of ionic catalyst in aqueous medium with a yield of 90%
`CH_2 = CH_2 + O_2 underset(Pd(II)//Cu(II) "(in water)")overset("Catalyst")to CH_3 CHO (90%)`
Discussion
35.

Give information on percentage proportion of element and mixtures present in Portland cement.

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SOLUTION :The average composition of Portland cement is :
`CaO:50-60%, SiO_(2):20-25%`
`Al_(2)O_(3):5-10%, MgO:2-3%`
`Fe_(2)O_(3):1-2%, SO_(3):1-2%`
For a good quality cement, the ratio of silica `(SiO_(2))` to alumina `(Al_(2)O_(3))` should be between 2.5 and 4 and the ratio of lime (CaO) to the TOTAL of the oxides of silicon `(SiO_(2))` aluminium `(Al_(2)O_(3))` and iron `(Fe_(2)O_(3))`should be as CLOSE as possible to 2.
Discussion
36.

Give information of oxidation numbers of elements present in periodic table ?

Answer»

Solution :The highest oxidation NUMBER of a representative ELEMENT is the group number for the first two groups and the group number minus 10, for the other groups.
Oxidation number increases FORM LEFT to right in the table.
Discussion
37.

Give information of silicon compound.

Answer»

Solution :FELDSPAR, ZEOLITES, mica and asbestos are SILICATES minerals. The basic structural unit of silicates is `SiO_4^(4-)`(Refer Fig.) in which silicon atom is bonded to four oxygen atoms in tetrahedron fashion.
In silicate compound, atoms are joined together via corners by sharing 1, 2, 3, 4 or 4 oxygen.
When silicate units are linked together, they form chain, ring, sheet or three-dimensional structures. Negative charge on silicate structure is neutralised by positively charged metal ions.
Two important man-made silicates are glass and cement.
Discussion
38.

Give information about universal solvent.

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Solution :WATER is universal solvent. It has high dielectric constant 80. It give facility to more ions in solution. Hydration of ions take PLACE.
Discussion
39.

Give increasing order for stability of alkene (Delta H^(@)). (CH_(3))_(2)C=C(CH_(3))_(2), CH_(3)CH_(2)CH_(2)CH=CH_(2), (CH_(3))_(2)C=CHCH_(3), CH_(3)CH_(2)C(CH_(3))=CH_(2)

Answer»

SOLUTION :`CH_(3)CH_(2)CH_(2)CH=CH_(2) LT CH_(3)CH_(2)C(CH_(3))=CH_(2) lt (CH_(3))_(2)C=CHCH_(3) lt (CH_(3))_(2)C=C(CH_(3))_(2)`
Discussion
40.

Give increasing order of group 13 elements for atomic radius.

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Solution :The increasing ORDER of atomic RADIUS of GROUP 13 elements is as FOLLOWS : B `lt` Al `gt` GA `lt` In `lt` Tl
Discussion
41.

Give industrial preparation for diborane.

Answer»

SOLUTION :`2BF_3 + 6NaH OVERSET"450 K"to B_2H_6 + 6NaF`
Discussion
42.

Give industrial preparation method of ethyne.

Answer»

Solution :(i) Quick lime can be obtained by heating limstone as shown in the following reactions :
`underset("Limestone")(CaCO_(3(s))) overset(Delta)rarr underset("Lime")(CaO_((s))) + CO_(2(G))`
(ii) Now heating calcium oxide with coke, calcium carbide and carbon monoxide are obtained.
`underset("Quick lime")(CaO) + 3C underset(Delta)overset("Closed furnace")rarr underset("Calcium carbide")(CaC_(2)) + CO`
(III) By reaction of calcium carbide with water, ethyne is obtained as follows.
`underset("carbide")underset("Calcium")(CaC_(2))+underset("Water")(2H_(2)O)rarr underset("hydroxide")underset("Calcium")(Ca(OH)_(2)) + underset("Ethyne")(C_(2)H_(2))`
By this way ethyne is PRODUCED in INDUSTRIES.
Discussion
43.

Give in the product by carboxylic acid method with examples ?

Answer»

Solution :* Sodium salts of carboxylic acids on heating with soda lime (mixture of sodium hydroxide and calcium atom LESS than the carboxylic acid. This process of elimination of carbon DIOXIDE from a caroxylic acid is known as decarboxylation.
`underset("Acid")(RCOOH)underset(+3 NaOH, -H_(2)O,-Na_(2)CO_(3))overset("Soda lime", Delta)rarr underset("Hydrocarbon")(RH)`
* Soda lime is a mixture of sodium hydroxide (NaOH) and calcium OXIDE (CaO). NaOH of soda lime reacts with acid and forms sodium salt of acid.
Example -1 : Methane from acetic acid.
`underset("Sodium ethanoate")(CH_(3)COO Na^(1))+NaOH underset(Delta)overset(CaO)rarr underset("Methane")(CH_(4))+Na_(2)CO_(3)`
Here, formation of sodium ethanote is from acetic acid.
Example -2 : Ethane from propanoic acid.
`underset("Propanoic acid")(CH_(3)CH_(2)COOH)underset((NaOH//CaO)Delta)overset("Soda lime")rarr underset("Sodium ethanoate")([C_(2)H_(5)COONa^(+)])overset(Delta)rarr underset("Ethane")(C_(2)H_(6))`
NOte : In decarboxylation process, `CO_(2)` is liberated and we get product of 1 carbon less. These reaction is useful when we get 1 carbon less in out product.
Discussion
44.

Give important trends and anomalous properties of boron.

Answer»

Solution :Certain important trends can be observed in the chemical behaviour of group-13 elements. The trichlorides, bromides and iodides of all these elements being covalent in NATURE are hydrolysed in WATER.
Species like tetrahedral `[M(OH)_4]^(-)`and octahedral `[M(H_2O)_6]^(3+)` , except in boron, exist in aqueous medium.
The monomeric trihalides, being electron DEFICIENT, are strong Lewis acids. Boron trifluoride easily reacts with Lewis bases such as `NH_3` to COMPLETE octet around boron.
It is due to the absence of d-orbitals that the maximum covalence of B is 4. Since the d orbitals are available with Al and other elements, the maximum covalence can be expected beyond 4.
Most of the other metal halides (e.g., `AlCl_3`) are dimerised through halogen BRIDGING (e.g., `Al_2Cl_6`). The metal species completes its octet by accepting electrons from halogen in these halogen bridged molecules.
Discussion
45.

Give incomplete combustion reaction of methane, hepetane and nonane.

Answer»

Solution :`"Alkane" + O_(2) overset("Incomplete combustion")rarrCO+H_(2)O`
`CH_(4) + (3)/(2)O_(2) rarr CO+2H_(2)O`
`underset("Heptane")(C_(7)H_(16)) + (15)/(2)O_(2) rarr 7CO + 8H_(2)O`
`underset("Nonane")(C_(9)H_(20)) + (19)/(2)O_(2) rarr 9CO + 10H_(2)O`
GENERAL reaction :
`C_(n)H_(2n+2) +((n+1)/(2)) O_(2)rarr nCO + (n)/(2)H_(2)O`
Discussion
46.

Give in detail about Viscosity.

Answer»

Solution :Definition of Viscosity and fluidity of liquid : It is one of the characteristic properties of liquids. ..Viscosity is a MEASURE of RESISTANCE to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows... Strong intermoleculr forces between molecules hold them together and resist movement of layers past one another.
Velocity of layers of liquids and laminar flow :When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layer increases as the distance of layers from the fixed layer increases.
Laminar flow :This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called .Laminar flow..
Mathematical Representation of viscosity : The figure given under is showing, the layer above it ACCELERATES the flow and the layer below retards its flow.

If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by the amount `(du)/(dz)`.
A force is required to maintain the flow of layers. This force is proportional to the area of contact of layers and velocity gradient. Mathematically .......
(i) `F alpha A`
(ii) `D alpha (du)/(dz)`
`therefore F alpha A((du)/(dz))`
`therefore F = eta A ((du)/(dz))`
where, A = The area of contact
`(du)/(dz)=` Velocity gradient
`eta =` Proportionality constant
= Viscosity co - efficient
= Measurement of viscosity
Viscosity, co - efficient constant and Unit :
Formula of Viscosity force (F ) : `F = eta A (du)/(dz)`
If surface are of liquid (A) = 1 unit
Velocity gradient `= (du)/(dz)=1` then `F = eta`
..Proportionality constant `(eta)` is measurement of viscosity and constant surface area and unit of velocity gradient is equal to force. Unit of Viscosity is co - efficient...
Intermolecular forces :
1 newton second per SQUARE metre = Ns `m^(-2) phi` Pascal per second = Pa `s^(-1)`
In CGS system the unit of coefficient of viscosity is poise (named after great scientist Jean Louise Poiseuille).
1 poise `= 1g cm^(-1)s^(-1)`
`= 1xx10^(-1)kg m^(-1)s^(-1)`
Factor affecting on Viscosity :
(i) Temperature : Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.
(ii) Intermolecular forces : Intermolecular forces kept molecules together, strong intermolecular forces between molecules hold them together and resist movement of layers past one another. so viscosity of liquid is depend upon intermolecular forces.
Intermolecular attraction force increases viscosity if liquid decreases and capacity to distribution of layers of liquid decreases.
Viscosity `alpha(1)/("Fluidity of liquid")`
Ex. Glass is an extremely viscous liquid.
However PROPERLY of flow of glass can be experienced by measuring the thickness of window pones of old buildings.
Discussion
47.

Give important physical properties of liquid ?

Answer»

SOLUTION :(i) VAPOUR Pressure
(II) SURFACE tension
(iii) VISCOSITY
Discussion
48.

Give important compounds of calcium.

Answer»

SOLUTION : Important compounds of calcium are CALCIUMOXIDE, calcium hydroxide, calcium sulphate, calcium CARBONATE and CEMENT.
Discussion
49.

Give importance of water for living organisms and write.

Answer»

Solution :A major part of all living organisms is made up of water. Human body has about 65% and some plants have as nuch as 95% water.
It is a crucial compound for the survival of all life forms.
It is a solvent of great importance.
The distribution of water over the earth.s SURFACE is not uniform. The ESTIMATED WORLD water supply is given in following Table.
The high heat of vaporisation and heat capacity are responsible for moderation of the climate and body temperature of living BEINGS.
It is an excellent solvent for transportation of ions and molecules required for plant and animal metabolism.
Discussion
50.

Give (i) structural formula (ii) dash formula and IUPAC name of (CH_(3))_(3) CH_(2)CH (CH_(3))_(2)

Answer»

Solution :(i) Structural formula: `H_(3)C - UNDERSET(underset(CH_(3))(|))OVERSET(overset(CH_(3))(|))(C ) - CH_(2) - underset(underset(CH_(3))(|))(CH_(3)) - CH_(3)`
(ii) Dash structure formula:
(iii) IUPAC Name: 2, 2 4-Trimethylpentane
`underset("but if write 2, 4, 4-It is wrong")(H_(3)overset(1)(C )-underset(underset(CH_(3))(|))overset(overset(CH_(3))(2|))(C ) - overset(3)(CH_(2)) - underset(underset(CH_(3))(|))overset(4)(CH)- overset(5)(CH_(3))`
Discussion