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.

Correct order f keto content for following compound is : Me_(3)C-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CM_(3) Me_(2)CH-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CHMe_(3) MeCH_(2)-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CH_(2)Me

Answer»

`RgtQgtP`
RltQltP
`QgtRgtP`
`QltRltP`

SOLUTION :N//A
Discussion
2.

Correct order for enolic content for a compound :

Answer»

`IGT II gt III`
`III gt II GTI`
`II gt I gtIII`
`I gt III gt II`

SOLUTION :N//A
Discussion
3.

Correct name for the given compound CH_(3)-CH_(2)-underset(CH_(2)CH_(3))underset(|)(C)H-CH_(2)-underset(CH_(3))underset(|)(C)H-CH_(2)-CH_(3) is

Answer»

3-ethyl-5-methylheptane
5-ethyl-3-methylheptane
1,1-diethyl-3-methylpentane
3-methyl-5,5-diethylpentane

Answer :A
Discussion
4.

Correct matching about composition of portland cement is {:("List - I","List - II"),("I) "Al_(2)O_(3),"A) "50-60%),("II) "CaO,"B) "20-25%),("III) "SiO_(2),"C) "5-10%),("IV) "MgO,"D) "2-3%):}

Answer»

`{:("UL(I),ul(II),ul(III),ul(IV)),(C,A,B,D):}`
`{:("ul(I),ul(II),ul(III),ul(IV)),(A,B,C,D):}`
`{:("ul(I),ul(II),ul(III),ul(IV)),(A,C,B,D):}`
`{:("ul(I),ul(II),ul(III),ul(IV)),(C,A,D,B):}`

Answer :A
Discussion
5.

Correct IUPAC name of the following compound is

Answer»

3-(Hepta-2,4,6-trientyl)-4-bromo cyclopenta -2,4, -dien-1-ol
7-(2-Bromo-4-hydroxy cyclopenta-1,4-dienyl) hepta-1,3,5-triene
7-(5-Bromo-3-hydroxy cyclopenta-1,4-dienyl) hepta-1,3,5-triene
3-Bromo-4-(hepta-2,4,6-trienyl) cyclopenta-2,4-dien-1-ol

Answer :D
Discussion
6.

Correct IUPAC name of{:(""CH_3),(""|),(CH_3-CH_2-C=CH-CH-CH_2-CH_3),(""|),(CH_3-CH_2-CH-CH_2-CH_2-CH_2-CH_3):}

Answer»

5,6 - DIETHYL - 3 - methyldec -4- ene
5,6 - Diethyl - 8 - methyldec - 6 - ene
6 - BUTYL - 5 - ETHYL - 3- methyl OCT - 4- ene
2,4,5 - Triethy -3- nonene

Answer :A
Discussion
7.

Correct IUPAC name of

Answer»

5,6-Diethyl-3-methyl dec-4-ene
5,6-Diethyl-8-methyl dec-6-ene
6-Butyl-5-ethyl-3-methyl oct-4-ene
2,4,5-triethyl-3-nonene

ANSWER :A
Discussion
8.

Correct IUPAC name of {:(CH_(3)-CH_(2)-C=CH-overset(CH_(3))overset("|")("C")H-CH_(2)-CH_(3)),("|"),(CH_(3)-CH_(2)-CH-CH_(2)-CH_(2)-CH_(2)-CH_(2)):} is

Answer»

5,6-Diethyl-3-methyl dec-4-ene
5,6-Diethyl-8-methyl dec-6-ene
6-Butyl-5-ethyl-3-methyl oct-4-ene
2,4,5-triethyl-3-nonene

Solution :The IUPAC NAME is 5,6-diethyl-3-methyl-dec-4-ene.
Discussion
9.

Correct IUPAC name for H_(3)C-underset(C_(2)H_(5))underset(|)(CH)-underset(C_(2)H_(5))underset(|)(CH)-CH_(3) is.............. .

Answer»

2-ethyl-3-methylpentane
3, 4-dimethylhexane
2-sec-butylbutane
2, 3-dimethylbutane

SOLUTION :
Discussion
10.

Correct IUPAC name for H_(3)C- underset(underset(C_(2)H_(5))(|))(CH)- underset(underset(C_(2)H_(5))(|))(CH)-CH_(3) is ……..

Answer»

2-ethyl-3-methylpentane
3, 4-dimethylhexane
2 sec-butylbutane
2, 3-dimethylbutane

Solution :Given STRUCTURE is written in longest simple carbon chain is as under:
`overset(1)(C )H_(3)- overset(2)(C )H_(2) - underset(underset(CH_(3))(|))overset(3)(CH)-underset(underset(CH_(3))(|))overset(4)(CH)-overset(5)(CH_(2))-overset(6)(CH_(3))`
Two methyl SUBSTITUTES are PRESENT so it is dimethyl and their positin is on `3^(rd) and 4^(th)` carbon `THEREFORE` Name is 3, 4-dimethylhexane
Discussion
11.

Correct gradation of basic character

Answer»

`NH_(3) GT CH_(3)NH_(2) gt NF_(3)`
`CH_(3)NH_(2) gt NH_(3) gt NF_(3)`
`NF_(3) gt CH_(3)NH_(2) gt NH_(3)`
`CH_(3)NH_(2) gt NF_(3) gt NH_(3)`

SOLUTION :BASIC NATURE `alpha+I`
Discussion
12.

Correct energy value order is

Answer»

`NS NP ND(n-1)F`
`ns np(n-1)d (n-2)f`
`ns np(n-1)d (n-1)f`
`ns(n-1)dn(n-1)f`

SOLUTION :The higher the `(n+1)` value the higher is energy associated with the orbitals.
Discussion
13.

Correct basic strength order is :

Answer»

`R GT q gt p gt s`
`r gt p gt q gt s`
`q gt r gt PGT s`
`r gt q gt s gt p`

Answer :B
Discussion
14.

Copper wire test for halogens is known as

Answer»

Liebig's TEST
LASSAIGNE's test
FUSION test
BEILSTEIN test

SOLUTION :Beilsdtein.s test.
Discussion
15.

Copper sulphate crystals contain 25.45% Cu and 36.07% H_2O. If the law of constant proportions is true, calculate the weight of Cu required to obtain 40g of crystalline copper sulphate.

Answer»


ANSWER :10.18 G
Discussion
16.

Copper reduces NO_3^(-) into NO_2 depending upon concentration of HNO_3 in solution Assuming [Cu^(2+)]=0.1M, "and" P_(NO)=P_(NO_2)=10^(-3)bar, at which concentration of HNO_3, Thermodynamic tendency for reduction of NO_3^(-) into NO and NO_2 by copper is same ? Given: E_(cu^(2+)|cu)^(@)=+0.34 "volt", E_(NO_3^(-)|NO)^(@)=+0.96 "volt",E_(NO_3^(-)|NO_(2))^(@)=+0.76 "volt"]

Answer»

`10^(1.23)M`
`10^(0.56)M`
`10^(0.66)M`
`10^(0.12)M

Answer :C
Discussion
17.

Copper reacts with nitric acid . A brown gas is formed and the solution turns blue. The equation may be written as : Cu+NO_3^(-) rarr NO_2+Cu^(2+) Balance the equation by oxidation number method.

Answer»

Solution :Step 1 : Skeleton EQUATION
`Cu+NO_3^(-) rarr NO_2+Cu^(2+)`
Step 2 : Writing oxidation numbers of each atom
`overset(0)(Cu)+overset(+5-2)(NO_3^(-))rarroverset(+4-2)(NO_2)+overset(+2)(Cu^(2+))`
Step 3.

The oxidation number copper has increased from 0 to 2 while that of nitrogen has decreased from +5 to +4.
Step 4 . Show the increases / DECREASE of oxidation number O.N. increases by 2 per atom

Step 5 : Balance the increased /decrease in oxidation number by multiplying `NO_3^(-)` by 2 and Cu by 1.
`Cu+2NO_3^(-) rarr NO_2+Cu^(2+)`
Step 6. Balance other atoms except H and O as
`Cu+2NO_3^(-) rarr 2NO_2+Cu^(2+)`
Step 7. Reaction TAKES place in acidic medium , so add `H^+` ions to the side deficient in `H^+` and blance H and O atoms :
`Cu+2NO_3^(-) +4H^(+) rarr 2NO_2+Cu^(2+) + 2H_2O`
Discussion
18.

Copper oxide test is used to detect ....

Answer»

Carbon & HYDROGEN
Oxygen & NITROGENS
Nitrogen & Sulphur
Phosphorous & Chlorine

SOLUTION :Carbon & Hydrogen
Discussion
19.

Copper oxide test is used to detect ............. .

Answer»

SOLUTION :CARBON &HYDROGEN
Discussion
20.

Copper is the most noble of first row transition metals and occurs in small deposits in several countries. Ores of copper include chalanthrite. (CuSO_(4).5H_(2)O), atacanrite [Cu_(2)Cl(OH)_(3)] cuprite (Cu_(2)O), copper glance (Cu_(2)S) nad malachite [Cu_(2)(OH)_(2)CO_(3)]. However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS_(2)). The extraction of ore chalcopyrite (CuFeS_(2)). The extraction of copper from chalcopyrite involves partial roasting, removal of iron nad selfreduction. In self reduction, hte reducing spent is

Answer»

S
`O^(2-)`
`S^(2-)`
`SiO_(2)`

SOLUTION :N//A
Discussion
21.

Copper is the most noble of first row transition metals and occurs in small deposits in several countries. Ores of copper include chalanthrite. (CuSO_(4).5H_(2)O), atacanrite [Cu_(2)Cl(OH)_(3)] cuprite (Cu_(2)O), copper glance (Cu_(2)S) nad malachite [Cu_(2)(OH)_(2)CO_(3)]. However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS_(2)). The extraction of ore chalcopyrite (CuFeS_(2)). The extraction of copper from chalcopyrite involves partial roasting, removal of iron nad selfreduction. Iron is removed from chalcopyrite as

Answer»

FeO
FeS
`Fe_(2)O_(3)`
`FeSiO_(3)`

SOLUTION :N//A
Discussion
22.

Copper is the most noble of first row transition metals and occurs in small deposits in several countries. Ores of copper include chalanthrite. (CuSO_(4).5H_(2)O), atacanrite [Cu_(2)Cl(OH)_(3)] cuprite (Cu_(2)O), copper glance (Cu_(2)S) nad malachite [Cu_(2)(OH)_(2)CO_(3)]. However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS_(2)). The extraction of ore chalcopyrite (CuFeS_(2)). The extraction of copper from chalcopyrite involves partial roasting, removal of iron nad selfreduction. Partial roasting of chalcopyrite produces

Answer»

`Cu_(2)S` and FEO
`Cu_(2)O` and FeO
CuS and `Fe_(2)S_(3)`
`Cu_(2)O` and `Fe_(2)O_(3)`

SOLUTION :N//A
Discussion
23.

Copper is purified by electrolysis in water solution and most of the impurities in the blister copper are left behind in solution. Why can not aluminium be unpaired in a similar manner ?

Answer»

Because AL ions are not PRODUCED in solution
Because Al ions are not reduced by electrolysis in a WATER solution
Because very few Al ions are reduced in water solution
None of the above

Answer :B
Discussion
24.

Copper (I) is diamagnetic whereas Copper (II) is paramagnetic. Give reason

Answer»

Solution :CU (I) has `d^(10)` CONFIGURATION, i.e., all ELECTRONS are paired WHEREAS Cu (II) has `d^(9)` configuration, i.e., has no unpaired electron
Discussion
25.

Copper forms two oxides following law of variable proportions. One gram of each oxide in hydrogen gas gave 0.799 g and 0.888 g of the metal respectively. Give the composition of these oxides.

Answer»

SOLUTION :OXIDE (A) : weight of copper =0.799g
weight of oxygen =0.201 G
weight of metal per gram of oxygen =3.97g
Oxide (B): WT of copper =0.888g
weight of oxygen =0.112 g
weight of metal per gram of oxygen =7.93g
The ratio of weights of METALS reacting with same weight of oxygen =1:2
The oxides are CuO and `Cu_(2)O`
Discussion
26.

Copper forms two oxide Cu_(2)O_(x) and Cu_(2)O_(y). For the same amount of metal,twice as much oxygen was used to form first oxide than to second oxide. What is the ratio of X and Y?

Answer»


ANSWER : 2:1
Discussion
27.

Copper dissolves in dilute nitric acis but not in dilute HCI Explain

Answer»

Solution :Since `E^(@)` of `Cu^(2+)//Cu` electrode (+0.34 V) is higher than that of `H^(+)//H_(2)` electrode (0.0V) therefore `H^(+)` ions cannot oxidise Cu to `Cu^(2+)` ions and hence Cu does not dissolve in dil HCI in CONTRAST the electrode potential of `NO_(3)^(-)//NO` electrode (+0.97 V) is HIGER than that of COPPER electrode and hence it can oxidise Cu to `Cu^(2+)` ions and hence Cu dissolve in dil `HNO_(3)` thus cu dissolves in dil `HNO_(3)` DUE to oxidation of Cu by `NO_(3)^(-)` ions and not by `H^(+)` ions
Discussion
28.

Copper dissolves in dilute HNO_3 but not in dilute HCl. Explain.

Answer»

Solution :Since `E^@` of `Cu^(2+)|Cu` electrode ,(+0.34V) is higher than that of `H^+|1/2H_2` (0.0V) , therefore , `H^+` IONS cannot oxidise Cu to `Cu^(2+)` ions and consequently, COPPER does not dissolve in dil . HCl on the other hand , the electrode potential of `NO_3^(-)` ion, `NO_3^(-) |NO(+0.97V)` is higher than that of copper electrode and hence , it can oxidise `Cu" to "Cu^(2+)` and hence copper DISSOLVES in dil . `NHO_3^(-)`.
Discussion
29.

Copper crystallizes into a fcc lattice with edge length 3.61xx10^(-8) cm. Show that the calculated density is in agreement with its measured value of 8.92 "g cm"^(-3).

Answer»

SOLUTION :`rho=(ZxxM)/(a^3xxV_0)`
For fcc LATTICE of copper , Z=4
Atomic mass of copper , `M=63.5 "g MOL"^(-1)`
`therefore rho=(4xx63.5 "g mol"^(-1))/((3.61xx10^(-8) cm)^3xx(6.022xx10^23 "mol"^(-1)))=8.97 "g cm"^(-3)`
which is in CLOSE agreement with the measured value.
Discussion
30.

Copper crystallizes into a fcc lattice with edge length3.61 xx 10^(-8) cm, show that the calculated density is in agrrement with its measured value of8.92 " g cm"^(-3) .

Answer»

Solution : ` p = ( Z xxM)/(a^(3) xx V_(0))`
For FCC lattice of copper , Z =4Atomic mass of copper, M = ` 63.5 " G mol"^(-1)`
` p = (4 xx 63.5 " g mol"^(-1))/((3.61 xx 10^(-8) "cm")^(3) xx ( 6.022 xx 10^(23)"mol"^(-1))) =8.97 " g cm"^(-3)`
which is in CLOSE agreement with the MEASURED with the measured value .
Discussion
31.

Copper crystallises into a fee lattice. Its edge length is 3.62 xx 10^(-8) cm. Calculate the density of copper (atomic mass of Cu=63-5 u, N_A= 6-022 xx 10^(23) mol^(-1)).

Answer»


SOLUTION :For FCC, `r=a/(2sqrt2) or a=2sqrt2r=2xx1.414xx128` pm =362 pm
`rho=(ZxxM)/(a^3xxN_0)=(4xx63.5 "g mol"^(-1))/((362xx10^(-10)cm)^3xx6.02xx10^23 mol^(-1))=8.9 "g cm"^(-3)`
Alternatively, `rho="Mass of unit CELL "/"Volume of unit cell "=(4xx(63.5 "g mol"^(-1)//6.02xx10^23 mol^(-1)))/((362xx10^(-10)cm)^3)=8.9 g cm^(-3)`
Discussion
32.

Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is

Answer»

108 pm
128 pm
157 pm
181 pm

Answer :B
Discussion
33.

Copper crystal has a face-centred cubic lattice structure. Atomic radius of copper atom is 128 pm. Calculate the density of copper. Atomic mass of copper=63.5

Answer»


Solution :For FCC, `r=a/(2sqrt2) or a=2sqrt2r=2xx1.414xx128` pm =362 pm
`RHO=(ZxxM)/(a^3xxN_0)=(4xx63.5 "g MOL"^(-1))/((362xx10^(-10)cm)^3xx6.02xx10^23 mol^(-1))=8.9 "g cm"^(-3)`
ALTERNATIVELY, `rho="MASS of unit cell "/"Volume of unit cell "=(4xx(63.5 "g mol"^(-1)//6.02xx10^23 mol^(-1)))/((362xx10^(-10)cm)^3)=8.9 g cm^(-3)`
Discussion
34.

Copper crystal has a face -centred cubic lattice structure. Atomic radius of copper atom is 128 pm. Calculate the density of copper . Atomic mass of copper = 63.5 orCopper crystallises into a fcc lattice, its edge lengthis3.62 xx 10^(-8)cm. Calculate the dinsity of copper(atomic mass of Cu= 63.5 u,N_(A) = 6.022 xx 10^(23)"mol"^(-1))

Answer»


Solution :For fcc , `r= a/sqrt2 or a = 2sqrt2r = 2xx 1.414 xx 128"pm"` = 362 pm
`p = (Z xx M)/(a^(3) xx N_(0)) = ( 4xx 63.5 "g MOL"^(-1))/((362xx10^(-10)"cm")^(3) xx (6.02 xx 10^(23)"mol"^(-1)))= 8.9"cm"^(-3)`
Alternatively, `p = ( "MASS of unit cell")/("volume of unit cell") = ( 4xx(63.5 "g mol"^(-1)//6.02xx10^(23)"mol"^(-1)))/((362xx 10^(-10)"cm")^(3))= 8.9 "g cm"^(-3)`
Discussion
35.

Copper becomes green when exposed to moist air for a long period. This is due to:

Answer»

the FORMATION of a layer of CUPRIC oxide on the surface of copper.
the formation of a layer of basic carbonate of copper on the surface of copper.
the formation of a lyer of cupric HYDROXIDE on the surface of copper.
the formation of basic copper sulphate layer on the surface of the metal.

Answer :B
Discussion
36.

Copper and chlorine compounds makes blue fire work. Why?

Answer»

Solution :To produce colours, fire works EXPERTS BURN the METAL and chlorine together in a vapour, where the two elements are gases instead of solid.
(ii) The burning excites the electron pushing them into a higher than normal the electrons returns to their normal level, they release their extra energy as a colour BURST of light.
(iii) True blue fireworks are the hardest to make since the compound copper breaks down in a HOT flame
Discussion
37.

Coordination number of Zn in ZnS (Zine blende) is

Answer»

4
6
2
None of these

Solution :In ZNS, `S^(2-)"ions ADOPT f.c.c. lattice in which " Zn^(2+)" ions are present in alternate tetrahedral sites "`.
`THEREFORE" Co-ordination number of " S^(2-) = " Co-ordination number of " Zn^(2+) = 4`
Discussion
38.

Coordination number of (K) is:

Answer»

2
4
3
6

Solution :(A) `RARR HCOONa` (B) `Na_(2)C_(2)O_(4)` (D) `Na_(2)CO_(3)`
(F) `Na_(2)S_(2)O_(3)` (H) `[Bi(S_(2)O_(3))_(3)]^(3-)`
(X) `rarr HCOOH (C) H_(2)` (E) `Na_(2)SO_(3)`
(G) `Bi_(2)(S_(2)O_(3))_(3)` (I) `Ag_(2)S_(2)O_(3)` (J) `Ag_(2)S` (K) `[Ag(S_(2)O_(3))_(2)]^(3-)`
Discussion
39.

Coordination number of cation is minimum in

Answer»

NaCI
CsCI
ZnO
KCI

Answer :C
Discussion
40.

underset(COOH)overset(COOH)(|) and underset(COOH)overset(COOK)(I) behave is acids as well as reduing agents. Then which of the following are correct statements?

Answer»

When behaves as reducing agent, then its equivalent weights are EQUAL to HALF of its MOLECULAR weight respectively
1000 ml of 1N SOLUTION of each is neutralized by 1000 mL 1N `CA(OH)_2`
1000 mL of 1M solution is neutralized by 1000 mL of IM `Ca(OH)_2`
1000 mL of 1M solution is neutralized by 200 mL 2M of `KMnO_4`in acidic medium.

Solution :For `{:(COOH),(|),(COOH):}&{:(COOK),(|),(COOH):}`when behve as reducing agent `(COOH)_2 rarr2CO_2` n - factor = 2, eq wt. `M//2 {:(COOK),(|),(COOH):} rarr 2CO_2` n - factor = 2, eq wt. = M/2 for neutralization meq of one reactant = meq of other reactant can be neutralized `1000 xx 2 = 1000 xx 2 `
Discussion
41.

Convert the following pressure in to atmosphere.(a) 735 torr (b) 985 mL bar (c ) 1.42xx10^(5)Nm^(-2)[Note : (a) 1 atm = 760 torr, (b) 1 atm = 1.013 bar(c ) 1 atm = 1.01325xx10^(5)N m^(-2)

Answer»


ANSWER :(a) 0.967 ATM(B) 0.972 atm (C ) 1.40 atm
Discussion
42.

Convert the following into specified units : (a) 100 s into ns

Answer»

Solution :(a) `therefore 1s = 10^(9)` ns
The CONSERVATION factor would be:
`1 = (10^(9) ns)/(ns)`
`100s = 100 s xx 1 = 100 s xx (10^(9)ns)/(1s) = 1.0 xx 10^(11) ns`
(b) `therefore 1 km = 10^(3) m`
`therefore` The conversion factor would be
`1 = (10^(3) m)/(1 km)`
`therefore 60 km = 60 km xx 1 = 60 km xx (10^(3) m)/(1 km)`
`=6.0 xx 10^(4) m`
( c) `therefore 1 kg = 10^(3) g` and `1 g = 10^(3) mg`
1kg = `10^(3) xx 10^(3) = 10^(6) mg`
Hence, the conversion factor would be
`1=(10^(6) mg)/(1 kg)`
`=2.00 xx 10^(8) mg`
(d ) `therefore 1 cm = 10^(-2) m`
`therefore 1 cm^(3) = 10^(-2) xx 10^(-2) xx 10^(-2) m^(3) = 10^(-6) m^(3)`
Hence, the conversion factor would be:
`1 = (10^(-6) m^(3))/(1 cm^(3))`
`1000 cm^(3) = 1000 cm^(3) xx1`
`=1000 cm^(3) xx (10^(-6) m^(3))/(1 cm^(3))`
(e ) `therefore 1 pm = 10^(-12)m` and `1m = 10^(8) nm`
`therefore 1pm = 10^(-12) xx 10^(9) = 10^(-3) nm`
( f) `therefore 1G = 10^(-3) g`
`therefore` The conversion factor would be:
`1=(10^(-3) kg)/(1kg)`
`therefore 75 g = 75 g xx 1 = 75 g xx (10^(-3) kg)/(1 kg)`
`=(7.5 xx 10^(-2)) kg`
(g) `therefore 1 mum = 10^(-6) m` and `1m = 10^(2) cm`
`1mu m =10^(-6) xx 10^(-2) = 10^(-4) cm`
Hence, the conversion factor is:
`1 = (10^(-4) cm)/(1 mum)`
`50 mum = 50 mum xx 1 = 50 mum xx (10^(-4) cm)/(1 mum)`
`=5.0 xx 10^(-3) cm`
Discussion
43.

Convert the following into specified units. (a) 10 years into hours (b) 15 metric tonnes into milligrams.

Answer»


ANSWER :(a) `8.766 xx 10^(4)` HOURS, (B) `.15 xx 10^(10)` MG
Discussion
44.

Convert the following into basic units: (i) 28.7 pm(ii) 15.15 pm (iii) 25365 mg

Answer»

SOLUTION :(i) `28.7 pm = 28.7 "pm" xx (10^(-12) m)/(1 "pm") = 2.87 xx 10^(-11) m`
(ii) `15.15 mu s = 15.15 mus xx (10^(-6) s)/(1 mu s) = 1.515 xx 10^(-5) s`
(III) 25365 MG = `25365 xx (1 g)/(1000 mg) xx (1 kg)/(1000 g) = 2.5365 xx 10^(-2) kg`
Discussion
45.

Convert the following into basic units : (i) 28.7 pm (ii) 15.15 pm(iii) 25365 mg

Answer»



SOLUTION :(i) `28.7 "pm" = 28.7 "pm" xx (10^(-12)m)/(1 "pm")`
`=2.87xx10^(-11)m`
(ii) `15.15 "pm"xx(10^(-12)s)/(1"pm")=1.515xx10^(-11)s`
(III) `25365 mgxx (1g)/("1000 mg")xx(1kg)/(1000g)=2.5365xx10^(-2)kg`
Discussion
46.

Convert the following into basic units : (1) 37.6pm(2) 25.50 pm(3)75325 mg

Answer»




ANSWER :(1) `3.76xx10^(-11)m, (2) 2.550xx10^(-5)s , (3) 7.5325xx10^(-2) KG`
Discussion
47.

Convert methane to methylene chloride

Answer»

Solution :`UNDERSET("Methane") (CH_(4)) overset(Cl_(2)//hv) underset(-HCl) (to) CH_(3) Cl overset(Cl_(2)//hv)underset(HCl) (to) underset("Methylene CHLORIDE") (CH_(2) Cl_(2))`
Discussion
48.

Convert ethyne to benzene and name the process

Answer»

SOLUTION :CONVERSIONOF Ethylne intoBenzene

thisprocesis ONEOF the cyclicpolymerisationprocess.
Discussion
49.

Convert bromoethane to ethane .

Answer»

Solution :`underset("BROMOETHANE") (CH_(3) CH_(2) Br) + H_(2) OVERSET("Ni (or) Pd") underset(523 K) (to) underset("Ethane") (CH_(3) - CH_(3)) + HBr`
Discussion
50.

Convert 40 calories into joules.

Answer»


SOLUTION :1 calorie = 4.184 J `:.` CONVERSION FACTOR `= ((4.184J))/(("1 calorie"))`
40 calories = (40 calories) `XX` (conversion factor) = (40 calories) `xx (("4.184 J"))/(("1 calorie"))=167.36J`.
Discussion