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28751.

The main product of the following reaction is C_(6)H_(5)CH_(2)CH(OH)CH(CH_(3))_(2)overset("conc. H_(2)SO_(4))to

Answer»




Solution :`C_(6)H_(5)-CH_(2)-underset(OH)underset(|)CH-overset(CH_(3))overset(|)CH-CH_(3)underset(-H_(2)O)overset("conc." H_(2)SO_(4)(H^(+)))to C_(6)H_(5)-CH_(2)-overset(o+)CH-overset(CH_(3))overset(|)CH-CH_(3)underset("(REARRANGEMENT)")overset(1,2H(Ө)"shift")to C_(6)H_(5)-overset(+)CH-CH_(2)-overset(CH_(3))overset(|)CH-CH_(3)to ` TRANS + cis.
Trans is major product due to thermodynamically more stable as well as carbocation show anti elimination.
Discussion
28752.

The molar conductances of NaOH, NaCl and BaCl_(2) at infinite dilution are 2.481xx10^(-2), 1.265xx10^(-2) and 2.800xx10^(-2)" S "m^(2)mol^(-1) respectively. Calculate wedge_(m)^(@)Ba(OH)_(2).

Answer»


SOLUTION :`wedge_(m)^(@)[Ba(OH)_(2)]=wedge_(m)^(@)(BaCl_(2))+2wedge_(m)^(@)(BaCl_(2))+2wedge_(m)^(@)(NAOH)-2wedge_(m)^(@)(NACL)`.
Discussion
28753.

The main product of the following reactions is C_6H_5CH_2CH( OH) CH(CH_3) _2 overset( conc. H_2SO_4) to

Answer»

` (##AKS_NEO_CAO_CHE_XII_V02_P03_APP_E02_044_O01.png" WIDTH="30%">
` (##AKS_NEO_CAO_CHE_XII_V02_P03_APP_E02_044_O02.png" width="30%">
` (##AKS_NEO_CAO_CHE_XII_V02_P03_APP_E02_044_O03.png" width="30%">

ANSWER :B
Discussion
28754.

The molar conductivitiesLambda_(NaOAc)^(0) and Lambda_(HCl)^(0) at infinitedilute in water at25^(@) Care 91.0 S cm^(2)//"mol" and 426.2 S cm^(2)//"mol" respectively . To calculate Lambda_(HOAc)^(0) the additional value required is

Answer»

`Lambda_(NaCl)^(@)`
`Lambda_(H_(2)O)^(@)`
`Lambda_(KCL)^(@)`
`Lambda_(NAOH)^(@)`
Discussion
28755.

The molar conductances of NaCl,HCl and CH_(3)COONa at infinite dilution are 126.45, 426.16 and 91ohm^(-1)cm^(2)mol^(-1) respectively. The molar conductance ofCH_(3)COOH at infinite dilution is:-

Answer»

`201.28ohm^(-1)cm^(2)mol^(-1)`
`390.71ohm^(-1)cm^(2)mol^(-1)`
`698.28ohm^(-1)cm^(2)mol^(-1)`
`540.48ohm^(-1)cm^(2)mol^(-1)`

Solution :`(^^)_(m)^(@)(CH_(3)COOH)`
`=(^^)^(@)(CH_(3)COONA)+(^^)^(@)(HCL)-(^^)^(@)(NaCl)`
`=91+426.16-126.45=390.71ohm^(-1)cm^(2)mol^(-1)`
Discussion
28756.

The main product of the following reaction R-COOH+CH_(2)N_(2)toProduct

Answer»

`R-CONH_(2)`
`R-CN`
`R-COOCH_(3)`
`R-COONH_(4)`

Solution :`RCOOH+CH_(2)N_(2) overset)(ether)toRCOOCH_(3)+N_(2)`
Discussion
28757.

The molar conductance of NzCl varies with the concentration as shown in the following table and all values follows the equation lamda_(m)^(C)=lamda_(m)^(infty)-bsqrt(C) where lamda_(m)^(C)= molar specific conductance lamda_(m)^(infty)= molar specific conductance at infinite dilution C= molar concentration When a cenrtain conductivity cell (C) was filled with 25xx10^(-4)(M) NaCl solution. The resistance of the cell was found to be 1000 ohm At infinite dilution, Conductance of Cl^(-) and SO_(4)^(-2) are 80ohm^(-1)cm^(2)"mole"^(-1) and 160ohm^(-1)cm^(2)"mole"^(-1) respectively. Q. If the cell (C) is filled with 5xx10^(-3)(N)Na_(2)SO_(4) the observed resistance was 400 ohm. what is the molar conductance of Na_(2)SO_(4)

Answer»

`19.25ohm^(-1)cm^(2)"mole"^(-1)`
`96.25ohm^(-1)cm^(2)"mole"^(-1)`
`385ohm^(-1)cm^(2)"mole"^(-1)`
`192.5ohm^(-1)cm^(2)"mole"^(-1)`

SOLUTION :For `Na_(2)SO_(4)` solution
`K=((l)/(a))xx(1)/(R)=(0.1925)/(400)=4.81xx10^(-4)ohm^(-1)cm^(-1)`
`lamda_(m)=(Kxx1000)/(M)=(4.81xx10^(-4)xx1000)/((5)/(2)xx10^(-3))`
`lamda_(m)(Na_(2)SO_(4))=192.4ohm^(-1)cm^(2)"mole"^(-1)`
Discussion
28758.

The main product of the following reaction is C_6H_5CH_2CH(OH)CH(CH_3)_2 overset("conc". H_2SO_4)to ?

Answer»




ANSWER :B
Discussion
28759.

The molar conductance of NaCl varies with the concentration as shown in the following table. And all values follows the equation lambda_m^C=lambda_m^oo-bsqrtC Where lambda_m^C=Molar specific conductance lambda_m^oo=Molar specific conductance at infinite direction C=Molar concentration {:("Molar Concentration of NaCl","Molar Conductance inohm"^(-1)cm^(2)"mole"^(1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):} When a certain conductivity cell (C ) was filled with 25xx10^(-4) (M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At infinite dilution, conductance of Cl^(-) and SO_(4)^(-2) are 800 "ohm"^(-1) cm^(2)"mole"^(-1) and 160 "ohm"^(-1) cm^(2)"mole"^(-1) respectively. If the cell ( C) is filled with 5xx10^(-3) (N)Na_2SO_4 the observed resistance was 400 ohm. What is the molar conductance of Na_2SO_4 ?

Answer»

`19.25 ohm^(-1) cm^(2) "mole"^(-1)`
`96.25 ohm^(-1) cm^(2) "mole"^(-1)`
`385 ohm^(-1) cm^(2) "mole"^(-1)`
`192.5 ohm^(-1) cm^(2) "mole"^(-1)`

SOLUTION :For `Na_2SO_4` solution
`K=(l/a)xx1/R=0.1925/400=4.81xx10^(-4) ohm^(-1)cm^(-1)`
`lambda_m=(Kxx1000)/M=(4.81xx10^(-4)xx1000)/(5/2xx10^(-3))`
`lambda_m(Na_2SO_4)=192.4 ohm^(-1) cm^2 "mole"^(-1)`
Discussion
28760.

The main product of the following reaction is C_(6)H_(5)CH_(2)CH(OH)CH(CH_(3))_(2) overset("conc. "H_(2)SO_(4))to?

Answer»




SOLUTION :
Dehydration occurs in such a way that the DOUBLE bond comes in conjugation with the `C_(6)H_(5)` group and further tans-alkenes are more stable than the cis-alkenes, therefore, OPTION (c) is correct.
Discussion
28761.

The molar conductance of NaCl varies with the concentration as shown in the following table. And all values follows the equation lambda_m^C=lambda_m^oo-bsqrtC Where lambda_m^C=Molar specific conductance lambda_m^oo=Molar specific conductance at infinite direction C=Molar concentration {:("Molar Concentration of NaCl","Molar Conductance inohm"^(-1)cm^(2)"mole"^(1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):} When a certain conductivity cell (C ) was filled with 25xx10^(-4) (M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At infinite dilution, conductance of Cl^(-) and SO_(4)^(-2) are 800 "ohm"^(-1) cm^(2)"mole"^(-1) and 160 "ohm"^(-1) cm^(2)"mole"^(-1) respectively. What is the cell constant of the conductivity cell ( C) ?

Answer»

`0.385 CM^(-1)`
`3.85 cm^(-1)`
`38.5 cm^(-1)`
`0.1925 cm^(-1)`

SOLUTION :For `25xx10^(-4)` (M) NaCl solution
`lambda_m=lambda_m^oo-bsqrtC`
`lambda_m=127-10^3(25xx10^(-4))^(1//2)`
`lambda_m=127-10^(3)xx5xx10^(-2)`
`lambda_m=77`
But `lambda_m=(Kxx1000)/M " " K=(l/a)xx1/R`
`lambda_m=(l/a)xx1/Rxx1000/M`
`lambda_m`=[Cell CONSTANT]`xx1000/(RxxM)`
`implies `77=[Cell constant]`xx1000/(1000xx25xx10^(-4))`
Cell constant `=77xx25xx10^(-4)=0.1925 cm^(-1)`
Discussion
28762.

The main product of the following reaction is

Answer»

`C_6H_5-OVERSET(OH)overset(|)CH_2CI`
`C_6H_5-overset(CI)overset(|)CH_2OH`
`C_6H_5CH_2OH` and `CH_3OH`
NONE of these

Solution :NA
Discussion
28763.

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation lambda_m^( C)=lambda_m^(oo)-bsqrtC where lambda_(m)^( C)=molar specific conductance lambda_(m)^(oo)=molar specific conductance at infinite dilution C=molar concentration {:("Molar Concentration of NaCl","Molar Conductance" "In" "ohm"^(-1)cm^(2)"mole"^(-1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):} When a certain conductivity cell (C) was filled with 25xx10^(-4)(M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At Infinite dilution, conductance of Cl^(-) and SO_4^(-2) are 80 ohm^(-1) cm^(2) "mole"^(-1) and 160 ohm^(-1) cm^2 "mole"^(-1) respectively. If the cell ( C) is filled with 5xx10^(-3)(N)Na_(2)SO_(4) the obserbed resistance was 400 ohm.What is the molar conductance of Na_(2)SO_(4).

Answer»

`19.25 "OHM"^(-1)cm^2"mole"^(-1)`
`96.25 "ohm"^(-1)cm^2"mole"^(-1)`
`385 "ohm"^(-1)cm^2"mole"^(-1)`
`192.5 "ohm"^(-1)cm^2"mole"^(-1)`

Solution :`SIGMA=(Mlambda)/(1000 G)=(MlambdaR)/1000implies lambda=(1000sigma)/(MR)`
`lambda=((1000xx0.1925)/(5XX10^(-)3))/2xx400=192.5`
Discussion
28764.

The main product of following reaction will be:

Answer»




SOLUTION :
Discussion
28765.

The molar concentration of HCl (aq.) is 10^(-5)M. Which of the following statement are correct (d_("solution") = 1 gm/cc)

Answer»

The MOLE fraction of `HCl ~= 1.8 xx 10^(-7)`
The CONCENTRATION of HCl in PPM is 3.65 ppm
The MOLALITY of HCl solution is approximately `10^(-5) m`
The (w/v)% of solution is `3.65 xx 10^(-5) %`

ANSWER :A::C::D
Discussion
28766.

The main product of the following reaction is

Answer»

`C_6H_5-overset(OH)overset(|)CH-overset(NHCH_(3))overset(|)CH_2`
`C_6H_5-overset(NHCH_(3))overset(|)CH-overset(OH)overset(|)CH_2`
`C_6H_5-overset(OH)overset(|)CH-overset(NHCH_(3))overset(|)CH_2`
`C_6H_5-overset(CH_(2)NH_2)overset(|)CH-overset(OH)overset(|)CH_2`

Solution :NA
Discussion
28767.

The molar conductance at infinite dilution of AgNO_(3), NaCland NaNO_(3)are 116.5, 110.3 and 105.2 ohm^(-1)cm^(2)"mole"^(-1) respectively. The molar conductance of at infinite dilution is

Answer»

111.4
130.6
121.6
150.2

Answer :C
Discussion
28768.

The main product of reduction of nitrobenzene with lithium aluminium hydride is

Answer»

Azoxybenzene
Aniline
Azobenzene
Diazonium salt

Solution :
Discussion
28769.

The molality of a urea solution in which 0.0100 g of urea, [(NH_(2))_(2)CO] is added to 0.3000 dm^(3) of water at STP is :

Answer»

`5.55xx10^(-4)m`
`33.3` m
`3.33xx10^(-2)`m
0.555 m

Solution :Density of water at `25^(@)C=1`g/ML
VOLUME of water `=0.3xx1000=300` mL
Mass of water `=300xx1=300g`
Moles of urea `=(0.01)/(60)`
MOLALITY `=(0.01xx1000)/(60xx300)=5.55xx10^(-4)`m
Discussion
28770.

The main product of bromination of acetanilide in glacial acetic acid is

Answer»

o-bromoaniline
p-bromoaniline
p-bromoacetanilide
2, 4, 6-tribromoacetanilide.

Answer :B
Discussion
28771.

The molality of a sulphuric acis solution in which the mole fraction of water is 0.86 is

Answer»

Solution :Mole FRACTION of `H_(2)SO_(4)` in the solution
`=1-0.86=0.14`
`(n_(2))/(n_(1)+n_(2))=0.14`
MOLALITY is `n_(2)` in 1000 g of water, i.e., in `n_(1)`
`=1000//18=55.55` moles
`THEREFORE (n_(2))/(55.55+n_(2))=0.14"or"n_(2)=0.14n_(2)+7.777`
`"or"0.86n_(2)=7.777 or n_(2)=9.0`
Discussion
28772.

The molality of a sulphuric acid solution is 0.6 'mol//kg'. The total weight of the solution which contains 1 kg of solvent.

Answer»

1000 g
980.3 g
1058.8 g
1013.3 g

Answer :C
Discussion
28773.

The main product obtained when a solution of sodium carbonate reacts with mercuric chloride is

Answer»

`Hg(OH)_(2)`
`HgCO_(3).HgO`
`HgCO_(3)`
`HgCO_(3).Hg(OH)_(2)`.

Solution :Mercuric chloride when HEATED with sodium carbonate solution, a yellow precipitate of mercuric oxide is formed.
`HgCl_(2)+Na_(2)CO_(3)toHgO+2NaCl+CO_(2)`
Discussion
28774.

The molality of a 1 molarsolution will be (Given, density of solvent = 1.5 kg .dm^(-3)

Answer»

1.5
1
0.66
10.72

Solution :`m XX d = M or m = M/d=1/1.5=0.66`
Discussion
28775.

The molality of a sulphuric acid aqueous solution in which the mole fraction of water is 0.85 is

Answer»

9.8
10.58
6.5
11.25

Solution :`(X_(H_(2)SO_(4)))/(X_(H_(2)O)) = (0.15)/(0.85) = (n_(H_(2)SO_(4)))/(n_(H_(2)O))`
`(n_(H_(2)SO_(4)))/(n_(H_(2)O)) = (0.15)/(0.85 XX 18)`
Molality `= (0.15 xx 1000)/(0.85 xx 18) = 9.8 M`
Discussion
28776.

The molality of a solution containing 0.1 mol of a substance in 100g of water is :

Answer»

0.5 m
0.1 m
1 m
0.0180 m

Solution :MOLARITY `=(0.1xx1000)/(100)=1M`
Discussion
28777.

The main product obtained by the electrolysis of the aqueous ethanolic solution of potassium bromide and sodium carbonate, is

Answer»

ETHYL bromide
Bromoform
Ethylene bromide
Ethylidene bromide

Answer :B
Discussion
28778.

The molarity of 98% by wt. H_(2)SO_(4) (d = 1.8 g/ml) is

Answer»

6 m
18 m
10 m
4 m

Answer :B
Discussion
28779.

The main product in the reactionHCONHRoverset (POCl_3) underset ("pyridine") rarr?

Answer»

RCN
RNC
RCNO
RNCO

Answer :B
Discussion
28780.

The molality of 3M solution os mehtanol if thedensity of the solution is 0.9g cm^(-3), is

Answer»

3.73
`3.0`
3.33
3.1

SOLUTION :`"Volume of the solution="1 L or 1000 cm^(3)`
`"Mass of the solution "=vxxd `
`= (1000CM(3)) xx (0.9 g cm^(3))=900 g`
`"Mass of methanol"(CH_(3)OH)=(3 molxx32 "g mol"^(-1))=96 g`
`"Mass of solvent" =(900-96g)=804 g=0.804 KG `
`"Molality of solution"=("No. of moles of methanol")/("Mass of solvent in kg")`
`(3mol)/(0.804kg)=0.373"mol kg"^(-1)`
=.373 MOLAL
Discussion
28781.

The main product in the dehydration of : (CH_(3))_(3) C underset(OH)underset(|)(CH)CH_(2)CH_(3) in the presence of conc. H_(2)SO_(4) at 170^(@)C is :

Answer»

`(CH_(3))_(3) C CH = CHCH_(3)`
`(CH_(3))_(2)C = underset(CH_(3))underset(|)(C)CH_(2)CH_(3)`
`(CH_(3))_(2)CH underset(CH_(3))underset(|)(C) = CHCH_(3)`
`(CH_(3))_(3)C CH_(2)CH = CH_(2)`

Solution :The intermediate CARBOCATION `(CH_(3))_(3)C - underset(oplus)(CH) - CH_(2) - CH_(2)` UNDERGOES REARRANGEMENT by methyl shift to form stable tertiary carbocation as :
`CH_(3) - underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(C) - underset(oplus)(CH) - CH_(2) - CH_(3) overset("Rearrangement")harr CH_(3) - underset(oplus)overset(CH_(3))overset(|)(C) - underset(CH_(3))underset(|)(CH) - CH_(2) - CH_(3)`
This carbonium ion loses `H^(+)` to form `(CH_(3))_(2) C = underset(CH_(3))underset(|)(C) - CH_(2) - CH_(3)` as the MAIN product.
Discussion
28782.

The molal freezing point depression constant of benzene (C_(6)H_(6)) is "4.90 K kg mol"^(-1). Selenium exists as a polymer of the type Se_(x). When 3.26 g of selenium is dissolved in226 g of benzene the obsereved freezing point is 0.112^(@)C lower than for pure benzene. Deduce the molecular formula of selenium ("At. mass of Se =78.8 g mol"^(-1)).

Answer»

SOLUTION :`Se_(8)`
Discussion
28783.

The main product in the following reaction is C_(6)H_(5)NH_(2) underset(H^(+))overset(Na_(2)Cr_(2)O_(7))(rarr) ?

Answer»

`C_(6)H_(5)CHO`
`C_(6)H_(5)COOH`
`C_(6)H_(5)CH_(2)OH`
p-benzoquinone.

Solution :
Discussion
28784.

The molal freezing point constant of water is 1.86^(@)C//M. Therefore the freezing point of 0.1 M NaCl solution in water is expected to be

Answer»

`-1.86^(@)C`
`-0.186^(@)C`
`-0.372^(@)C`
`+0.372^(@)C`

SOLUTION :`NaCl rarr Na^(+)+CL^(-)`, I = 2
`DeltaT_(f)=iK_(f)m=2xx1.86xx0.1=0.372^(@)C`,
`T_(f)=-0.372^(@)C`.
Discussion
28785.

The main product in the reaction of : CH_(3) - CH_(2) - underset(C1)underset(|)(CH) - CH_(3) + underset((alc.))(KOH) is :

Answer»

`CH_(3) - CH = CH - CH_(3)`
`CH_(2) = CH - CH_(2) - CH_(3)`
`CH_(3) - CH_(2) - underset(C1)underset(|)(CH) - CH_(2)OH`
`CH_(3) - CH_(2) - underset(OH)underset(|)(CH) - CH_(3)`

Solution :`CH_(3) - CH_(2) - underset(Cl)underset(|)(CH) - CH_(3) OVERSET(alc.KOH)RARR CH_(3)CH = CH - CH_(3)`
Discussion
28786.

The molal freezing point constant of water is 1.86C/M. Therefore the freezing point of 0.1 M NaCl solution in water is expected to be

Answer»

`-1.86^@C`
`-0.186^@C`
`-0.372^@C`
`+0.372^@C`

ANSWER :C
Discussion
28787.

The main process for the manufacture of sodium carbonate is :

Answer»

SOLVAY process
Lead CHAMBER process
Down process
Nelson cell

Answer :A
Discussion
28788.

The molal freezing point constant for water is 1.86^(@)C//m. Therefore, the freezing point of 0.1 M NaCl solution in water is expected to be

Answer»

`-272.628K`
`+272.628K`
`-0.372^(@)C`
`+0.372^(@)C`

ANSWER :B::C
Discussion
28789.

The main problem encountered during electrophilic subsitution reactions of aromatic amines is that they have __________.

Answer»


ANSWER :HIGH REACTIVITY
Discussion
28790.

The molal freezing point constant for water is 1.86^(@)C//m. Therefore, the freezing point of 0.1 M NaCl solution in water is expected to be:

Answer»

`-1.86^(@)C`
`-0.186^(@)C`
`-0.372^(@)C`
`+0.372^(@)C`

ANSWER :C
Discussion
28791.

The molal freezing point constant for water is 1.86 K . Kg "mole"^(-1). The freezing point of 0.1 m NaCl solution is

Answer»

`-1.86^(@)C`
`-0.372^(@)C`
`-0.186^(@)C`
`0.372^(@)C`

ANSWER :B
Discussion
28792.

The main point of difference between buna-N and buna-S is

Answer»

the formeris homopolymerwhereas the later is copolymer
formercontainsbuta-1,3-diene but laterdoes not
former contains acrylonitrilebut latercontains
All of the above

Solution :Buna-N and buna-S both are synthetic rubber and copolymers. The DIFFERENCE LIES in their composition . Buna-N constituentsare buta-1,3-diene and ACRYLONITRILE. Buna-S CONSTITUENTS are buta-1, 3-dieneand styrene .
Discussion
28793.

The main point of difference between DNA and RNA is :

Answer»

PRESENCE of THYMINE in DNA and RNA
Presence of DEOXYRIBOSE and thymine
in DNA, RIBOSE and uracil in RNA
Presence of ribose and thymine in DNA,
deoxyribose and uracil in RNA
Presence of deoxyribose in DNA
and ribose in RNA

Answer :B
Discussion
28794.

The main oxides formed on combustion of Li,Na and K in excess of air respectively are

Answer»

`Li_(2)O, Na_(2)O_(2) and KO_(2)`
`Li_(2)O, Na_(2)O and KO_(2)`
`LiO_(2), Na_(2)O_(2) and K_(2)O`
`Li_(2)O_(2), Na_(2)O_(2) and KO_(2)`

Answer :A
Discussion
28795.

The main ore of zinc is:

Answer»

ZINC blende
Zincite
Calamine
All

Answer :D
Discussion
28796.

The main group elements i.e. representative elements complete their electron configuration using s and p electrons in the periodic table. These elements range from the most metallic to the most non-metallic with intermediate properties, the semi-metals, in between .The elements which occur at the two extremes of the periodic table are highly reactive and therefore, these highly reactive elements do not occur in free state : they usually occur in the combined forms. Beryllium and aluminium exhibit many properties which are similar, but, the two elements differ in :

Answer»

FORMING covalent HALIDES
Forming polymeric hydrides
Exhibiting maximum COVALENCY in compounds
Exhibiting AMPHOTERIC nature in their oxides

Solution :Beryllium has the maximum vallency +2 while aluminium exhibits its valency as +3.
Discussion
28797.

The mainore ofironis its

Answer»

CHLORIDE
SULPHATE
NITRATE
OXIDE

ANSWER :D
Discussion
28798.

The main ore of iron is:

Answer»

Chloride
Sulphate
Nitride
Oxide

Answer :D
Discussion
28799.

The main group elements i.e. representative elements complete their electron configuration using s and p electrons in the periodic table. These elements range from the most metallic to the most non-metallic with intermediate properties, the semi-metals, in between .The elements which occur at the two extremes of the periodic table are highly reactive and therefore, these highly reactive elements do not occur in free state : they usually occur in the combined forms. Select the correct statement (this is with respect to the elements of main group).

Answer»

Chemical reactivity of the elements is generally lowest in the centre of a period
Oxides of the elements in the centre of period in general are amphoteric or neutral
Metallic character increases with increasing atomic number in a GROUP where as decreases from left to right in a period
All of these

Solution :(A)Have intermediate values of ionisation energy and electron affinity.
(B)Generally the elements found in the centre of period for main group elements are metalloids.They generally FORM amphoteric oxides.
(C )Down the group size of ATOM increases and ELECTROPOSITIVE character increase.So metallic character increases. ACROSS the period , size of atom decreases and electropositive character decreases.So metallic character decreases and non-metallic character increases.
Discussion
28800.

The main function of the salt bridge is:

Answer»

To allow IONS to GO from ONE cell to another
To provide link between two half cells
To KEEP the emf of the cell positive
To maintain electrical NEUTRALITY of the solution in two half cells

Answer :D
Discussion