datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following precedes re-formation of the nuclear envelope during M-phase of the cell cycle?

Options:
A Decondensation from chromosome and reassembly of the nuclear lamina
B Transcription from chromosome and reassembly of the nuclear lamina
C Formation of the contractile ring and formation of the phragmoplast
D Formaltion of the contractile ring and transcription from chromosomes

The correct answer is:
Decondensation from chromosome and reassembly of the nuclear lamina

Solution:
At telophase stage, nuclear membrane vesicles associate with the surface of individual chromosomes and fuse to reform the nuclear membranes, which partially enclose clusters of chromosomes before coalescing to reform the complete nuclear envelope. During this process, the nuclear pores reassemble and (the dephosphorylated reassociate to form the nuclear lamina.) One of the lamina proteins (lamina-B) remains with the nuclear membrane fragments throughout mitosis and may help nucleate ressembly. After the nucleus reforms, the pores pump in nuclear proteins, the chromosome decondense and RNA synthesis resumes, causing the nucleolus to reappear.

Topic: Biology, Cell Cycle and Cell Division, JIPMER, JIPMER 2011
Which one of the following process is non-spontaneous?

Options:
A Dissolution of \( \mathrm{CuSO}_{4} \) in water.
B Reaction between \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \) to form water.
C Water flowing down hill.
D Flow of electric current from low potential to high potential

The correct answer is:
Flow of electric current from low potential to high potential

Solution:
The physical or chemical processes which do not proceed by its own in a particular direction and requires outside help is called a non-spontaneous process.

Therefore, flow of heat from a cold body to hot body is a non-spontaneous process as it requires external work.

Topic: Chemistry, Thermodynamics (C), JEE Main
Which one of the following processes will produce hard water?

Options:
A
Saturation of water with $\mathrm{MgCO}_3$

B
Saturation of water with $\mathrm{CaSO}_4$

C
Addition of $\mathrm{Na}_2 \mathrm{SO}_4$ to water

D
Saturation of water with $\mathrm{CaCO}_3$

The correct answer is:
Saturation of water with $\mathrm{CaSO}_4$

Solution:
Permanent hardness of water is due to chlorides and sulphates of calciumand magnesium.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2003
Which one of the following product is formed when calcium salt of adipic acid is heated?

Options:
A
B
C
D

The correct answer is:

Solution:
During dry distillation of calcium salts of carboxylic acid ,it loose calcium carbonate and generate carbonyl compounds.

carbonyl compounds formed contain one carbon less from the respective calcium carboxylate.

Calcium adipate on dry distillation gives cyclopentanone Which have one carbon less Than reactant

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which one of the following products obtained when diethyl ether is boiled with water in presence of dilute acid?

Options:
A Glycol
B Ethyl alcohol
C Ethylene oxide
D Peroxide

The correct answer is:
Ethyl alcohol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which one of the following properties is not shown by NO ?

Options:
A It is diamagnetic in gaseous state
B It is a neutral oxide
C It combines with oxygen to form nitrogen dioxide
D It's bond order is 2.5

The correct answer is:
It is diamagnetic in gaseous state

Solution:
As per MOT, electronic configuration of NO is
Ï1 s 2 < Ï * 1 s 2 <Ï2 s 2 < Ï * 2 s 2 <Ï2 p z 2 <Ï2 p x 2 =Ï2 p y 2 < Ï * 2 p x 1 = Ï * 2 p y 0
It is paramagnetic due to Ï 2 * py 1 (unpaired electron)
Bond order of NO (i.e. O 2 + )= 10â5 2 =2.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which one of the following properties is not shown by NO ?

Options:
A It is diamagnetic in gaseous state
B It is a neutral oxide
C It combines with oxygen to form nitrogen dioxide
D It's bond order is 2.5

The correct answer is:
It is diamagnetic in gaseous state

Solution:
As per MOT, electronic configuration of NO is
Ï1 s 2 < Ï * 1 s 2 <Ï2 s 2 < Ï * 2 s 2 <Ï2 p z 2 <Ï2 p x 2 =Ï2 p y 2 < Ï * 2 p x 1 = Ï * 2 p y 0
It is paramagnetic due to Ï 2 * py 1 (unpaired electron)
Bond order of NO (i.e. O 2 + )= 10â5 2 =2.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2014 (06 Apr)
Which one of the following property is not correct for ozone?

Options:
A It oxidises lead sulphide
B It oxidises potassium iodide
C It oxidises mercury
D It cannot act as bleaching agent

The correct answer is:
It cannot act as bleaching agent

Solution:
Ozone acts as bleaching agent because of its oxidising properties.
Other statement are true.
Option 4 is correct.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one of the following proved effective for biological control of nematodal disease in plants?

Options:
A Paecilomyces lilacinus
B Pisolithus tinctorius
C Pseudomonas cepacia
D Gliocladium virens

The correct answer is:
Pseudomonas cepacia

Solution:
No Solution Available

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Mains)
Which one of the following reactions does not come under hydrolysis type reaction?

Options:
A SiCl4(l)+2H2O(l)→SiO2(s)+4HCl(aq)
B Li3N(s)+3H2Ol→NH3(g)+3LiOH(aq)
C 2F2(g)+2H2Ol→4HF(aq)+O2(g)
D P4O10(s)+6H2Ol→4H3PO4(aq)

The correct answer is:
2F2(g)+2H2Ol→4HF(aq)+O2(g)

Solution:
2F2(g)+2H2O(ℓ)⟶4HF(aq)+O2(g)
It's a type of Redox reaction.

Topic: Chemistry, Redox Reactions, JEE Main
Which one of the following reactions does not occur during extraction of copper ?

Options:
A 2Cu2 S+3O2→2Cu2O+2SO2
B 2FeS+3O2→2FeO+2SO2
C CaO+SiO2→CaSiO3
D FeO+SiO2→FeSiO3

The correct answer is:
CaO+SiO2→CaSiO3

Solution:
The following changes occur during smelting.

(a) Most of the ferrous sulphide gets oxidised to ferrous oxide which combines with silica (flux) to form fusible slag.

2FeS + 3O2 → 2FeO + 2SO2

FeO + SiO2 → FeSiO3 (Ferrous silicate)

The slag being lighter floats and forms the upper layer. It is removed through the slag hole from time to time.

(b) During roasting or in the blast furnace if any oxide of copper is formed, it combines with FeS and is changed back into its sulphide.

2Cu2S + 3O2 → 2Cu2O + 2SO2

Cu2O + FeS → Cu2S + FeO

Ferrous oxide thus formed again combines with silica to form more slag.

FeO+SiO2→FeSiO3

So, no formation of calcium silicate CaSiO3 in extraction of Cu.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Which one of the following reactions does not occur ?

Options:
A

B

C

D

The correct answer is:

Solution:
(1) Aniline is Lewis base give acid base reaction with AlCl3 and form Anilinium ion
(2) Anilinium ion has strongest deactivated ring so further Friedel Craft alkylation not occurs.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Which one of the following reactions does not represent correct combination of substrate and product under the given conditions?

Options:
A

B

C

D

The correct answer is:

Solution:
Cyanide on reduction with DIBAL-H gives aldehyde. Acid halide with rosenmund reduction reaction gives aldehyde. So, all these reaction are correct except benzyl alcohol with oxidation with strong oxidising agents sodium dichromate will give carboxylic acid not the aldehyde.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Which one of the following reactions does not yield an alkyl halide

Options:
A Diethyl ether $+\mathrm{Cl}_2$
B Diethyl ether $+H I$
C Diethyl ether and $\mathrm{PCl}_5$
D Diethyl ether $\xrightarrow{\text { Reduction }} X \xrightarrow{\mathrm{SO}_2 \mathrm{Cl}_2}$

The correct answer is:
Diethyl ether $+\mathrm{Cl}_2$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?

Options:
A HOCl+H2O2→H3O++Cl-+O2
B PbS+4H2O2→PbSO4+4H2O
C 2MnO4-+3H2O2→2MnO2+3O2+2H2O+2OH-
D Mn2++H2O2→Mn4++2OH-

The correct answer is:
2MnO4-+3H2O2→2MnO2+3O2+2H2O+2OH-

Solution:
In option A and C reducing action of hydrogen peroxide is shown.

For reducing ability H2O2 changes to O2, i.e. oxidize, so in option 'A' & C''O2is formed but 'A' is in acidic medium so option - C is correct.

2MnO4-+3H2O2→2MnO2+3O2+2H2O+2OH-

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Which one of the following reactions is an example for calcination process?

Options:
A 2 Ag+2 HCl+(O)â2 AgCl+H2O
B 2 Zn+O2â2 ZnO
C 2 ZnS+3 O2â2 ZnO+2 SO2
D MgCO3âMgO+CO2

The correct answer is:
MgCO3→MgO+CO2

Solution:
The process of heating the concentrated ore in the absence of air is known as calcination. The carbonate ore gets decomposed to form oxide of metal on calcination.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Which one of the following reactions is expected to readily give a hydrocarbon product in good yields?

Options:
A $\mathrm{RCOOK} \frac{\text { Electrolytic }}{\text { oxidation }}$
B $\mathrm{RCOO}^{-} \mathrm{Ag}^{+} \stackrel{\mathrm{Br}_{2}}{\longrightarrow}$
C $\mathrm{CH}_{3} \mathrm{CH}_{3} \frac{\mathrm{Cl}_{2}}{\mathrm{hu}}$
D $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \stackrel{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\longrightarrow}$

The correct answer is:
$\mathrm{RCOOK} \frac{\text { Electrolytic }}{\text { oxidation }}$

Solution:
Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated carboxylic acids yields higher alkane at anode.

$$

2 \mathrm{RCOOK} \underset{\text { Oxidation }}{\text { Electrolytic }}{\longrightarrow} \underset{\text { Anode }}{2 \mathrm{RCOO}^{-}}+\underset{\text { Cathod }}{2 \mathrm{~K}^{+}}

$$

At anode $2 \mathrm{RCOO}^{-} \rightarrow 2 \mathrm{RCOO}+2 \mathrm{e}^{-} \longrightarrow$

$$

\mathrm{R}-\mathrm{R}+2 \mathrm{CO}_{2}

$$

At Cathode $2 \mathrm{~K}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}$

$$

2 \mathrm{~K}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_{2} \uparrow

$$

(Kolbe's Method)

Topic: Chemistry, Hydrocarbons, JEE Main
Which one of the following reactions is NOT possible?

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Which one of the following reactions of Xenon compound is not feasible?

Options:
A 3 XeF4+6 H2Oâ2 Xe+XeO3+12 HF+1.5 O2
B 2 XeF2+2 H2Oâ2 Xe+4 XF+4 HF+O2
C XeF6+RbFâRb+[XeF7]-
D XeO3+6HFâXeF6+2H2O

The correct answer is:
XeO3+6HF→XeF6+2H2O

Solution:
Here Remaining reactions are feasible except the last one which is not possible .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which one of the following reactions of Xenon compounds is not feasible ?

Options:
A
$\mathrm{XeO}_3+6 \mathrm{HF} \rightarrow \mathrm{Xe}_6+3 \mathrm{H}_2 \mathrm{O}$

B
$3 \mathrm{Xe}_4+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Xe}+\mathrm{XeO}_3+12 \mathrm{HF}+1.5 \mathrm{O}_2$

C
$2 \mathrm{XeF}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Xe}+4 \mathrm{HF}+\mathrm{O}_2$

D
$\mathrm{XeF}_6+\mathrm{RbF} \rightarrow \mathrm{Rb}\left(\mathrm{XeF}_7\right]$

The correct answer is:
$\mathrm{XeO}_3+6 \mathrm{HF} \rightarrow \mathrm{Xe}_6+3 \mathrm{H}_2 \mathrm{O}$

Solution:
Remaining are feasible

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2009
Which one of the following reactions of xenon compounds is not feasible?

Options:
A $3 \mathrm{Xe} \mathrm{F}_{4}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow$
$2 \mathrm{Xe}+\mathrm{XeO}_{3}+12 \mathrm{HF}+1.5 \mathrm{O}_{2}$
B $2 \mathrm{Xe} \mathrm{F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Xe}+4 \mathrm{HF}+\mathrm{O}_{2}$
C $\mathrm{XeF}_{6}+\mathrm{RbF} \longrightarrow \mathrm{Rb}\left[\mathrm{XeF}_{7}\right]$
D $\mathrm{Xe} \mathrm{O}_{3}+6 \mathrm{HF} \longrightarrow \mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{Xe} \mathrm{O}_{3}+6 \mathrm{HF} \longrightarrow \mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O}$

Solution:
The products of the concerned reaction react each other forming back the reactants.
$$
\mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{XeO}_{3}+6 \mathrm{HF}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one of the following reactions will not form acetaldehyde?

Options:
A CH3CH2OH⟶CrO3-H2SO4
B CH2=CH2+O2PdII/CuIIH2O
C CH3CN i) DIBAL-H ii )H2O>
D CH3CH2OHCu573 K>

The correct answer is:
CH3CH2OH⟶CrO3-H2SO4

Solution:
Ethyl alcohol on oxidation with acidified chromium oxide forms acetic acid. The reaction is as follows:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Which one of the following reactions will not result in the formation of carbon-carbon bond ?

Options:
A Friedel Craft's acylation
B Reimer-Tieman reaction
C Cannizaro reaction
D Wurtz reaction

The correct answer is:
Cannizaro reaction

Solution:
In cannizaro reaction carbon-carbon bond not formed
Friedel Craft's acylation

Reimer-Tieman reaction
Wurtz reaction

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which one of the following reactions will not result in the formation of carbon-carbon bond ?

Options:
A Friedel Craft's acylation
B Reimer-Tieman reaction
C Cannizaro reaction
D Wurtz reaction

The correct answer is:
Cannizaro reaction

Solution:
In cannizaro reaction carbon-carbon bond not formed
Friedel Craft's acylation

Reimer-Tieman reaction
Wurtz reaction

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2014 (09 Apr Online)
Which one of the following reactions will not yield propanoic acid?

Options:
A CH3CH2CH2Br+Mg,CO2 dry ether /H3O+
B CH3CH2CH3+KMnO4(Heat),OH-/H3O+
C CH3CH2CCl3+OH-/H3O+
D CH3CH2COCH3+OI-/H3O+

The correct answer is:
CH3CH2CH2Br+Mg,CO2 dry ether /H3O+

Solution:
CH3CH2CH2Br⟶MgCH3CH2CH2MgBr⟶CO2/H3O+CH3CH2CH2COOHButanoic acid

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Which one of the following reactions would be the best for the formation of 2-bromobutane
(1) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_3 \xrightarrow{\mathrm{HBr}}$
(2)$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}}$
(3)$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \xrightarrow{\mathrm{Br}_2}$
(4)$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\text { Peroxide }]{\mathrm{HBr}}$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Which one of the following reactions would produce secondary alcohol

Options:
A
B
C CH3CHO→ LiAlH4
D

The correct answer is:

Solution:
Aldehydes on reduction by LiAlH4 gives primary alcohol while ketons gives secondary alcohol. With grignard reagent ketone gives tertiary alcohol.
Also specify - successive members have different structural formula.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which one of the following reacts with glass?

Options:
A $\mathrm{H}_{2} \mathrm{SO}_{4}$
B HF
C $\mathrm{HNO}_{3}$
D $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

The correct answer is:
HF

Solution:
Silica $\left(\mathrm{SiO}_{2}\right)$ is present in the glass. This silica react with hydrofluoric acid.
$$
\begin{aligned}
\mathrm{SiO}_{2}+4 \mathrm{HF} \longrightarrow & \mathrm{SiF}_{4}+2 \mathrm{H}_{2} \mathrm{O} \\
\mathrm{SiF}_{4}+2 \mathrm{HF} \longrightarrow & \mathrm{H}_{2} \mathrm{SiF}_{6} \\
\text { fluorosilicic acid }
\end{aligned}
$$
Note : HF is used for the etching of glass.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one of the following reacts with $H C N$ and Tollen's reagent, but is not oxidised by Fehling's solution

Options:
A Methanal
B Ethanal
C Benzaldehyde
D Acetone

The correct answer is:
Acetone

Solution:
Acetaldehyde have $\mathrm{CH}_3 \mathrm{CO}-$ group so it give positive iodoform test with $I_2$ and $\mathrm{NaOH}$ while formaldehyde does not have $-\mathrm{CH}_3 \mathrm{CO}$ group so it will not give the positive haloform test.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which one of the following reagents is not suitable for the elimination reaction?

Options:
A NaI
B NaOEt/EtOH
C NaOH/H2O, ∆
D NaOH/H2O-EtOH

The correct answer is:
NaI

Solution:
To get the desired product, an elimination reaction needs to occur.

For an elimination reaction, a strong base ( O − Et,    O − H ) needed. Because I − is not basic, NaI is not suitable for the following reaction.
With NaI, substitution takes place.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2016 (10 Apr Online)
Which one of the following reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon?

Options:
A
Ethyl acetate

B
Butan -2-one

C
Acetamide

D
Acetic acid

The correct answer is:
Butan -2-one

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2004
Which one of the following relations is correct with respect to first (I) and second (II) ionization potentials of sodium and Magnesium?

Options:
A \( \mathrm{I}_{\mathrm{Na}}>\mathrm{I}_{\mathrm{Mg}} \)
B \( \mathrm{I}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}} \)
C \( \mathrm{II}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}} \)
D \( \mathrm{II}_{\mathrm{Na}}>\mathrm{II}_{\mathrm{Mg}} \)

The correct answer is:
\( \mathrm{II}_{\mathrm{Na}}>\mathrm{II}_{\mathrm{Mg}} \)

Solution:
After losing one electron, Na+ ion gets noble gas configuration.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Which one of the following represents a palindromic sequence in DNA?

Options:
A 5'-GAATTC-3'
3'-CTTAAG-5'
B 5'CCAATG-3'
3'GAATCC-5'
C 5'-CATTAG-3'
3'-GATAAC-5'
D 5'-GATACC-3'
3'-CCTAAG-5'

The correct answer is:
5'-GAATTC-3'
3'-CTTAAG-5'

Solution:
In a palindrome, the base sequence in the second half of a DNA strand is the mirror image of sequence in first half. But in palindrome with rotational symmetry the base sequence in first half of one strand is mirror immage of second half of complementary strand, e.g.,
5'-G A A T T C-3
3-C T T A A G-5

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2012 (Mains)
Which one of the following represents the correct dimensions of the coefficient of viscosity?

Options:
A
$\mathrm{ML}^{-1} \mathrm{~T}^{-2}$

B
$\mathrm{MLT}^{-1}$

C
$\mathrm{ML}^{-1} \mathrm{~T}^{-1}$

D
$\mathrm{ML}^{-2} \mathrm{~T}^{-2}$

The correct answer is:
$\mathrm{ML}^{-1} \mathrm{~T}^{-1}$

Solution:
Dimensions of $\eta$ (coefficient of viscosity)
$$
=\frac{\mathrm{MLT}^{-2}}{\mathrm{M}^0 \mathrm{~L}^0 \cdot \mathrm{M}^0 \mathrm{LT}^{-1}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}
$$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2004
Which one of the following represents the graph between $\log p$ (on $Y$-axis) and $\frac{1}{T}$ (on $X$-axis)?
( $p=$ vapour pressure of a liquid, $T=$ absolute temperature)

Options:
A

B

C

D

The correct answer is:

Solution:
According to Clausis and Clapeyron, if a graph is plotted between $\log p$ and $\frac{1}{T^{\prime}}$ a straight line is obtained with - ve slope.

Topic: Chemistry, Solutions, JEE Main
Which one of the following represents the graph between \( \log \mathrm{p} \) (on \( \mathrm{Y}- \) axis)and \( \frac{1}{\mathrm{~T}} \) (on \( \mathrm{X}- \) axis)?
( \( \mathrm{p}= \) vapour pressure of a liquid, \( \mathrm{T}= \) absolute temperature)

Options:
A
B
C
D

The correct answer is:

Solution:
According to Clausius-Clapeyron, if a graph is plotted between log⁡p and1T, a straight line is obtained with negative slope.

Topic: Chemistry, States of Matter, JEE Main
Which one of the following salt is most acidic in water

Options:
A $\mathrm{NiCl}_2$
B $\mathrm{BeCl}_2$
C $\mathrm{FeCl}_3$
D $\mathrm{AlCl}_3$

The correct answer is:
$\mathrm{AlCl}_3$

Solution:
$\mathrm{AlCl}_3$ on hydrolysis gives weak base and strong acid among all.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Which one of the following scientist's name is correctly matched with the theory put forth by him?

Options:
A Mendel - Theory of Pangenesis
B Weismann - Theory of continuity of Germplasm
C Pasteur - Inheritance of acquired characters
D de Vries - Natural selection

The correct answer is:
Weismann - Theory of continuity of Germplasm

Solution:
Following are the right matching pairs.
de Vries - Mutation theory
Darwin - Theory of pangenesis and Natural selection
Pasteur - Discarded the theory of spontaneous generation of life and proved that life arises from pre-existing life.

Topic: Biology, Evolution, NEET, NEET 2008 (Mains)
Which one of the following sequences was proposed by Darwin and Wallace for organic evolution?

Options:
A overproduction, variations, constancy of population size, natural selection
B variations, constancy of population size, overproduction, natural selection
C overproduction, constancy of population size, variations, natural selection
D variations, natural selection, overproduction,constancy of population size

The correct answer is:
overproduction, constancy of population size, variations, natural selection

Solution:
Overproduction, variations, the constancy of population size, natural selection sequence was proposed by Darwin and Wallace for organic evolution.
Overproduction: There is an overproduction of organisms due to high reproductive rate, even when the resources are limited. Even though the population increases enormously, the number of individuals in every generation remains constant.
Variation: means differences, like you have black and white peppered moths, red, green capsicum and hundreds of colours of butterflies. And they are all different because they are adapting themselves, to be better than each other.
Natural selection: It states that nature creates struggle and selects those animals which win in the competition. The organisms which have useful variations will win in the competition. Whereas the organisms which have no such useful variations will die.

Topic: Biology, Evolution, NEET, NEET 2003
Which one of the following set of elements can be detected using sodium fusion extract?

Options:
A Sulfur, Nitrogen, Phosphorus, Halogens
B Phosphorus, Oxygen, Nitrogen, Halogens
C Nitrogen, Phosphorus, Carbon, Sulfur
D Halogens, Nitrogen, Oxygen, Sulfur

The correct answer is:
Sulfur, Nitrogen, Phosphorus, Halogens

Solution:
By sodium fusion extract we can detect sulphur, nitrogen, Phosphorus and halogens, because they are converted in to their ionic form with sodium metal.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Which one of the following set of metals deposit as anode mud during the process of electrolytic refining of copper?

Options:
A Sn and \( \mathrm{Ag} \)
B \( \mathrm{Pb} \) and \( \mathrm{Zn} \)
C \( \mathrm{Ag} \) and \( \mathrm{Au} \)
D Fe and \( \mathrm{Ni} \)

The correct answer is:
\( \mathrm{Ag} \) and \( \mathrm{Au} \)

Solution:
During electrolytic refining of copper, impurities of silver Ag , gold Au, platinum (Pt) and selenium Se being less electropositive are not affected by electrolytic solution of acidified copper sulphate. Thus, these settle down under the anode as anode mud or anode sludge.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main