datavorous/entrance-exam-dataset huggingface dataset:
The compound that would produce a nauseating smell/odour with a hot mixture of chloroform and ethanolic potassium hydroxide is

Options:
A PhCONH $_{2}$
B PhNHCH $_{3}$
C $\mathrm{PhNH}_{2}$
D PhOH

The correct answer is:
$\mathrm{PhNH}_{2}$

Solution:
When primary amine react with chloroform and ethanolic potassium hydroxide, they produce - isocyamides as main product. The reaction is also known as carbyl amine reaction. $\mathrm{Ph}-\mathrm{NH}_{2} \frac{\mathrm{CHCl}_{3}}{\mathrm{KOH} \text { (ethanolic) }} \mathrm{Ph} \cdot \mathrm{N} \longrightarrow \mathrm{C}$
(Nauseating smell)

Topic: Chemistry, Amines, WBJEE, WBJEE 2017
The compound used as an explosive is

Options:
A 2,4, 6-tribromoaniline
B 2,4,6-trinitrotoluene (TNT)
C 1,3, 5-trinitrobenzene
D 2,4, 6-trichlorotoluene

The correct answer is:
2,4,6-trinitrotoluene (TNT)

Solution:
2,4,6-trinitrotolucne is used as an explosive.

Topic: Chemistry, Hydrocarbons, JEE Main
The compound used in the treatment of lead poisoning is:

Options:
A Desferrioxime B
B EDTA
C Cis-platin
D D-penicillamine

The correct answer is:
EDTA

Solution:
(1) Sodium calcium edetate, a salt of EDTA (Ethylene Diamine Tetra-acetic Acid), is used to bind metals in the chelation therapy, such as for treating mercury and lead poisoning.
(2) Cis-platin is used as an anti-cancer drug.
(3) D- penicillamine is used for treating copper poisoning.
(4) Desterrioxime B is used for treating iron poisoning.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The compound which acts as antioxidant in packed food is

Options:
A Benzoic acid
B Butylated hydroxy Toluene (BHT)
C Para chlorometaxylenol
D sodium bi carbonate

The correct answer is:
Butylated hydroxy Toluene (BHT)

Solution:
BHT added to food as preservative.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The compound which can give Amine with KOH and Br2 , is -

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The compound which does not contain an asymmetric carbon atom is

Options:
A Glycolaldehyde
B Glyceraldehyde
C Glucose
D Galactose

The correct answer is:
Glycolaldehyde

Solution:
Glycolaldehyde is $\mathrm{CH}_2 \mathrm{OH}-\mathrm{CHO}$ which does not have an asymmetric carbon atom. It is the smallest possible molecule that contains both an aldehyde and hydroxyl group.

Topic: Chemistry, Biomolecules, JEE Main
The compound which forms acetaldehyde when heated with dilute $\mathrm{NaOH}$, is

Options:
A 1, 1-dichloroethane
B 1,1 , 1-trichloroethane
C 1-chloroethane
D 1, 2-dichloroethane

The correct answer is:
1, 1-dichloroethane

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The compound which gives turbidity immediately with Lucas reagent at room temperature is

Options:
A butan-1-ol
B butan-2-ol
C 2-methyl propan-2-ol
D 2-methyl propan-1-ol

The correct answer is:
2-methyl propan-2-ol

Solution:
Tertiary alcohols gives turbidity immediately with Lucas reagent.
2-methyl propan-2-ol is an example of tertiary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The compound which is not formed during the dry distillation of a mixture of calcium formate and calcium acetate is

Options:
A methanal
B propanal
C propanone
D ethanal

The correct answer is:
propanal

Solution:
Propanal is not formed during the dry distillation of a mixture of calcium formate and calcium acetate.
$$
\begin{aligned}
(\mathrm{HCOO})_{2} \mathrm{Ca} \stackrel{\text { Dry distillation }}{\longrightarrow} \mathrm{HCHO} &+\mathrm{CaCO}_{3} \\
\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Ca} \stackrel{\Delta}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COCH}_{3} &+\mathrm{CaCO}_{3} \\
(\mathrm{HCOO})_{2} \mathrm{Ca}+\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Ca} \stackrel{\Delta}{\longrightarrow} & 2 \mathrm{CH}_{3} \mathrm{CHO} +2 \mathrm{CaCO}_{3}
\end{aligned}
$$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH– is

Options:
A
B
C
D

The correct answer is:

Solution:
Aromatic nucleophilic substitution reaction takes place via the formation of a carbanion intermediate. Hence, the rate of reaction depends on the stability of the carbanion.

Electron withdrawing effect can stabilise the carbanion. The Electron withdrawing nature of the nitro group will be minimum at the meta position.

Hence the correct answer is

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The compound with more covalent character in the following is

Options:
A FeF3
B VF5
C VF2
D TiF2

The correct answer is:
VF5

Solution:
The magnitude of polarization or increased covalent character depends upon a number of factors. These factors are,

(1) Small size of cation: Smaller the size of cation, greater is its polarizing power i.e. greater will be the covalent nature of the bond.
(2) Large size of anion: Larger the size of anion, greater is its polarizing power i.e. greater will be the covalent nature of the bond.
(3) Large charge on either of the two ions: As the charge on the ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases with the result its ability for forming the covalent bond increases.
(4) Electronic configuration of the cation: For the two ions of the same size and charge, one with a pseudo noble gas configuration (i.e. 18 electrons in the outermost shell) will be more polarizing than a cation with noble gas configuration (i.e., 8 electron in the outer most shell).

In VF5 the oxidation state of vanadium is +5.Where as in other compounds given oxidation states of the metal ions are +2 and +3.Hence, due to the greater charge on the cation VF5 has greater covalent character.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The compound $X$, in the reaction, is
\(\mathrm{X} \xrightarrow{\mathrm{CH}_3 \mathrm{MgI}} \mathrm{Y} \xrightarrow{\text { hydrolysis }} \mathrm{Mg}(\mathrm{OH}) \mathrm{I}+\mathrm{CH}_3 \mathrm{COOH}\)

Options:
A $\mathrm{CH}_3 \mathrm{CHO}$
B $\mathrm{CO}_2$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{CO}$
D $\mathrm{HCHO}$

The correct answer is:
$\mathrm{CO}_2$

Solution:
$\mathrm{CO}_2$ adds to Grignard's reagent to yield acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound $X$, in the reaction, is
$\mathrm{X} \xrightarrow{\mathrm{CH}_3 \mathrm{MgI}} \mathrm{Y} \xrightarrow{\text { hydrolysis }} \mathrm{Mg}(\mathrm{OH}) \mathrm{I}+\mathrm{CH}_3 \mathrm{COOH}$

Options:
A $\mathrm{CH}_3 \mathrm{CHO}$
B $\mathrm{CO}_2$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{CO}$
D $\mathrm{HCHO}$

The correct answer is:
$\mathrm{CO}_2$

Solution:
$\mathrm{CO}_2$ adds to Grignard's reagent to yield acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound (X) is

Options:
A
B
C
D

The correct answer is:

Solution:
The compound (X) is which is formed as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The compound X is

Options:
A

B

C

D All of these

The correct answer is:

Solution:
Hydroboration - oxidation reaction of alkenes leads to anti-Markovnikov's hydration. Further addition of water adds in syn-manner, i.e., H and OH are added to the same face of the double bond leading to formation of trans-product. In short, hydroboration-oxidation of alkenes is regioselective as well as stereoselective.

Topic: Chemistry, Hydrocarbons, JEE Main
The compounds A and B in the following reaction are, respectively:

Options:
A A= Benzyl alcohol, B= Benzyl isocyanide
B A= Benzyl chloride, B= Benzyl isocyanide
C A= Benzyl alcohol, B= Benzyl cyanide
D A= Benzyl chloride, B= Benzyl cyanide

The correct answer is:
A= Benzyl chloride, B= Benzyl isocyanide

Solution:
The final product is an isocyanide as the AgCN bond has covalent nature, due to which C is unavailable for bond. Hence, N bonds with the benzylic carbon leading to formation of isocyanide.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The compounds \( \left[\mathrm{Co}\left(\mathrm{SO}_{4}\right)\left(\mathrm{NH}_{3}\right)_{5}\right] \mathrm{Br} \) and \( \left[\mathrm{Co}\left(\mathrm{SO}_{4}\right)\left(\mathrm{NH}_{3}\right)_{5}\right] \mathrm{Cl} \) represent

Options:
A linkage isomerism
B ionisation isomerism
C coordination isomerism
D no isomerism

The correct answer is:
no isomerism

Solution:
Isomerism is the phenomenon in which more than one compounds have the same chemical formula but different chemical structures. But given molecules does not have same formula. So, they are not isomers.

Topic: Chemistry, Coordination Compounds, JEE Main
The compound(s) that is(are) removed as slag during the extraction of copper is :

(1) CaO

(2) FeO

(3) Al2O3

(4) ZnO

(5) NiO

Choose the correct answer from the options given below

Options:
A 3,4 only
B 2 only
C 1,2,5 Only
D 1,2 only

The correct answer is:
2 only

Solution:
Copper ores contain iron as the main impurity. Iron react with silica and is removed as slag. Therefore, the slag formed in the blast furnace is FeSiO3.

The reaction that takes place in the blast furnace is as follows:

FeO+SiO2→FeSiO3(slag)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The compound/s which will show significant intermolecular H-bonding is/are :

Options:
A (a), (b) and (c)
B (b) only
C (c) only
D (a) and (b) only

The correct answer is:
(b) only

Solution:
(a) Shows intra molecular H-bonding.
(b) Shows significant intermolecular H-bonding.
(c) It does not show intermolecular H-bonding due to steric hindrance.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The compressibility factor for a definite amount of van der Waal's gas at \( 0^{\circ} \mathrm{C} \) and \( 100 \) atm is found to be \( 0.5 \). Assuming the volume of gas molecules negligible, the van der Waals' constant 'a' in \( \mathrm{L}^{2} \mathrm{~mol}^{-2} \) atm for the gas is

Options:

The correct answer is:
1.26

Solution:
Z=PVnRT=0.5
Now, P+n2aV2V-nb=nRT
P+n2aV2V=nRT (b is negligible)
PV2-nRTV+n2a=0
â´ V=nRTÂ±n2R2T2-4n2aÃb2P
Since, V is constant at given P and T, thus, discriminant is 0.
â´n2R2T2=4n2aP or a=R2T24P
=0.08212Ã27324Ã100
=1.26 L2 mol-2 atm

Topic: Chemistry, States of Matter, JEE Main
The compressibility factor for a real gas at high pressure is :

Options:
A
$1+\mathrm{RT} / \mathrm{pb}$

B
$1$

C
$1+\mathrm{pb} / \mathrm{RT}$

D
$1-\mathrm{pb} / \mathrm{RT}$

The correct answer is:
$1+\mathrm{pb} / \mathrm{RT}$

Solution:
At high pressure $Z=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2012 (Offline)
The compressibility factor of a gas is less than 1 at STP. Its molar volume $V_m$ will be

Options:
A $V_m\gt22.42$
B $V_m \lt 22.42$
C $V_m=22.42$
D None

The correct answer is:
$V_m \lt 22.42$

Solution:
If $Z \lt 1$ then molar volume is less than $22.4 L$

Topic: Chemistry, States of Matter, JEE Main
The compressibility of water is $4 \times 10^{-5}$ per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atomsphere will be

Options:
A $0.4 c c$
B $4 \times 10^{-5} \mathrm{cc}$
C $0.025 c c$
D $0.004 c c$

The correct answer is:
$0.4 c c$

Solution:
$C=\frac{1}{K}=\frac{\Delta V / V}{\Delta P} \Rightarrow \Delta V=C \times \Delta P \times V$
$=4 \times 10^{-5} \times 100 \times 100=0.4 c c$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The compressibility of water is 4Ã10-5 per unit atmospheric pressure. The decrease in volume of 100 cm3 of water under a pressure of 100 atm will be

Options:
A 0.4 cm3
B 4Ã10-5Â cm3
C 0.025Â cm3
D 0.004Â cm3

The correct answer is:
0.4 cm3

Solution:
K=1B=∆VVP .
Here, P=100 atm ,
K=4×10-5 and
V=100 cm3
Hence, ∆V=0.4 cm3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The compressibility of water is $6 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}$. If one litre of water is subjected to a pressure of $4 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}$, then the decrease in its volume in millilitre will be

Options:
A 10
B 20
C 24
D 15

The correct answer is:
24

Solution:
Bulk modulus $B=-\frac{P}{\Delta V / V}$
$-$ ve sign shows that with an increase in pressure, a decrease in volume occurs
Compressibility $\mathrm{K}=\frac{1}{\mathrm{~B}}=-\frac{\Delta \mathrm{V}}{\mathrm{PV}}$
Decrease in volume,
$\Delta \mathrm{V}=\mathrm{PVK}=4 \times 10^{7} \times 1 \times 6 \times 10^{-10}$
$=24 \times 10^{-3}$ litre $=24 \times 10^{-3} \times 10^{3} \mathrm{~cm}^{3}$
$=24 \mathrm{~cc}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The concentrated sulphuric acid that is peddled commercial is $95 \% \mathrm{H}_2 \mathrm{SO}_4$ by weight. If the density of this commercial acid is $1.834 \mathrm{~g} \mathrm{~cm}^{-3}$, the molarity of this solution is

Options:
A
$17.8 \mathrm{M}$

B
$12.0 \mathrm{M}$

C
$10.5 \mathrm{M}$

D
$15.7 \mathrm{M}$

The correct answer is:
$17.8 \mathrm{M}$

Solution:
$95 \% \mathrm{H}_2 \mathrm{SO}_4$ by weight means $100 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4$ solution contains $95 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4$ by mass. Molar mass of $\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$
Moles in $95 \mathrm{~g}=\frac{95}{98}=0.969$ mole
Volume of $100 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4$
$$
\begin{aligned}
& =\frac{\text { mass }}{\text { density }}=\frac{100 \mathrm{~g}}{1.834 \mathrm{~g} \mathrm{~cm}^{-3}} \\
& =54.52 \mathrm{~cm}^3=54.52 \times 10^{-3} \mathrm{~L}
\end{aligned}
$$
$$
\begin{aligned}
\text { Molarity } & =\frac{\text { Moles of solute }}{\text { Volme of solute in } \mathrm{L}} \\
& =\frac{0.969}{54.52 \times 10^{-3}}=17.8 \mathrm{M}
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (07 May Online)
, the concentration and time is shown as
Calculate the value of \( K_{C} \) for the reaction

Options:

The correct answer is:
4

Solution:
⇒KfKb=0.30.6=KC=12=CA
2C⇌2A
KC'=1K2=4

Topic: Chemistry, Chemical Equilibrium, JEE Main
The concentration in g/L of a solution of cane sugar (Molecular weight = 342) which is isotonic with a solution containing 6 g of urea (Molecular weight= 60) per litre is

Options:
A 3.42
B 34.2
C 5.7
D 19

The correct answer is:
34.2

Solution:
The two solution should have same molar concentration for being isotonic.
Molar concentration of cabe sugar = Molar concentration of urea = 6 6 0 = 0.1 mol/L
concentration of cane sugar in g/L
= molar concentration × molecular weight
= 0.1 × 342 = 34.2

Topic: Chemistry, Solutions, JEE Main
The concentration of an organic compound in chloroform is $6.15 \mathrm{~g}$ per $100 \mathrm{~mL}$ of solution. A portion of this solution in a $5 \mathrm{~cm}$ polarimeter tube causes an observed rotation of $-1.2^{\circ}$. What is the specific rotation of the compound?

Options:
A $+12^{\circ}$
B $-3.9^{\circ}$
C $-39^{\circ}$
D $+61.5^{\circ}$

The correct answer is:
$-39^{\circ}$

Solution:
$[\alpha]=\frac{[\alpha]_{\text {observed }}}{l \times C}=\frac{-1.2}{5 \times \frac{6.15}{1000}}=-39^{\circ}$

Topic: Chemistry, General Organic Chemistry, JEE Main
The concentration of dissolved oxygen (DO) in cold water

can go upto

Options:
A $14 \mathrm{ppm}$
B 8 ppm
C $10 \mathrm{ppm}$
D $16 \mathrm{pom}$

The correct answer is:
$10 \mathrm{ppm}$

Solution:
Dissolved oxygen (DO) in cold water can reach a concentration upto $10 \mathrm{ppm}$.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The concentration of dissolved Oxygen in water for growth of fish should be more than X¯ ppm and Biochemical Oxygen Demand in clean water should be less than Y¯ ppm. X and Y in ppm are, respectively.

Options:
A XY65
B XY48
C XY415
D XY612

The correct answer is:
XY65

Solution:
The concentration of dissolved oxygen in water is very important for aquatic life. If the concentration of dissolved oxygen of water is below 6 ppm, the growth of fish gets inhibited. The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water, is called Biochemical Oxygen Demand (BOD). The amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically. Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of:

Options:
A Nitrate
B Iron
C Fluoride
D Lead

The correct answer is:
Nitrate

Solution:
Parameters Maximum prescribed concentration in drinking water
Iron 0.2 ppm
Fluroide 1.5 ppm
Lead 50 ppb
Nitrate 50 ppm
Hence the concentration of nitrate in a given
water sample exceeds from the upper limit as
given above.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2016 (03 Apr)
The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of

Options:
A Iron
B Fluoride
C Lead
D Nitrate

The correct answer is:
Nitrate

Solution:
Parameters	Maximum prescribed conc. in drinking water
Iron	0.2 ppm
Fluoride	1.5 ppm
Lead	50 ppb
Nitrate

Hence the concentration of nitrate in a given water sample exceeds from the upper limit as given above.

Topic: Chemistry, Environmental Chemistry, JEE Main
The concentration of oxalic acid is ' $x$ ' $\mathrm{mol} \mathrm{L}^{-1}$. $40 \mathrm{~mL}$ of this solution reacts with $16 \mathrm{~mL}$ of $0.05 \mathrm{M}$ acidified $\mathrm{KMnO}_4$. What is the $\mathrm{pH}$ of ' $x$ ' $\mathrm{M}$ oxalic acid solution ? (Assume that oxalic acid dissociates completely)

Options:
A 1.3
B 1.699
C 1
D 2

The correct answer is:
1.3

Solution:
Oxalic acid $=x \mathrm{~mol} / \mathrm{L}$
Oxalic acid $\mathrm{KMnO}_4$
$M_1 V_1=M_2 V_2$
$40 \mathrm{~mL} \times x=16 \mathrm{~mL} \times 0.05$
$x=\frac{16 \times 0.05}{40}=\frac{1}{50}$
$x=\frac{1}{50} \mathrm{M}$
Now convert molarity into normality
$N \times$ eq. wt. $=M \times$ mol. wt. of oxalic acid
$N \times 45=\frac{1}{50} \times 90$
$N=\frac{1}{25}$
This normality represents the hydrogen ion concentration.
So, $\quad\left[\mathrm{H}^{+}\right]=\frac{1}{25}$
$\mathrm{pH}=\log \frac{1}{\left[\mathrm{H}^{+}\right]}=\log 25=1.3$

Topic: Chemistry, Redox Reactions, JEE Main
The concentration vs time graph for the reaction \( \mathrm{A} \rightleftharpoons \mathrm{nB} \), is presented in the figure. Determine the value of \( \mathrm{n} \).

Options:

The correct answer is:
2

Solution:
Initially in the container, only the reactant A is present. After 5 hours equilibrium is established.

A ⇌ nBInitially (t=0) 0.6 M 0At equilibrium (t=5 hr) 0.3 M 0.6 M

At equilibrium, the rate of forward reaction becomes equal to the rate of backward reaction and the net rate of the reaction is zero.

-∆A∆t=+1n×∆B∆t-0.3-0.65-0=+1n×0.6-05-0n=2

Topic: Chemistry, Chemical Equilibrium, JEE Main
The concentrations of the reactant $A$ in the reacntion $A \rightarrow B$ at different times are given below
$\begin{array}{ll}\text { Concentration }(\boldsymbol{M}) & \text { Time (Minutes) } \\ 0.069 & 0 \\ 0.052 & 17 \\ 0.035 & 34 \\ 0.018 & 51\end{array}$
The rate constant of the reaction according to the correct order of reaction is

Options:
A $0.001 \mathrm{M} / \mathrm{min}$
B $0.001 \mathrm{~min}^{-1}$
C $0.001 \mathrm{~min} / \mathrm{M}$
D $0.001 \mathrm{M}^{-1} \mathrm{~min}^{-1}$

The correct answer is:
$0.001 \mathrm{M} / \mathrm{min}$

Solution:
\begin{array}{c|c|c}
\hline Interval & Conc.change & Rate \\
\hline 0-17 \mathrm{~min} & 0.069-0.052=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline 17-34 \mathrm{~min} & 0.052-0.035=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline 34-51 \mathrm{~min} & 0.035-0.018=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline
\end{array}
$\because$ Rate remains constant. So, it is independent of concentration, the reaction is of zero order.
According to rate law
Rate $=\mathrm{K}(\text { conc. })^{0}=0.001 \mathrm{M} / \mathrm{min}$

Topic: Chemistry, Chemical Kinetics, VITEEE, VITEEE 2015
The concept of stationary orbit was given by:

Options:
A Neil Bohr
B JJ Thomson
C Ruther Ford
D More than one of the above

The correct answer is:
Neil Bohr

Solution:
Bohr's Atomic Model of the stationary orbit:
- Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge $\mathrm{Ze}$ (called hydrogen-like atom).

Topic: Physics, Atomic Physics, JEE Main
The concept that explains how the large quantity of substrate is acted upon by a small concentration enzyme is

Options:
A Coenzyme theory
B Enzyme collision theory
C Induced fit theory
D Lock and key theory

The correct answer is:
Lock and key theory

Solution:
Emil Fisher (1894) proposed a 'lock and key' theory to explain the mechanism of enzyme action. According to this theory, the active sites of the enzyme have a specific geometric shape, wherein the substrate molecule gets fitted to form an enzymeâsubstrate complex. This is like the fitting of a key to a particular lock. According to this theory,
Enzyme + Substrate â Enzymeâsubstrate complex â Enzyme + Product
This concept can explain how a large quantity of substrates is acted upon by a small concentration enzyme. It also explains, why enzymes remain unaffected after the completion of a bioreaction.

Topic: Chemistry, Biomolecules, NEET
The condensers of capacity ${ }^{C_1}$ and ${ }^{C_2}$ are connected in parallel, then the equivalent capacitance is

Options:
A $C_1+C_2$
B $\frac{C_1 C_2}{C_1+C_2}$
C $\frac{C_1}{C_2}$
D $\frac{C_2}{C_1}$

The correct answer is:
$C_1+C_2$

Solution:
Let the potential across the capacitor be $\mathrm{V}$
$\begin{aligned} & q=q_1+q_2 \\ & C V=C_1 V+C_2 V \\ & C=C_1+C_2\end{aligned}$

Topic: Physics, Capacitance, JEE Main
The condition for the coaxial system $x^2+y^2+2 \lambda x+c=0$, where $\lambda$ is a parameter and $c$ is a constant, to have distinct limiting points, is

Options:
A $c=0$
B $c < 0$
C $c=-1$
D $c>0$

The correct answer is:
$c>0$

Solution:
For $c>0$, the coaxial system of circles represents a non-intersecting coaxial system for which 2 distinct limiting points.

Topic: Mathematics, Circle, JEE Main
The condition for which the straight line $y=m x+c$ touches the parabola $y^2=4 a x$ is

Options:
A $a=c$
B $\frac{a}{c}=m$
C $m=a^2 c$
D $m=a c^2$

The correct answer is:
$\frac{a}{c}=m$

Solution:
It is a formula.

Topic: Mathematics, Parabola, JEE Main
The condition that indicates a polluted environment is:

Options:
A eutrophication
B 0.03% of CO2 in the atmosphere
C BOD value of 5 ppm
D pH of rain water to be 5.6

The correct answer is:
eutrophication

Solution:
(2) Clean water would have B.O.D value of less than 5 ppm whereas highly polluted water coluld have a
B.O.D value of 17 ppm or more.
(3) The process in which nutrient enriched water bodies support a dense plant population which kill animal life by depriving it of oxygen results in subsequent loss of biodiversity is known as Eutrophication.
(4) If the concentration of dissolved oxygen of water is below 6 ppm, the growth of fish get inhibited.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (05 Sep Shift 1)