datavorous/entrance-exam-dataset huggingface dataset:
Among the statements (1) – (4), the correct ones are
(1) Lithium has the highest hydration enthalpy among the alkali metals.
(2) Lithium chloride is insoluble in pyridine.
(3) Lithium cannot form ethynide upon its reaction with ethyne
(4) Both lithium and magnesium react slowly with H2O

Options:
A (1), (2) and (4) only
B (1) and (4) only
C (2) and (3) only
D (1), (3) and (4)

The correct answer is:
(1), (3) and (4)

Solution:
Lithium has the highest hydration enthalpy among the alkali metals due to its small size.
Lithium chloride is covalent in nature so it’s soluble in non-polar solvent.
Lithium and Magnesium react slowly with H2O .

Topic: Chemistry, s Block Elements, NEET
Among the statements given below, which statement/s is/are correct regarding the relationship between enzyme and substrate?

I. Above a certain concentration of substrate, an enzyme reaches its maximum rate of reaction
II. Increasing the substrate concentration to a higher level doesn't reverse the effects of a competitive inhibitor.
III. At high substrate concentration, a non-competitive inhibitor no longer affects the enzyme activity.
IV. The higher the concentration of substrate the faster an enzyme can catalyse a reaction.

Options:
A I and II only
B I, II and IV
C I only
D III and IV only

The correct answer is:
I only

Solution:
Substrate concentration is a factor that plays a role in enzyme activity. More the substrate concentration, more is the enzyme activity till it reaches saturation where even on increasing substrate will not increase the activity. Statement I agrees to it but statement IV is incomplete. Statement III is stating the presence of non-competitive inhibitor, which binds to a separate site on enzyme than the substrate and alters the active site of enzyme thus even if substrate concentration is increased it will not lead to any change. Lastly, about competitive inhibitor has a competition with the substrate for the same position of the active site. If substrate conc. is higher than it will replace competitive inhibitor for the active site and its effect on enzyme activity.

Topic: Chemistry, Biomolecules, NEET
Among these canonical structures, the correct order of stability is

Options:
A I > II > III
B III > II > I
C I > III > II
D II > I > III

The correct answer is:
I > III > II

Solution:
I > III > II

Neutral molecule is more stable than changed. -ve charge stable at more stable electronegative atom and the charge stable at more electropositive atom and positive charge is stable on electronegative atom if its octet is complete.

Topic: Chemistry, General Organic Chemistry, NEET
Among two plots, plot 'H' is having less species compared to plot 'K'. Identify the statement that is correct for plot H and plot K.

Options:
A Plot 'X' shows less year to year variation in total biomass
B Plot 'K' is less resistant to natural disturbance
C Plot 'H' has more productivity than plot 'K'
D Plot 'K' has more productivity than plot 'H'

The correct answer is:
Plot 'K' has more productivity than plot 'H'

Solution:
Biodiversity is can be said to be the measure of the complexity of interactions between organisms and the ecosystem. The greater the number of species results in more interactions which supports biodiversity. Also, helps an ecosystem or habitat to survive and adapt to environmental stress. These will increase the biomass of the particular ecosystem thus, making productivity high. Hence, the plot K will have more productivity than plot H.

Topic: Biology, Biodiversity and its Conservation, NEET
Amongst H2O,H2S,H2Se and H2Te, the one with the highest boiling point is –

Options:
A H2O because of hydrogen bonding
B H2Te because of higher molecular weight
C H2S because of hydrogen bonding
D H2Se because of lower molecular weight

The correct answer is:
H2O because of hydrogen bonding

Solution:
Boiling point trend is H2O>H2Te>H2Se>H2S.
H2O is highest due to H- bonding

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Amongst [NiCO)4,[NiCN)42- and [NiCl4]2-

Options:
A [Ni(CO)4] and [NiCl4]2- are diamagnetic and [Ni(CN)4]2- is paramagnetic
B [NiCl4]2- and [Ni(CN)4]2- are diamagnetic and [Ni(CO)4] is paramagnetic
C [Ni(CO)4] and [Ni(CN)4]2- are diamagnetic and [NiCl4]2- is paramagnetic
D [Ni(CO)4] is diamagnetic and [NiCl4]2- and [Ni(CN)4]2- are paramagnetic

The correct answer is:
[Ni(CO)4] and [Ni(CN)4]2- are diamagnetic and [NiCl4]2- is paramagnetic

Solution:
[Ni(CO)4] and [Ni(CN)4]2- are both diamagnetic as they have no unpaired electrons on central atom Ni while [NiCl4]2- is paramagnetic as Ni atom has two unpaired electrons.

Topic: Chemistry, Coordination Compounds, NEET
Amongst NO3-, AsO33-, CO32-, ClO3-, SO32- and BO33- , the non-planar species are

Options:
A CO32-, SO32- and BO32-
B AsO33-, ClO3- and SO32-
C NO3-,Â CO32-Â andÂ BO33-
D SO32-,Â Â NO3-Â andÂ BO33-

The correct answer is:
AsO33-, ClO3- and SO32-

Solution:
NO3-,CO32- and BO33- have sp2 hybridisation are hence are planar species while AsO33-, ClO3- and SO32- have sp3 hybridisation and hence are non-planar species.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Amongst the elements with following electronic configurations, which one of them may have the highest ionisation energy ?

Options:
A $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$
B $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^2$
C $[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2 4 \mathrm{p}^3$
D $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^1$

The correct answer is:
$[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3$

Solution:
Key Idea Ionisation energy usually increases from left to right in a period with decrease in atomic size and decreases from upto down in a group with increase in atomic size.
$\begin{array}{lc}\text { Electronic configuration } & \text { Group } \\ {[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3} & \mathrm{~V} \\ {[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^2} & \text { IV } \\ {[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2, 4 \mathrm{p}^3} & \mathrm{~V} \\ {[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^1} & \text { III }\end{array}$
Since IE increases in a period and decreases in a group, [Ne] $3 s^2 3 \mathrm{p}^3$ configuration has the highest ionisation energy among the given.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2009 (Screening)
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform and (vi) pentanol.

Options:
A Toluene and Formic acid are water insoluble ; Phenol and Pentanol are partially soluble; Chloroform and Ethylene glycol are water soluble
B Phenol and Chloroform are water insoluble ; Toluene and Pentanol are partially soluble; Formic acid and Ethylene glycol are water soluble
C Toluene and Ethylene glycol are water insoluble ; Phenol and Pentanol are partially soluble; Formic acid and Chloroform are water soluble
D Toluene and Chloroform are water insoluble ; Phenol and Pentanol are partially soluble; Formic acid and Ethylene glycol are water soluble

The correct answer is:
Toluene and Chloroform are water insoluble ; Phenol and Pentanol are partially soluble; Formic acid and Ethylene glycol are water soluble

Solution:
Correct Option is : (D)
Toluene and Chloroform are water insoluble ; Phenol and Pentanol are partially soluble; Formic acid and Ethylene glycol are water soluble

Topic: Chemistry, Solutions, NEET
Amongst the following four statements (A-D), some are wrong.

(a) Unarmed pathogen vectors are also used in the transfer of DNA into the host.
(b) Restriction enzymes are used in isolation of DNA from other macro-molecules.
(c) Downstream processing is one of the steps of biotechnology.
(d) The first transgenic sheep, Rosie, produced enriched milk with human alpha-lactalbumin.

Which of these statements are with mistakes?

Options:
A Statements a and c
B Statements a and b
C Statements b and d
D Statements c and d

The correct answer is:
Statements b and d

Solution:
First transgenic cow was named Rosie that produced milk with human alpha-lactalbumin. It was 2.4 gm per litre. Restriction enzymes are used to cut DNA at specific sequences. Downstream processing is an important step in biotechnology. It includes collection and purification of materials. The pathogens to be used in transfer of DNA should be unarmed or disarmed which means it should not be able to cause any disease.

Topic: Biology, Biotechnology and Its Applications, NEET
Amongst the following hydroxide, the one which has the lowest value of solubility product (Ksp) at ordinary temperature (about 25°C ) is

Options:
A Mg(OH)2
B Ca(OH)2
C Sr(OH)2
D Be(OH)2

The correct answer is:
Be(OH)2

Solution:
The solubility of hydroxides of group 2 elements increases on moving down the group. Be(OH)2 has minimum solubility in water. The lesser the solubility of the hydroxides, the smaller will be is solubility product. Hence, Be(OH)2 has the lowest solubility product amongst the group 2 hydroxides.

Topic: Chemistry, s Block Elements, NEET
Amongst the following, the form of water with the lowest ionic conductance at 298 K is

Options:
A sea water
B distilled water
C saline water used for intravenous injection
D water from a well

The correct answer is:
distilled water

Solution:
Sol. The form of H2O with the lowest ionic conductance at 298 K is distilled water.

Topic: Chemistry, s Block Elements, NEET
Amongst the following, the most stable complex is

Options:
A [Fe(H2O)6]3+
B [Fe(NH3)6]3+
C [Fe(C2O4)3]3-
D [FeCl6]3-

The correct answer is:
[Fe(C2O4)3]3-

Solution:
In all these complexes, Fe is in +3 oxidation state. However, the complex ([Fe(C2O4)3]3-) is a chelate because three C 2 O 4 2 - ions act as the chelating ligands. Thus, the most stable complex is [Fe(C2O4)3]3-.

Topic: Chemistry, Coordination Compounds, NEET
Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is

NH3, AlCl3, BeCl2, CCl4, PCl5

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Among the given species NH3 and CCl4 is having eight electrons at the central atoms. In ammonia, one lone pair and three bond pairs at nitrogen. In carbon tetrachloride four bond pairs are present.

In AlCl3, aluminium has three bond pairs, i.e., it has incomplete octet.

In BeCl2, beryllium has two bond pairs, i.e., it has incomplete octet.

In PCl5, phosphorus has five bond pairs, i.e, it has expanded octet.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2023 (All India)
Amongst the following, which characteristic feature is not applicable to Bt cotton?

Options:
A Bt is the abbreviated term for botulinum toxin.
B Such cotton is resistant to armyworm and beetles.
C The toxin is activated in the body of the insect.
D The toxin is coded by a gene called 'cry'.

The correct answer is:
Bt is the abbreviated term for botulinum toxin.

Solution:
Bt is the abbreviated term for Bacillus thuringiensis and not botulinum toxin.

Topic: Biology, Biotechnology and Its Applications, NEET
Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?

Options:
A IF5
B SF4
C XeF2
D ClF3

The correct answer is:
XeF2

Solution:
As we can see that maximum number of lone pairs are present in XeF2. So maximum 'lone pair-lone pair' electrons repulsions will be in XeF2. Hence option 3 is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2022 (Phase 1)
Amongst the given options which of the following molecules/ion acts as a Lewis acid?

Options:
A OH-
B NH3
C H2O
D BF3

The correct answer is:
BF3

Solution:
Lewis Acids are the chemical species which have empty orbitals and are able to accept electron pairs from Lewis bases. BF3 has 6 electrons in the central metal atom, boron. Hence, it is electron deficient and is a Lewis acid. The other species OH-, NH3 and H2O are not having any empty orbitals in them, instead they can donate lone pairs(Lewis bases).

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2023 (All India)
Amongst the three isotopes of Neon - 1020Ne, 1021Ne and 1022Ne the nucleus with the lowest n/p ratio is

Options:
A 1022Ne
B 1021Ne
C 1020Ne
D All of these

The correct answer is:
1020Ne

Solution:
1020Ne, Z=10, A=20, N=A-Z, =10, n/p=1
1021Ne, Z=10, A=21, N=A-Z=11, np=1110=1.1
1022Ne, Z=10, A=22, N=A-Z=12, np=1210=1.2

Topic: Physics, Nuclear Physics, NEET
Amoxicillin is semi-synthetic modification of

Options:
A Penicillin
B Streptomycin
C Tetracycline
D Chloramphenicol

The correct answer is:
Penicillin

Solution:
Amoxicillin is semi-synthetic modification of penicillin

Topic: Chemistry, Chemistry in Everyday Life, NEET
Amphetamines and caffeine are respectively

Options:
A

Amphetamines	Caffeine

Stimulant	Depressant

B

Amphetamines	Caffeine

Depressant	Depressant

C

Amphetamines	Caffeine

Depressant	Stimulant

D

Amphetamines	Caffeine

Stimulant	Stimulant

The correct answer is:

Amphetamines	Caffeine

Stimulant	Stimulant

Solution:
Drugs like barbiturates (depressant), amphetamines (stimulant) and benzodiazepines (depressant) are normally used as medicines to treat patients suffering from mental illnesses such as depression and insomnia. These drugs are often abused. Caffeine also acts as a stimulant of CNS.

Topic: Biology, Human Health and Diseases, NEET
Amphibians share with reptiles all of the following characters except

Options:
A ventral heart.
B indirect development.
C dioecious, oviparous.
D poikilothermic.

The correct answer is:
indirect development.

Solution:
Amphibians and Reptiles show direct development. Both of them have a ventral heart. They both are cold blooded (poikiolothermic). They also have both male and female reproductive organs in separate individuals. They produce young ones by laying eggs which are hatched after they have been laid by the parent.

Topic: Biology, Animal Kingdom, NEET
Amylopectin is composed of

Options:
A α -D-glucose, C1—C4 and C1—C6 linkage
B α -D-glucose, C1—C4 and C2—C6 linkage
C β -D-glucose, C1—C4 and C2—C6 linkage
D β -D-glucose, C1—C4 and C1—C6 linkage

The correct answer is:
α -D-glucose, C1—C4 and C1—C6 linkage

Solution:
Amylopectin is a homopolymer of α -D-glucose having C1—C4 β -glycocidic linkage (for linear chain formation) and C1—C6 linkage (for branch chain formation).

Topic: Chemistry, Biomolecules, NEET
An 80 kg person is parachuting and is experiencing a downward acceleration of 2.8 m s-2. The mass of the parachute is 5 kg. The upward force on the open parachute is (Take g=9.8 m s-2)

Options:
A 595 N
B 675 N
C 456 N
D 925 N

The correct answer is:
595 N

Solution:
Mass of the person M=80 kg
Mass of the parachute m=5 kg
â´ Total mass of the system =M+m=85 kg
Downward acceleration a=2.8 m s-2
Let upward force =F
Applying Newton's II law of motion to this system
F=m+Mg-m+Ma
or F=m+Mg-a
F=859.8-2.8 N
=85Ã7 N
F=595 N

Topic: Physics, Laws of Motion, NEET
An AB solid has CsCl type of structure, where A occupies corner. If the atoms from corners along one of the body diagonal are removed, the formula of a the solid would be

Options:
A A7B8
B A5B6
C A3B4
D A4B3

The correct answer is:
A3B4

Solution:
Along a body diagonal two corners will be removed i.e. 2 atoms decrease
So A atoms =6×18=34
B atoms =1 (m body centre)
So A:B=34:1=3:4
Hence formula A3B4 .

Topic: Chemistry, Solid State, NEET
An abnormal human baby with 'XXX' sex chromosomes was born due to:

Options:
A Formation of abnormal sperms in the father
B Formation of abnormal ova in the mother
C Fusion of two ova and one sperm
D Fusion of two sperms and ovum

The correct answer is:
Formation of abnormal ova in the mother

Solution:
Formation of abnormal ova i.e., 22 + XX in the mother will lead to birth of baby with 'XXX' genotype, due to chromosomal non disjunction in ova. Nondisjunction is the failure of the chromosomes to disjoin and move to opposite poles.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2015 (Phase 1)
An ac source is connected in the given circuit. The value of $\phi$ will be

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
$\begin{aligned} & X_L=\omega L \\ & X_L=100 \pi \times \frac{1}{\pi}=100 \Omega\end{aligned}$
$\begin{aligned} & \tan \phi=\frac{X_L}{R}=\frac{100}{100}=1 \\ & \phi=45^{\circ}\end{aligned}$

Topic: Physics, Alternating Current, NEET, NEET 2023 (Manipur)
An A.C. source is connected to a resistive circuit. Which of the following is true?

Options:
A Current leads ahead of voltage in phase
B Current lags behind voltage in phase
C Current and voltage are in same phase
D Any of the above may be true depending upon the value of resistance

The correct answer is:
Current and voltage are in same phase

Solution:
When resistance is connected to A.C source, then current & voltage are in same phase.

Topic: Physics, Alternating Current, NEET
An A.C. source of voltage V=100âsin( 100âÏt ) is connected to a resistor of resistance 20ââÎ© . The rms value of current through resistor is

Options:
A 10 A
B 10 2 A
C 5 2 A
D None of these

The correct answer is:
5 2 A

Solution:
I 0 = V 0 R = 100 20 = 5 A
â´ I rms = I 0 2 = 5 2 A

Topic: Physics, Alternating Current, NEET
An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive reactance are both equal to $3 \Omega$, the phase difference between the applied voltage and the current in the circuit is

Options:
A $\pi / 4$
B $\pi / 2$
C zero
D $\pi / 6$

The correct answer is:
$\pi / 4$

Solution:
$$
\begin{aligned}
\tan \phi=\frac{X_L}{R} & =\frac{L \omega}{R} \\
\tan \phi & =\frac{3 \Omega}{3 \Omega} \\
\tan \phi & =1 \\
\phi & =\tan ^{-1}(1) \\
\phi & =45^{\circ} \\
\phi & =\frac{\pi}{4} \mathrm{rad}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2011 (Screening)
An AC voltage source of variable angular frequency Ï and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When Ï is increased

Options:
A The bulb glows dimmer
B The bulb glows brighter
C Total impedance of the circuit is unchanged
D Total impedance of the circuit increases

The correct answer is:
The bulb glows brighter

Solution:
Z=R2+XC 2 Irms=VrmsZP=Irms2R
Where XC=1ωC
As ω is increased, XCwill decrease or Z will decrease. Hence Irms or P will increase. Therefore, bulb glows brighter.

Topic: Physics, Alternating Current, NEET
An accurate pendulum clock is mounted on ground floor of a high building. How much time will it lose or gain in one day if it is transferred to top storey of a building which is h = 200 m higher than the ground floor? Radius of earth is 6.4Ã 10 6 m .

Options:
A It will lose 6.2 s
B It will lose 2.7 s
C It will gain 5.2 s
D It will gain 1.6 s

The correct answer is:
It will lose 2.7 s

Solution:
T = 2 Ï lâ¡ g or T â 1 g
â´ T â² T = g g â²
But g â² = g 1 + h R 2 or g â² g = 1 1 + h R 2
â´ T â² T = 1 + h R
or T â² = T 1 + h R
Since, T â² > T , the clock will lose the time.
â´ Î T = T â² - T = T h R
â´ Time lost in t=1 day is
= 2 4 Ã 3 6 0 0 2 0 0 6.4 Ã 1 0 6 s = 2.7 s

Topic: Physics, Oscillations, NEET
An acid base indicator has Ka=3×10-5 . The acid form of the indicator is red and the basic form is blue. By how much must the pH change in order to change the indicator from 75% red to 75% blue (log 3 = 0.4770)

Options:
A 0.65
B 1.3
C 0.954
D 1.9

The correct answer is:
0.954

Solution:
pH change = pH of 75 % blue – pH of 75% red
=  pKa + log 75 25 − pKa + log 25 75
=log3+log3
pH=0.954

Topic: Chemistry, Ionic Equilibrium, NEET
An acid chloride and a carboxylic acid can be distinguished with

Options:
A aq. AgNO3
B aq. NaOH
C C2H5OH
D dil. HCl

The correct answer is:
aq. AgNO3

Solution:
RCOCl reacts with H2O to release Cl- ions, which react with AgNO3 to form AgCls , insoluble in HNO3 . The acid forms a precipitate of RCOOAg, soluble in dil. HNO3 .

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
An acid solution of 0.005 M has a pH of 5. The degree of ionisation of acid is

Options:
A 0.1×10-2
B 0.2×10-2
C 0.5×10-4
D 0.6×10-6

The correct answer is:
0.2×10-2

Solution:
Acid, [H+] = 10-pH=10-5
α=H+Concentration=10-50.005=0.2×10-2

Topic: Chemistry, Ionic Equilibrium, NEET
An acid solution of pH=6 is diluted 1000 times, the pH of final solution becomes

Options:
A 6.95
B 4
C 6
D 9

The correct answer is:
6.95

Solution:
pH=6,[H+]=10-pH=10-6M
When diluted to 1000 times
[H+]=10-610-3=10-9
[H+]=[H+]acid+[H+]water=10-9+10-7
pH=-log[H+]=-log[10-9+10-7]=6.95

Topic: Chemistry, Ionic Equilibrium, NEET
An acid solution of pH = 6 is diluted 1000 times, the pH of the final solution becomes

Options:
A 6.01
B 9
C 3.5
D 6.99

The correct answer is:
6.99

Solution:
âµ pH = 6
â´H+=10-6 M
After dilution H+=10-61000=10-9 M
â´[H+] from H2O cannot be neglected.
Total H+=10-9+10-7 (neglect common ion effect on H2O)
=10-710-2+1
=10-7(1.01)
pH= -logâ¡1.01Ã10-7
=7-0.0043
=6.9957

Topic: Chemistry, Ionic Equilibrium, NEET
An acid type indicator, Hln differes in colour from its conjugate base (In-) . The human eye is sensitive to colour differences only when the ratio [In-]/[HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change (Ka=1.0×10-5) ?

Options:
A 4
B 2
C 6
D 1

The correct answer is:
2

Solution:
The two conditions when colour of indicator will be visible are derived by
pH=pKa+log[In-][HIn]
(i) pH = 5 + log 10 = 6
(ii) pH = 5 + log 0.1 = 4
Thus, minimum change in pH = 2.

Topic: Chemistry, Ionic Equilibrium, NEET
An acidic buffer is prepared by mixing:

Options:
A Strong acid and it's salt with strong base.
B Strong acid and it's salt with weak base.
C Weak acid and it's salt with strong base.
D Equal volumes of equimolar solutions of weak acid and weak base.
(The $\mathrm{pK}_{\mathrm{a}}$ of acid $=\mathrm{PK}_{\mathrm{b}}$ of the base)

The correct answer is:
Weak acid and it's salt with strong base.

Solution:
Acidic buffer is a mixture of weak acid and its conjugate base or salt of that weak acid with another strong base in a molar ratio ranging from $\frac{1}{10}$ to 10

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2023 (Manipur)
An acidic solution of hydrogen peroxide behaves as an oxidising as well as reducing agent. Illustrate it with the help of a chemical equation.

Options:
A H2O2+ 2H++ 2e-→ 2H2O
B 2H2O2→ 2H2O + O2
C 2H2O2→ 2H2O + 2O2+ 2H+
D 2H2O2+ 2H++ 2e-→ 2H2O + O2

The correct answer is:
2H2O2→ 2H2O + 2O2+ 2H+

Solution:
Correct Option is : (C)
2H2O2→ 2H2O + 2O2+ 2H+

Topic: Chemistry, Hydrogen, NEET
An acoelomate animal with bilateral symmetry is:

Options:
A Jelly Fish
B Liver Fluke
C Pleurobrachia
D Ancylostoma

The correct answer is:
Liver Fluke

Solution:
Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates. In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g.,
aschelminthes. The animals in which the body cavity is absent are called acoelomates, e.g., Platyhelminthes.
They have dorso-ventrally flattened body, hence are called flatworms. Flatworms are bilaterally symmetrical, triploblastic and acoelomate animals with organ level of organisation.
Examples: Taenia (Tapeworm), Fasciola (Liver fluke).

Topic: Biology, Animal Kingdom, NEET