datavorous/entrance-exam-dataset huggingface dataset:
Justify that the reaction: 2 Na(s) + H₂ (g) → 2 NaH (s) is a redox change.

Options:
A Sodium is oxidized, and hydrogen is reduced.
B Sodium is reduced, and hydrogen is oxidized.
C Both sodium and hydrogen are oxidized.
D Both sodium and hydrogen are reduced.

The correct answer is:
Sodium is reduced, and hydrogen is oxidized.

Solution:
Correct Option is : (B)
Sodium is reduced, and hydrogen is oxidized.

Topic: Chemistry, Redox Reactions, NEET
Justify that the reaction: 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂ (g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

Options:
A Cu₂O is oxidized, Cu₂S is reduced, Cu acts as an oxidant, and SO₂ acts as a reductant.
B Cu₂O is reduced, Cu₂S is oxidized, Cu acts as a reductant, and SO₂ acts as an oxidant.
C Cu₂O is oxidized, Cu₂S is reduced, SO₂ acts as an oxidant, and Cu acts as a reductant.
D Cu₂O is reduced, Cu₂S is oxidized, SO₂ acts as a reductant, and Cu acts as an oxidant.

The correct answer is:
Cu₂O is oxidized, Cu₂S is reduced, Cu acts as an oxidant, and SO₂ acts as a reductant.

Solution:
Correct Option is : (A)
Cu₂O is oxidized, Cu₂S is reduced, Cu acts as an oxidant, and SO₂ acts as a reductant.

Topic: Chemistry, Redox Reactions, NEET
Justify the given statement with suitable examples— "the Properties of the elements are a periodic function of their atomic numbers."

Options:
A Elements in the same period exhibit similar chemical properties due to their similar valence electron configurations.
B Elements in the same group have different properties because they have different atomic numbers.
C The metallic character of elements decreases across a period as the atomic number increases.
D Atomic size increases down a group because of the increase in atomic number.

The correct answer is:
Elements in the same period exhibit similar chemical properties due to their similar valence electron configurations.

Solution:
Correct Option is : (A)
Elements in the same period exhibit similar chemical properties due to their similar valence electron configurations.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Justify the placement of O, S, Se, Te, and Po in the same group of the periodic table in terms of electronic configuration, oxidation state, and hydride formation.

Options:
A They all have the same number of valence electrons.
B They exhibit similar bonding patterns and tendencies to form compounds.
C They all belong to the same period.
D They all form diatomic molecules in their elemental forms.

The correct answer is:
They all have the same number of valence electrons.

Solution:
Correct Option is : (A)
They all have the same number of valence electrons.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Juvenile state of moss is

Options:
A protonema
B capsule
C prothallus
D all

The correct answer is:
protonema

Solution:
The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages, the first stage is the protonema stage, which develops directly from a spore. The second stage is the leafy stage, which develops from the protonema as a lateral bud.

Topic: Biology, Plant Kingdom, NEET
Juxta glomerular apparatus is formed by

Options:
A Modified cuboidal cells of DCT
B Modified cuboidal cells of thick segment of descending limb of loop of Henle
C Smooth muscle cells of vasa recta
D Cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact

The correct answer is:
Cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact

Solution:
The Juxtaglomerular apparatus is a specialized structure seen in the kidneys. It is present near the glomerulus. ('Juxta; means near). These are modified cells at the junction of distal convoluted tubule and afferent arteriole. This apparatus is responsible for regulating the blood pressure and filtration rate of glomerulus.

Topic: Biology, Excretory Products and their Elimination, NEET
K1 and K2 for oxalic acid are 6.5×102- and 6.1×10-5 respectively. What will be [HO-] in a 0.01 M solution of sodium oxalate?

Options:
A 9.6×10-6
B 1.4×10-1
C 1.28×10-6
D 1.3×10-8

The correct answer is:
1.28×10-6

Solution:
The hydrolysis of C2O42- is represented by
C2O42-+H2O⇌HC2O4-+OH-
Kh=KwK2=1.0×10-146.1×10-5=x20.01
So x=1.28×10-6

Topic: Chemistry, Ionic Equilibrium, NEET
K1 be the maximum kinetic energy of photoelectrons emitted when metal surface is illuminated by light of wavelength Î» 1. Similarly K2 is corresponding to wavelength Î» 2 for same metal. If Î» 1 = 2 Î» 2 then

Options:
A 2K1=K2
B K1=2K2
C K1<K2/2
D K1>2K2

The correct answer is:
K1<K2/2

Solution:
K 1 = hc λ 1 - W ....(i)
and K 2 = hc λ 2 - W ....(ii)
Substituting λ 1 = 2 λ 2 in Eq. (i), we get
K 1 = hc 2 λ 2 - W
= 1 2 hc λ 2 - W = 1 2 K 2 + W - W
∴ K 1 = K 2 2 - W 2 or K 1 < K 2 2

Topic: Physics, Dual Nature of Matter, NEET
K2CO3 cannot be prepared by Solvay's process because

Options:
A KHCO3 is less soluble than NaHCO3
B KHCO3 is too soluble to get separated from solution
C K2CO3 is too soluble to be precipitated by KCl
D K2CO3 is less soluble than Na2CO3

The correct answer is:
KHCO3 is too soluble to get separated from solution

Solution:
KHCO3 is too soluble and can not be separated by fractional crystallization like NaHCO 3 .

Topic: Chemistry, s Block Elements, NEET
K 4 [Fe(CN) 6 ] +  M x+ (aq)→ M 4 [Fe(CN) 6 ] x ↓ Coloured precipitate
Which of the following cation does not respond to the above reaction?

Options:
A Cu2+(aq)
B Fe3+(aq)
C Zn2+(aq)
D None of these

The correct answer is:
Zn2+(aq)

Solution:
Here only Zn+2 do not form coloured precipitate here
K 4 [Fe(CN) 6 ] +  Zn 2+ (aq)→ Zn 2 [Fe CN) 6 ↓ white precipitate

Topic: Chemistry, Coordination Compounds, NEET
Kb for weak base B is 1.0×10-6 then hydrolysis constant for the reaction
B++H2O⇌BOH+H+ is

Options:
A 10-6
B 10-7
C 10-8
D 10-9

The correct answer is:
10-8

Solution:
B++H2O⇌BOH+H+
For hydrolysis of B+ the hydrolysis constant can be given as
KH=KwKb
=10-1410-6=10-8 .

Topic: Chemistry, Ionic Equilibrium, NEET
KCl.MgCl2.6H2O is a

Options:
A mixed salt
B double salt
C basic salt
D complex salt

The correct answer is:
double salt

Solution:
It is a double salt as it ionises in aqueous solution as
KCl.MgCl2.6H2O⇌K++Mg2++3Cl-+6H2O

Topic: Chemistry, Coordination Compounds, NEET
$\mathrm{K}_{\mathrm{H}}$ value for some gases at the same temperature ' $\mathrm{T}$ ' are given:
$\begin{array}{c|c}
\text {gas } & \mathrm{K}_{\mathrm{H}} / \mathbf{k} \text { bar } \\
\hline \mathrm{Ar} & 40.3 \\
\mathrm{CO}_2 & 1.67 \\
\mathrm{HCHO} & 1.83 \times 10^{-5} \\
\mathrm{CH}_4 & 0.413
\end{array}$
where $\mathrm{K}_{\mathrm{H}}$ is Henry's Law constant in water. The order of their solubility in water is :

Options:
A $\mathrm{Ar} < \mathrm{CO}_2 < \mathrm{CH}_4 < \mathrm{HCHO}$
B $\mathrm{Ar} < \mathrm{CH}_4 < \mathrm{CO}_2 < \mathrm{HCHO}$
C $\mathrm{HCHO} < \mathrm{CO}_2 < \mathrm{CH}_4 < \mathrm{Ar}$
D $\mathrm{HCHO} < \mathrm{CH}_4 < \mathrm{CO}_2 < \mathrm{Ar}$

The correct answer is:
$\mathrm{Ar} < \mathrm{CO}_2 < \mathrm{CH}_4 < \mathrm{HCHO}$

Solution:
$\mathrm{K}_{\mathrm{H}}=$ Henry Law constant of gas in water, released to solubility as:
Solubility of a gas $\propto \frac{1}{\mathrm{~K}_{\mathrm{H}} \text { Value }}$
This means higher is the value of $\mathrm{K}_{\mathrm{H}}$, lower will be the solubility of gas in water.
$\therefore$ Solubility order of given gas in water
$=\mathrm{HCHO}>\mathrm{CH}_4>\mathrm{CO}_2>\mathrm{Ar}$

Topic: Chemistry, Solutions, NEET, NEET 2022 (Phase 2)
KMnO4 can be prepared from K2MnO4 as per the reaction,
3MnO42-+2H2O⇌2MnO4-+MnO2+4OH-
The reaction can go to completion by removing OH- ions by adding

Options:
A CO2
B SO2
C HCl
D KOH

The correct answer is:
CO2

Solution:
To remove OH- ion a weak acid (H2CO3) is required because a strong acid can reverse the reaction. So here HCl is a strong acid so it reverse the reaction.
KOH is strong base which will increase OH- concentration.
SO2 will react with water to form a strong acid H2SO4 .
CO2 combines with OH- ion to give carbonate which is easily removed.

Topic: Chemistry, d and f Block Elements, NEET
KMnO4 can be prepared from K2MnO4 as per the reaction
3MnO42-+2H2O⇌2MnO4-+MnO2+4OH-
The reaction can go to completion by removing OH- ions by adding

Options:
A KOH
B CO2
C SO2
D HCl

The correct answer is:
CO2

Solution:
As MnO4- can oxidise SO2 , HCl but not CO2 as in CO2 C-atom has its highest oxidation state (+4) it means CO2 can be used here.

Topic: Chemistry, d and f Block Elements, NEET
KMnO4 (mol. wt. = 158) oxidizes oxalic acid in acidic medium to CO2 and water as follows
5C2O42-+2MnO4-+16H+→10CO2+2Mn2++8H2O
What is the equivalent weight of KMnO4 ?

Options:
A 158
B 31.6
C 39.5
D 79

The correct answer is:
31.6

Solution:
MnO4-+8H++5e-→Mn2++4H2O
Equivalent weight of KMnO4 in acidic medium =molecularweight5=1585=31.6

Topic: Chemistry, Redox Reactions, NEET
K max ( in eV )=3, and frequency of light (in Hz )=1× 10 15 for a metal used as a cathode in a photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is -

Options:
A 1× 10 14 Hz
B 1.5× 10 14  Hz
C 2.1× 10 14 Hz
D 2.7× 10 14  Hz

The correct answer is:
2.7× 10 14  Hz

Solution:
Given, K max =3ââeV
h=4.125Ã 10 â15 âeVâ Hz -1
hÎ½= K max +W
4.125Ã10-15Ã1015=3+W
or 4.125= 3 +W
or W=1.125
The threshold frequency
Î½ 0 = W h
Î½ 0 = 1.125 4.125Ã 10 â15
Î½ 0 = 1.125Ã 10 â15 4.125
Î½ 0 =2.72Ã 10 14

Topic: Physics, Dual Nature of Matter, NEET
KO2 is used in space capsules, submarines and breathing masks, because

Options:
A it removes carbon dioxide
B it produce oxygen
C it reacts with moisture
D both A and B

The correct answer is:
both A and B

Solution:
KO2 is used in space capsules, submarines and breathing masks, because it both removes CO2 and produces O2 .
4KO2+2CO2→2K2CO3+3O2
4KO2+4CO2+2H2O→More4KHCO3+3O2

Topic: Chemistry, s Block Elements, NEET
K O 2 (potassium super oxide) is used in oxygen cylinders in space and submarines because it.

Options:
A

Eliminates nitrogen

B

Produce moisture

C

Absorbs CO2

D

Produce ozone

The correct answer is:

Absorbs CO2

Solution:
4K O 2 +2C O 2 â 2 K 2 C O 3 +3 O 2

4K O 2 +4C O 2 +2 H 2 O â 4KHC O 3 +3 O 2 âââââââââââââââââExcess

K O 2 absorbs C O 2 and produces oxygen.

Topic: Chemistry, s Block Elements, NEET
$K_P$ for the following reaction is 3.0 at $1000 \mathrm{~K}$.
$\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$
What will be the value of $\mathrm{K}_C$ for the reaction at the same temperature?
(Given: $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )

Options:
A 0.36
B $3.6 \times 10^{-2}$
C $3.6 \times 10^{-3}$
D 3.6

The correct answer is:
$3.6 \times 10^{-2}$

Solution:
$\begin{aligned}
& \text { } \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \\
& \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \cdot(\mathrm{RT})^{\Delta n_{\mathrm{g}}} \\
& \mathrm{K}_{\mathrm{p}}=3.0 . \mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \\
& \begin{aligned}
\Delta n_g & =2-1=1 \\
\therefore \quad 3.0 & =\mathrm{K}_{\mathrm{c}}(0.083 \times 1000)^1 \\
& \mathrm{~K}_{\mathrm{c}}=\frac{3}{83}=3.6 \times 10^{-2}
\end{aligned}
\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2022 (Phase 2)
Kα wavelength emitted by an atom of atomic number Z=11 is λ. Find the atomic number for an atom that emits Kα radiation with wavelength 4λ.

Options:
A Z=6

B Z=4

C Z=11

D Z=44

The correct answer is:
Z=6

Solution:
1 λ ∝ z-1 2
∴ λ 1 λ 2 = z 2 - 1 2 z 1 - 1 2
1 4 = Z 2 - 1 1 1 - 1 2
Solving this, we get, Z 2 = 6

Topic: Physics, Atomic Physics, NEET
Kangaroo rat is an inhabitant of:

Options:
A North American forests
B North American deserts
C East American grasslands
D East American deserts

The correct answer is:
North American deserts

Solution:
Even in the absence of water, Kangaroo rat in North American deserts is capable of meeting all its water requirements from its internal fat oxidation in which water is produced as a by product. Also it can concentrate its urine.

Topic: Biology, Organisms and Populations, NEET
Kaolinite is an ore of:

Options:
A Aluminium
B Iron
C Copper
D Zinc

The correct answer is:
Aluminium

Solution:
Correct Option is : 1
Aluminium

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Keel is characteristic of the flowers of

Options:
A gulmohur
B Cassia
C Calotropis
D bean

The correct answer is:
bean

Solution:
The bean or legume family is one of the most common plant families. Flowers in bean family typically have their two bottom petals grown together along one side forming a structure a bit like a narrow but deep scoop. This special Bean-family kind of two-in-one petal is called the keel, like the keel of a boat. Bean blossoms with this configuration are said to be papilionaceous.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2010 (Screening)
Keel is the characteristic feature of flower of:

Options:
A Tulip
B Indigofera
C Aloe
D Tomato

The correct answer is:
Indigofera

Solution:
Indigofera belongs to Family Fabaceae. This family was earlier called Papilionoideae, a sub-family of family Leguminosae. Keel is a characteristic of Family Fabaecae, enclosing stamens and pistil. Vexillary aestivation is the characteristic aestivation of Fabaceae.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2015 (Phase 1)
Keel petal is a characteristic feature of the flowers of:

Options:
A Gulmohur
B Cassia
C Calotropis
D Bean

The correct answer is:
Bean

Solution:
The keel is the characteristic of the flowers of family Papilionaceae, for example, bean, gram, pea, etc.
The flowers of this family have butterfly-shaped corolla. The outermost petals of these flowers are the largest (and called as vexillum); the two largest petals are similar and have wings, and the two anterior petals called keel are fused enclosing stamens and carpels.

Topic: Biology, Morphology of Flowering Plants, NEET
Keeping the angle of banking of the road constant, the maximum speed of the vehicles is to be increased by 10% . The radius of curvature of the road will have to be changed from 20 m to

Options:
A 22Â m
B 40Â m
C 24.2 m
D 14.4Â m

The correct answer is:
24.2 m

Solution:
tanâ¡Î¸=v2rg âr Î± v2
r2r1 =v2v12
or r220=1.1vv2âr2=24.2 m

Topic: Physics, Laws of Motion, NEET
Keeping the angle of banking of the road constant, the maximum speed of the vehicles is to be increased by 10% . The radius of curvature of the road will have to be changed from 20 m to

Options:
A 22Â m
B 40Â m
C 24.2 m
D 14.4Â m

The correct answer is:
24.2 m

Solution:
tanâ¡Î¸=v2rg
⇒ r∝v2
r2r1=v2v12
or r220=1.1vv2
â r2=24.2 m

Topic: Physics, Laws of Motion, NEET
Kekulé's initial structure of benzene indicated the possibility of two isomeric 1,2-dibromobenzenes. What was the difference between these isomers?

Options:
A The number of bromine atoms
B The position of bromine atoms
C The presence of hydrogen atoms
D The color of the compounds

The correct answer is:
The position of bromine atoms

Solution:
In one isomer, the bromine atoms are attached to the doubly bonded carbon atoms, while in the other, they are attached to the singly bonded carbons.

Topic: Chemistry, Hydrocarbons, NEET
Kelps and rockweed are examples of

Options:
A Green algae
B Brown algae
C Red algae
D Golden brown algae

The correct answer is:
Brown algae

Solution:
The large (giant) parenchymatous forms of brown algae are called kelps or seaweed or trees of seas or forest of the sea. (e.g., Macrocystis 30–60m; Nereocystis 20–30m, Laminaria 2–12m). Both contain fucoxanthin, which is brown in colour. Due to which both are grouped under brown algae.

Topic: Biology, Plant Kingdom, NEET
Kepler's third law states that the square of period of revolution T of a planet around the sun is proportional to the third power of average distance, r between the sun and the planet i.e T2=Kr3 .Here, K is constant.
If masses of the sun and the planet are M and m respectively, then as per Newton's law of gravitation force of attraction between them is F=GMmr2 , where G is gravitational constant. The relation between G and K is described as

Options:
A GK=4π2
B GMK=4π2
C K=G
D K=1G

The correct answer is:
GMK=4π2

Solution:
As, GMmr2=mv2r= centripetal force
⇒v2=GMr⇒T=2πrv
⇒T2=4π2r2v2
Out putting the value of v2 , we get
T2=4π2r2GMr⇒T2=4π2r3GM
⇒T2=Kr3
Here, K is given by,
4π2GM=K⇒GMK=4π2

Topic: Physics, Gravitation, NEET
Kepler's third law states that the square of the period of revolution (T) of a planet around the sun is proportional to the third power of the average distance r between sun and planet i.e., T2=Kr3.
Here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is
F=GMmr2, here G is gravitational constant.
The relation between G and K is described as:

Options:
A GK=4π2
B GMK=4π2
C K=G
D K=1G

The correct answer is:
GMK=4π2

Solution:
Here gravitational force is equal to centripetal force required for motion of planet.
Hence,
GMmr2=mv2r
v2=GMr
Time period is given by,
T=2πrv
T2=4π2r2v2
T2=4π2r2Gmr=4π2r3GM

We have given,
T2=kr3
On comparing,
k=4π2GM⇒GMk=4π2

Topic: Physics, Gravitation, NEET, NEET 2015 (Phase 1)
Ketone monosaccharides with 5 C-atoms is known as:

Options:
A Ketotriose
B Ketotetrose
C Ketopentose
D ketohexose

The correct answer is:
Ketopentose

Solution:
Correct Option is : 3
Ketopentose

Topic: Chemistry, Biomolecules, NEET
KF has NaCl structure. The edge length of its unit cell has been found to be 537.6 pm. The distance between K+ and F- in KF is

Options:
A 537.6 pm
B 1075.2 pm
C 268.5 pm
D None of these

The correct answer is:
268.5 pm

Solution:
For Nacl type crystals
a=2 r cation +  r anion
⇒  r cation +   r anion = a 2 = 537 2 =268.5 pm.

Topic: Chemistry, Solid State, NEET
KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as

Options:
A P > Q > R > S
B S > P > R > Q
C P > R > Q > S
D R > P > S > Q

The correct answer is:
S > P > R > Q

Solution:
SN2 reactions are through back side attack of attacking Nu- and not on the basis of stability of carbocation. Here less hindered site is more favourable for attack hence the correct order is

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
KI when heated with conc. H2SO4 , it forms

Options:
A HIO3
B KIO3
C I2
D HI

The correct answer is:
I2

Solution:
HI is obtained by the reaction of KI with conc. H2SO4 , is further oxidized by H2SO4 , because HI is strong reducing agent.
2KI+H2SO4→K2SO4+2HI
2HI+H2SO4→I2+SO2+2H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Kidney stones are produced due to deposition of uric acid and

Options:
A Silicates
B Minerals
C Calcium carbonate
D Calcium oxalate

The correct answer is:
Calcium oxalate

Solution:
Deposition of crystallized chemical lead to kidney stone like uric acid, calcium oxalate and calcium phosphate .

Topic: Biology, Excretory Products and their Elimination, NEET
Kingdom Plantae includes

Options:
A Algae, Bryophytes and Pteridophytes only.
B Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms only.
C Algae, Fungi, Pteridophytes, Gymnosperms and Angiosperms only.
D Algae, Pteridophytes, Gymnosperms and Angiosperms only.

The correct answer is:
Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms only.

Solution:
The Kingdom Plantae includes algae, bryophytes, pteridophytes, gymnosperms and flowering plants (angiosperms). They are common on land, on sea shore and in freshwater.

Topic: Biology, Plant Kingdom, NEET
Kirchhoff's first and second laws of electrical circuits are consequences of:

Options:
A conservation of energy and electric charge respectively
B conservation of energy
C conservation of electric charge and energy respectively
D conservation of electric charge

The correct answer is:
conservation of electric charge and energy respectively

Solution:
Kirchhoff's first and second law of electrical circuits are consequence of conservation of energy.

Topic: Physics, Current Electricity, NEET, NEET 2006
Klinefelter's syndrome can be easily identified by

Options:
A PCR
B Karyotype
C ELISA
D RFLP

The correct answer is:
Karyotype

Solution:
Klinefelter's syndrome is a chromosomal disorder in which the suffer has 47 chromosome due to an extra X chromosome. Genotype of such individual is 44A +XXY . This additional chromosome can easily be easily observed in karyotype study.

Topic: Biology, Principles of Inheritance and Variation, NEET
$\mathrm{KMnO}_4$ can be prepared from $\mathrm{K}_2 \mathrm{MnO}_4$ as per reaction \(3 \mathrm{MnO}_2^{2-}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+4 \mathrm{OH}^{-}\). The reaction can go to completion by removing $\mathrm{OH}^{-}$ions by adding

Options:
A $\mathrm{HCl}$
B $\mathrm{KOH}$
C $\mathrm{CO}_2$
D $\mathrm{SO}_2$

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

Solution:
Since, $\mathrm{OH}^{-}$are generated from weak acid $\left(\mathrm{H}_2 \mathrm{O}\right)$, a weak acid $\left(\right.$ like $\left.\mathrm{CO}_2\right)$ should be used to remove it because of strong acid $(\mathrm{HCl})$ reverse the reaction. $\mathrm{KOH}$ increases the concentration of $\mathrm{OH}^{-}$, thus again shifts the reaction in backward side.
$\mathrm{CO}_2$ combines with $\mathrm{OH}^{-}$to give carbonate which is easily removed.
$\mathrm{SO}_2$ reacts with water to give strong acid, so it cannot be used.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2013 (All India)
Knee and elbow joint is an example of

Options:
A Hinge joint
B Gliding joint
C Ball and socket joint
D Pivot joint

The correct answer is:
Hinge joint

Solution:
Elbow joint is an example of hinge joint. The elbow is a hinge joint; it can open and close like a door. Hinge joint is a form of diarthrosis (freely movable joint) that allows angular movement in one plane only, increasing or decreasing the angle between the bones e.g. elbow joint, knee joint etc.

Topic: Biology, Locomotion and Movement, NEET
Knee joint and elbow joints are examples of

Options:
A saddle joint
B ball and socket joint
C pivot joint
D hinge joint.

The correct answer is:
hinge joint.

Solution:
A hinge joint allows movement in one plane. Example: Knee and elbow joints.

Topic: Biology, Locomotion and Movement, NEET
Knowing the electron gain enthalpy values for O → O–and O → O2–as –141 and 702 kJ mol–1 respectively, how can you account for the formation of a large number of oxides having O2–species and not O–? (Hint Consider lattice energy factor in the formation of compounds).

Options:
A The formation of O2–species is more exothermic.
B The lattice energy of O2–compounds is higher.
C O2–ions form more stable compounds.
D O–ions are less reactive.

The correct answer is:
The formation of O2–species is more exothermic.

Solution:
Correct Option is : (A)
The formation of O2–species is more exothermic.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes?

Options:
A Yes, D2O can be used for drinking purposes because it is chemically similar to H2O.
B No, D2O is not suitable for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and cause health risks.
C D2O can be used for drinking but in limited quantities and under specific conditions.
D D2O is more suitable for drinking purposes than H2O due to its heavy properties.

The correct answer is:
No, D2O is not suitable for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and cause health risks.

Solution:
Correct Option is : (B)
No, D2O is not suitable for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and cause health risks.

Topic: Chemistry, Hydrogen, NEET