54317.Among the following which one is not paramagnetic? [Atomic numbers; Be = 4, Ne = 10, As
= 33, Cl = 17]
= 33, Cl = 17]
Ne2+
Be+
Cl-
As+
Explanation:
Paramagnetic character is based upon presence of unpaired electron.
.17Cl−=1s2,2s2 2p6,3s2 $ 3p_2^x 3p_2^y 3p_2^z $
In Cl− no unpaired electron , so it is in nature diamagnetic
4Be=1s2,2s1 $ 2p_1^x $
10Ne2+=1s2,2s2 $ 2p_2^x 2p_1^y 2p_1^z $
33As+=1s2,2s2 2p6,3s23p63d10,4s2 $ 4p_1^x 4p_1^y 4p_0^z
While all others have unpaired electron, so they are paramagnetic in nature.
Paramagnetic character is based upon presence of unpaired electron.
.17Cl−=1s2,2s2 2p6,3s2 $ 3p_2^x 3p_2^y 3p_2^z $
In Cl− no unpaired electron , so it is in nature diamagnetic
4Be=1s2,2s1 $ 2p_1^x $
10Ne2+=1s2,2s2 $ 2p_2^x 2p_1^y 2p_1^z $
33As+=1s2,2s2 2p6,3s23p63d10,4s2 $ 4p_1^x 4p_1^y 4p_0^z
While all others have unpaired electron, so they are paramagnetic in nature.
54318.0.24 g of a volatile gas, upon vaporisation, gives 45 mL vapour at NTP. What will be the
vapour density of the substance? (Density of H2 = 0.089)
vapour density of the substance? (Density of H2 = 0.089)
95.93
59.93
95.39
5.993
Explanation:
Weight of gas = 0.24 g, Volume of gas = 45 mL = 0.045 litre and density of H2 = 0.089.
We know that weight of 45 mL of H2 = Density x Volume = 0.089 x 0.045 = 4.005 x 10^-3 g
Therefore vapour density
=$ \dfrac{weight of certain value of substance}{weight of same volumn of hydrogen} $
=$ \dfrac{0.24}{4.005 \times 10^{-3}} $
= 59.93
Weight of gas = 0.24 g, Volume of gas = 45 mL = 0.045 litre and density of H2 = 0.089.
We know that weight of 45 mL of H2 = Density x Volume = 0.089 x 0.045 = 4.005 x 10^-3 g
Therefore vapour density
=$ \dfrac{weight of certain value of substance}{weight of same volumn of hydrogen} $
=$ \dfrac{0.24}{4.005 \times 10^{-3}} $
= 59.93
54319.The amount of zinc required to produce 224 mL of H2 at STP on treatment with dilute
H2SO4 will be
H2SO4 will be
65 g
0.065 g
0.65 g
6.5 g
Explanation:
Zn(65g) + H2SO4 (22400ml) -> ZnSO4 + H2
Since 65 g of zinc reacts to liberate 22400 mL of H2, at STP, therefore amount of zinc needed to produce 224 mL of H2 at STP
= 65/22400 x 224 = 0.65 g
Zn(65g) + H2SO4 (22400ml) -> ZnSO4 + H2
Since 65 g of zinc reacts to liberate 22400 mL of H2, at STP, therefore amount of zinc needed to produce 224 mL of H2 at STP
= 65/22400 x 224 = 0.65 g
54320.The dimensions of pressure are the same as that of
force per unit volume
energy per unit volume
force
energy.
Explanation:
Pressure = Force/Area
Therefore dimensions of Pressure = $ \dfrac{MLT^{-2}}{L^{2}} $
= ML-1T-2 and dimensions of energy per unit volume
$ \dfrac{Energy}{Volumn} = \dfrac{ML^{2}T^{-2}}{L^{3}} $
== ML-1T-2
Pressure = Force/Area
Therefore dimensions of Pressure = $ \dfrac{MLT^{-2}}{L^{2}} $
= ML-1T-2 and dimensions of energy per unit volume
$ \dfrac{Energy}{Volumn} = \dfrac{ML^{2}T^{-2}}{L^{3}} $
== ML-1T-2
54321.The number of moles of oxygen in one litre of air containing 21% oxygen by volume, under
standard conditions, is
standard conditions, is
0.0093 mol
2.10 mol
0.186 mol
0.21 mol.
Explanation:
21% of oxygen by volume means for 1 L of air volume of oxygen will be 0.21L and as we know 1 mole of ideal gas occupy 22.4L volume at STP i.e. Standard temperature and pressure.
Then considering air as mixture of ideal gases and oxygen as ideal gas we can say 1L of oxygen will have 1/22.4 moles of oxygen
It implies 0.21L of oxygen will have 0.21*(1/22.4) moles of oxygen i.e. 0.009375 moles of oxygen and that is your answer.
21% of oxygen by volume means for 1 L of air volume of oxygen will be 0.21L and as we know 1 mole of ideal gas occupy 22.4L volume at STP i.e. Standard temperature and pressure.
Then considering air as mixture of ideal gases and oxygen as ideal gas we can say 1L of oxygen will have 1/22.4 moles of oxygen
It implies 0.21L of oxygen will have 0.21*(1/22.4) moles of oxygen i.e. 0.009375 moles of oxygen and that is your answer.
54322.The total number of valence electrons in 4.2 g of NH3 ion is (NHA is the Avogadro’s number)
2.1 NA
4.2 NA
1.6 NA
3.2 NA
54323.A 5 molar solution of H2SO4 is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be
1 N
0.1 N
5 N
0.5 N
54324.The number of gram molecules of oxygen in 6.2 x 1024 CO molecules is (1990)
10 g molecules
5 g molecules
1 g molecules
0.5 g molecules
54326.The molecular weight of O2 and SO2 are 32 and 64 respectively. At 15°C and 150 mmHg pressure, one litre of O2 contains N molecules. The number of molecules in two litres of SO2 under the same conditions of temperature and pressure will be
N/2
N
2 N
4 N
54327.A nucleoside differs from a nucleotide. It lacks the
base
sugar
phosphate group
hydroxyl group
Explanation:
Nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a
nucleoside,i.e., Nucleoside = Nitrogen base + Pentose sugar. On the other hand a phosphate group is attached o the 5 -OH of a nucleoside through phosphodiester linkage to form a nucleotide,
i.e.,Nucleotide = Nitrogen base + Pentose sugar +Phosphate (PO4)
Nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a
nucleoside,i.e., Nucleoside = Nitrogen base + Pentose sugar. On the other hand a phosphate group is attached o the 5 -OH of a nucleoside through phosphodiester linkage to form a nucleotide,
i.e.,Nucleotide = Nitrogen base + Pentose sugar +Phosphate (PO4)
54328.Two genes A and B are linked. In a dihybrid cross involving these two genes, the F1 heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?
1 : 1 : 1 : 1
9 : 3 : 3 : 1
3 : 1
1 : 1
Explanation:
1:1 It can be explained by the following test cross.The other options are incorrect.
1:1 It can be explained by the following test cross.The other options are incorrect.
54329.In sickle-cell anaemia glutamic acid is replaced by valine. Which one of the following triplet codes for valine?
G G G
A A G
G A A
G U G
Explanation:
Sickle-cell anaemia is an autosome linked recessive trait. This disease is controlled by a single pair of allele HbA and Hbs only the homozygous individuals for Hbs , i.e., Hbs Hbs shows the diseased phenotype. The heterozygous individuals are carriers (HbA Hbs), Due to point mutation,
glutamic acid (Glu) is replaced by valine (Val) at sixth position of (5-chain of haemoglobin
molecule. This substitution occurs due to the single base substitution of the beta globin gene
from GAG (Glu) to GUG (Val). Whereas, the other codes GGG, AAG, GAA do not codes for valine.
Sickle-cell anaemia is an autosome linked recessive trait. This disease is controlled by a single pair of allele HbA and Hbs only the homozygous individuals for Hbs , i.e., Hbs Hbs shows the diseased phenotype. The heterozygous individuals are carriers (HbA Hbs), Due to point mutation,
glutamic acid (Glu) is replaced by valine (Val) at sixth position of (5-chain of haemoglobin
molecule. This substitution occurs due to the single base substitution of the beta globin gene
from GAG (Glu) to GUG (Val). Whereas, the other codes GGG, AAG, GAA do not codes for valine.
54330.In the F2-generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are
phenotypes-4, genotypes-16
phenotypes-9, genotypes-4
phenotypes-4, genotypes-8
phenotypes-4, genotypes-9
Explanation:
Mendel s dihybrid cross
Mendel s dihybrid cross
54331.ZZ/ZW type of sex determination is seen in
platypus
snails
cockroach
peacock
Explanation:
In ZZ/ZW case, the female has heteromorphic (ZW) sex chromosomes and the male has homomorphic (ZZ) sex chromosomes. Thus, peacock shows ZZ/ZW sex determination type.In platypus the sex determination is of XX-XY type. Both male and females has ten sex chromosome each. The male has XY, XY, XY, XY, XY and female has XXXXXXXXXX. In snails the sex determination is environmentally induced, while in cockroaches it is of XX-XO types.
In this type Y-chromosome is completely lacking. In this the presence of unpaired X-chromosomes determines the masculine sex.
In ZZ/ZW case, the female has heteromorphic (ZW) sex chromosomes and the male has homomorphic (ZZ) sex chromosomes. Thus, peacock shows ZZ/ZW sex determination type.In platypus the sex determination is of XX-XY type. Both male and females has ten sex chromosome each. The male has XY, XY, XY, XY, XY and female has XXXXXXXXXX. In snails the sex determination is environmentally induced, while in cockroaches it is of XX-XO types.
In this type Y-chromosome is completely lacking. In this the presence of unpaired X-chromosomes determines the masculine sex.
54332.All genes located on the same chromosome
form different groups depending upon their relative distance
form one linkage group
will not from any linkage group
form interactive groups that affect the phenotype
Explanation:
All the genes, present on a particular chromosome form a linkage group. The number of linkage groups of a species correspond to the total number of different chromosomes of that species.
All the genes, present on a particular chromosome form a linkage group. The number of linkage groups of a species correspond to the total number of different chromosomes of that species.
54333.In a DNA strand the nucleotides are linked together by
glycosidic bonds
phosphodiester bonds
peptide bonds
hydrogen bonds
Explanation:
In a DNA strand the nucleotides are linked together by 3’-5’ phosphodiester linkage.
In a DNA strand the nucleotides are linked together by 3’-5’ phosphodiester linkage.
54334.It is said that Mendel proposed that the factor controlling any character is discrete and independent. This proposition was based on the
results of regeneration of a cross
observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending
self-pollination of F, offsprings
cross-pollination of F,-generation with recessive parent
Explanation:
During dihybrid cross, Mendel observed that when two pairs of contrasting characters were selected for analysis; it was found that a particular character behaved independently from another character
During dihybrid cross, Mendel observed that when two pairs of contrasting characters were selected for analysis; it was found that a particular character behaved independently from another character
54335.Occasionally, a singlegene may express more than one effect. The phenomenon is called
multiple allelism
mosaicism
pleiotropy
polygeny
Explanation:
Occasionally, a single gene may express more than one trait. This phenomenon is called pleiotropy. Sometimes, one trait will be very evident and others will be less evident, e.g., a gene for white eye in Drosophila
also affect the shape of organs in male responsible for sperm storage as well as other structures.Similarly, sickle -cell anaemic individuals suffer from a number of problems, all of which are pleiotropic effects of the sickle-cell alleles.
Multiple allelism is a seies of three or more alternative or allelic forms of a gene, only two of
which can exist in any normal diploid individual,e.g., genes of blood groups in humans.Mosaicism describes the occurrence of cells that differ in their genetic component from other cells of the body.Polygeny refers to a single characteristic that is controlled by more than two genes. (it is also known as multifactorial inheritance).
Occasionally, a single gene may express more than one trait. This phenomenon is called pleiotropy. Sometimes, one trait will be very evident and others will be less evident, e.g., a gene for white eye in Drosophila
also affect the shape of organs in male responsible for sperm storage as well as other structures.Similarly, sickle -cell anaemic individuals suffer from a number of problems, all of which are pleiotropic effects of the sickle-cell alleles.
Multiple allelism is a seies of three or more alternative or allelic forms of a gene, only two of
which can exist in any normal diploid individual,e.g., genes of blood groups in humans.Mosaicism describes the occurrence of cells that differ in their genetic component from other cells of the body.Polygeny refers to a single characteristic that is controlled by more than two genes. (it is also known as multifactorial inheritance).
54336.Conditions of a karyotype 2n ± 1 and 2n ± 2 are called
aneuploidy
polyploidy
allopolyploidy
monosomy
Explanation:
Aneuploidy involves changes in chromosome number by additions or deletions of less than a
whole set, In this case organism gains or loses one or more chromosomes but not a complete set.
Polyploidy is defined as the addition of entire set of chromosome. Allopolyploidy is the polyploidy in which chromosome sets are non-homologous.
Monosomy is the process in which one chromosome Is removed from diploid set of chromosome (2n-1).
Aneuploidy involves changes in chromosome number by additions or deletions of less than a
whole set, In this case organism gains or loses one or more chromosomes but not a complete set.
Polyploidy is defined as the addition of entire set of chromosome. Allopolyploidy is the polyploidy in which chromosome sets are non-homologous.
Monosomy is the process in which one chromosome Is removed from diploid set of chromosome (2n-1).