Option (i) is the answer. Anomers are cyclic configurations of monosaccharides that differ in structure at carbon-1. I and II are anomers in this case because they differ solely in carbon-1.
Option (iii) is the answer. The absence of a free -CHO group is indicated by the fact that glucose pentaacetate does not react with hydroxylamine. Only the cyclic nature of glucose may explain this...
Option (i) is the answer. The -OH group is on the lowest asymmetric carbon on the right side of the I, II, and III structures, which is similar to (+) glyceraldehyde.
Option (iii) is the answer. Anomeric carbon is carbon that is next to an oxygen atom in the cyclic structure of glucose or fructose. 'a' and 'b' are next to the oxygen atom, as illustrated in the...
Option (iii) is the answer (A) and (C) are in the Cl-C4 range, while (B) is in the Cl-C6 range.
Option (i) and (iii) are the answers Globulular protein is the structure that develops when a chain of polypeptides coils around to form a spherical shape. Insulin and albumin, for example, are...
Option (i) and (iv) are the answers. Amylopectin and glycogen are both glucose branching polymers.
Option (ii) and (iv) are the answers. Acidic amino acids have more than one -COOH group one against the –NH2 group.
Option (i), (ii) and (iii) are the answers. (a)Lysine is a kind of amino acid with the structural formula . (b) Because the number of NH2 groups (2) is more than the number of COOH groups, it is a...
Option (i) and (iii) are the answers. The five-membered cyclic structure of ribose and fructose is shown (furanose structures). They have a five-membered ring, similar to the foran compound....
Option (ii) and (iv) are the answers. Disulphide linkage and hydrogen bonding hold polypeptide chains together in fibrous proteins.
Option (i) and (ii) are the answers. Purines are made up of a six-membered nitrogen-containing ring fused together with a five-membered nitrogen-containing ring. Purine bases guanine and adenine...
Option (i) and (iv) are the answers. Enzymes are protein molecules that act as biocatalysts in the body's chemical reactions.
Lactose is the sugar found in milk. Glucose and galactose are two monosaccharides found in lactose. Disaccharides are oligosaccharides that include two monosaccharide units.
When glucose is heated with HI for a long time, n-hexane develops, implying that all six carbon atoms are connected in a straight chain.
When a nitrogenous base is connected to the 1' position of a five-carbon sugar, a nucleoside is produced. The 5' carbon of the sugar in a nucleoside molecule is bonded to the 5' carbon of the sugar...
Glycosidic linkages connect the monosaccharide units of polysaccharides. When an oxide bond is created between two monosaccharide units with the loss of a water molecule, it is called a glycosidic...
When glucose is treated with a mild oxidising agent like Br2 water, it is transformed to gluconic acid, a six-carbon carboxylic acid. When glucose is treated with nitric acid, it is transformed to...
On the left side of the C5 carbon atom, the –OH group is connected. As a result, the provided compound is in the 'L' configuration.
Invert sugar is another name for sucrose. It comes from sugarcane and sugarbeet and is a naturally occurring sugar. When sucrose is hydrolyzed, the sign of rotation changes from Dextro (+) to laevo...
The sort of amino acids that make up a polypeptide chain are -amino acids and alpha-amino acids, where the amino acid is linked to the -carbon in the molecule.
The –NH group of each amino acid residue hydrogen is bound to the –C=O of an adjacent turn of the helix, forming a right-handed screw helix shape.
Enzyme oxidoreductases is the name given to a group of enzymes that catalyse redox processes. Alcohol Dehydrogenase, for example, aids in the reduction of alcohol levels in the human body when...
The sugar found in milk, lactose, is transformed to lactic acid during curdling, which is produced by bacteria.
When glucose is acetylated using acetic anhydride (CH3CO)2O in the presence of ZnCl2, glucose pentaacetate is formed, confirming the presence of five –OH groups.
The chemical in question is glucose pentaacetate. The presence of a free –C=O group in glucose indicates the presence of a free carbonyl group, as does the synthesis of oxime from glucose. Because...
Because vitamin C is a water-soluble vitamin, any excess is eliminated from the body on a regular basis. Vitamin C cannot be stored in the body, thus it must be consumed on a regular basis.
Sucrose's aqueous solution is dextrorotatory, rotating plane-polarized light entering the solution 66.5 degrees to the right. When sucrose is hydrolyzed with dilute acids or invertase enzyme, two...
An amino acid has both a –NH2 and a –COOH group. The –COOH group loses a proton [H]+ in aqueous solution of the amino acid, while the –NH2 acquires a proton to create a zwitterion, which is a...
Hydrogen bonds and other intermolecular interactions connect the amino acid residues in proteins. The hydrogen bonds are disrupted when a physical or chemical change occurs, and the native protein...
Bromine water can be used to treat glucose, which results in the carboxylic acid gluconic acid, which verifies the presence of an aldehyde group.
cosecx dx= = = = =
tanx dx= = = = = =
e^x dx= = =
sin2x dx= = = = 0
sin2x dx= = = = =
4x^3-5x^2 +6x+9 dx= = = = = = = = =
1/x dx= = = = =
(x+1)dx= = = = = =2
(x+e^2x)dxwhere and = = = = = = =
e^x dxwhere and [ The series within brackets is a G.P. and ] = =
(x^2-x)dxwhere Here, and = = = = = =
x^2 dxwhere Here, and = = = = =
xdxwhere Here, and = = = = = = =
(x+1)dxwhere and = = = = =
xdxSince, where and = = = = = = =
= = = = Therefore, The correct option is option (D) .
(A) (B) (C) (D) SOLUTION: = = = Therefore, The correct option is option (A) .
= = = = =
= = = =
= = = = = = =
= = = =
= = = = = = =
= = = =
= = = =
= = = =
= = =
-axis, 
and 
when 
Solution: Given equations are $y=\cos x \ldots(1)$ And, $y=\sin x \ldots$ $\text { Required area }=\text { Area (ABLA) }+\text { area (OBLO) }$ $\begin{array}{l} =\int_{1}^{1} x d...
exterior to the parabola 
is 
Solution: Given equations are $x^{2}+y^{2}=16 \dots \dots (1)$ $y^{2}=6x \dots \dots (2)$ The area bounded by the circle and parabola $=2[\text { Area }(\mathrm{OADO})+\text { Area...
-axis and the ordinates 
and 
is given by [Hint: 
if 
and 
if 
] 
Solution: $\text { Required area }=\int_{-1}^{1} y d x$ $\begin{array}{l} =\int_{-1}^{1} x|x| d x \\ =\int_{-1}^{0} x^{2} d x+\int_{0}^{1} x^{2} d x \end{array}$ $\begin{array}{l}...
, the 
-axis and the ordinates 
and is 
Solution: $\begin{array}{l} \text { Required area }=\int_{-2}^{1} y d x \\ =\int_{-2}^{1} x^{3} d x \\ =\left[\frac{x^{4}}{4}\right]_{-2}^{1} \end{array}$ $\begin{array}{l}...
Solution: Eq. of parabola is $y^{2}=4 x$......(i) Eq. of circle is $4 x^{2}+4 y^{2}=9 \ldots \ldots$ (ii) From the diagram, the points of intersection of parabola (i) and circle (ii) are...
and 
Solution: Given eqs. of lines are $\begin{array}{l} 2 x+y=4 \ldots(1) \\ 3 x-2 y=6 \ldots(2) \end{array}$ And, $x-3 y+5=0 \ldots$ (3) Area of the region bounded by the lines is the area of...
and 
Solution: Vertices of $\triangle A B C$ are $A(2,0), B(4,5)$, and $C(6,3)$. Eq. of line segment $A B$ is $\begin{array}{l} y-0=\frac{5-0}{4-2}(x-2) \\ 2 y=5 x-10 \\ y=\frac{5}{2}(x-2) \end{array}$...
and 
Solution: Area bounded by the curves, $\left\{(x, y): y \geq x^{2}\right.$ and $\left.y=|x|\right\}$, is represented by the shaded region as It can be observed that the required area is symmetrical...
[Hint: the required region is bounded by lines 
and 
![-y=11]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5fcefea7abf57385227f72cfb8cccde1_l3.png "Rendered by QuickLaTeX.com")
Solution: Area bounded by the curve, $|x|+|y|=1$, is represented by the shaded region $\mathrm{ADCB}$ as The curve intersects the axes at points $A(0,1), B(1,0), C(0,-1)$, and $D(-1,0)$. It can be...
, the line 
and 
axisSolution: Area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented bv the shaded reaion $\mathrm{OABCO}$ as Point of intersection of the parabola,...
and the line 
Solution: Area of the smaller region bounded by the ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the line, $\frac{x}{a}+\frac{y}{b}=1$ is represented by the shaded region BCAB as...
and the line, 
Solution: Area of the smaller region bounded by the ellipse, $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and the line, $\frac{x}{3}+\frac{y}{2}=1$ is represented by the shaded region BCAB as $\therefore$...
and the line 
Solution: Area enclosed between the parabola, $4 y=3 x^{2}$, and the line, $2 y=3 x+12$, is represented by the shaded area OBAO as Points of intersection of the given curves are $A(-2,3)$ and...
and the line 
Solution: Area enclosed between the parabola, $y^{2}=4 a x$, and the line, $y=m x$, is represented by the shaded area $\mathrm{OABO}$ as Points of intersection of both the curves are $(0,0)$ and...
between 
and 
Solution: Graph of $y=\sin x$ can be drawn as $\therefore$ Required area $=$ Area OABO $+$ Area BCDB $\begin{array}{l} =\int_{0}^{\pi} \sin x d x+\left|\int_{k}^{2 \pi} \sin x d x\right| \\ =[-\cos...
and evaluate 
Solution: Given eq. is $y=|x+3|$ Corresponding values of $x$ and $y$ are given in the following table. $$\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline$x$ & $-6$ & $-5$ & $-4$ & $-3$ & $-2$ & $-1$ & 0 \\...
and 
Solution: The area in the first quadrant is bounded by $y=4 x^{2}, x=0, y=1$, and $y=4$ is represented by the shaded area ABCDA as $\begin{aligned} \therefore \operatorname{Area} A B C D...
and Solution: The area required is represented by the shaded area OBAO as The point of intersection of the curves $y=x$ and $y=x^{2}$ is A(1,1) Draw AC perpendicular to x-axis $\therefore$ Area...
, , 
and x-axis (ii) ,, 
and x-axis.Solution: (i)The area required is represented by the shaded area ADCBA as $\begin{aligned} \text { Area } \mathrm{ADCBA} &=\int^{2} y d x \\ &=\int^{2} x^{2} d x \\...
and 
is: 
Solution: The area lying between the curve, $y^{2}=4 x$ and $y=2 x$, is represented by the shaded area OBAO as Points of intersection of these curves are $O(0,0)$ and $A(1,2)$. Draw AC perpendicular...
and the line 
is 
Solution: The smaller area enclosed by the circle, $x^{2}+y^{2}=4$, and the line, $x+y=2$, is represented by the shaded area ACBA as It is observed that, $\text { Area } A C B A=\text { Area } O A C...
and 
Solution: The eqs. of the sides of the triangle are $y=2 x+1, y=3 x+1$, and $x=4$. On solving these eqs., we get the vertices of triangle as $\mathrm{A}(0,1), \mathrm{B}(4,13)$, and $\mathrm{C}$...
and 
Solution: $\mathrm{BL}$ and $\mathrm{CM}$ are drawn perpendicular to $x$-axis. It is observed in the following figure that, Area $(\triangle \mathrm{ACB})=$ Area (ALBA) $+$ Area (BLMCB) - Area...
, , and 
.Solution: The area of the region bounded by the curves $y=x^{2}+2$, $y=x$, $x=0$ and $x=3$ is represented by the shaded area OCBAO as Therefore, Area of OCBAO = Area of ODBAO - Area of ODCO...
and 
Solution: The area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$ is reprsented by the shaded area as On solving the equations, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, we get the...
which is interior to the parabola 
.Solution: Required area is represented by the shaded area OBCDO On solving the given equation of circle, $4 x^{2}+4 y^{2}=9$, and parabola, $x^{2}=4 y$, we get $\mathrm{B}\left(\sqrt{2},...
, y-axis and the line 
is Solution: Area bounded by the curve, $y^{2}=4 x, y$-axis, and $y=3$ is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{3} x d y \\ &=\int_{0}^{3}...
and the lines and 
is 
Solution: The area bounded by the circle and the lines, $x=0$ and $x=2$, in the first quadrant is represented as $\begin{aligned} \therefore \text { Area } \mathrm{OAB} &=\int_{0}^{2} y d x \\...
Solution: The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$. Area of OACO is symmetrical about $x$-axis. $\therefore$ Area of $O A C O=2$ (Area of...
and the line 
Solution: The area bounded by the curve, $x^{2}=4 y$, and line, $x=4 y-2$, is represented by the shaded area OBAO. Let's say $A$ and $B$ be the points of intersection of the line and parabola....
and the 
.Solution: The area bounded by the parabola, $x^{2}=y$, and the line, $y=|x|$, can be represented as The given area is symmetrical about $y$-axis. $\therefore$ Area $\mathrm{OACO}=$ Area...
and is divided into two equal parts by the line 
, find the value of a.Solution: The line, $x=a$, divides the area bounded by the parabola and $x=4$ into two equal parts. $\therefore$ Area $O A D=$ Area $A B C D$ It is observed that the given area is symmetrical about...
cutt off by the line 
Solution: The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cutt off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the ABCDA It can be observed that the area $A B C D$ is...
and the circle Solution: The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is...
Solution: The given eq. of the ellipse can be represented as $\begin{array}{l} \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \\ \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}} \end{array}$ It can be observed that the...
Solution: The given eq. of the ellipse, $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$ can be reprsented by It is observed that the ellipse is symmetrical about x-axis and y-axis. $\therefore$ Area...
, 
, and the y-axis in the first quadrant.Solution: The area of the region bounded by $x^{2}=4y$, $y=2$, $y=4$ and the y-axis is the area of ABCD
, , and the x-axis in the first quadrant.Solution: Area of the region bounded by the curve $y^{2}=9x$, $x=2$, $x=4$, \and the x-axis is the area of ABCD
and the lines x=1, x=4 and the x-axis in the first quadrant.Solution:
equals: (A) (B) (C) (D) SOLUTION: Let I = = It is in the form of since here and I = Therefore, The correct option is option (B) .
equals to (A) (B) (C) (D) Ans. Let I = = ……….(i) Putting From eq. (i), I = = = Therefore, The correct option is option (B)...
Putting = = = = = = = = = =
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
Let I = = I = I = I = = 5I = I =
Let I = = = I = It is in the form of since here and = = . I =
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
Let I = It is in the form of since here and I = =
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
(a) Assertion and Reason both are true, Reason is the correct explanation of Assertion. (b) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (c) Assertion...
Let I = = = = = It is in the form of since here and =
Let I = = = I = It is in the form of since here and I =
Let I = It is in the form of since here and I =
Solution: \[\left( c \right)\text{ }\left( A\text{ }\to \text{ }5 \right),\text{ }\left( B\text{ }\to \text{ }1 \right),\text{ }\left( C\text{ }\to \text{ }4 \right),\text{ }\left( D\text{...
= = = = = =
Solution: \[\left( d \right)\text{ }\left( A\text{ }\to 4 \right),\text{ }\left( B\to \text{ }1 \right),\text{ }\left( C\text{ }\to 2 \right),\text{ }\left( D\text{ }\to 3 \right)\] Isomerism...
= = = = = = = =
Solution: \[\left( a \right)\text{ }\left( A\text{ }\to \text{ }3 \right),\text{ }\left( B\text{ }\to 1 \right),\text{ }\left( C\text{ }\to \text{ }5 \right),\text{ }\left( D\to \text{ }2...
Let I = = = = = =
Solution: \[\left( b \right)\text{ }\left( A\text{ }\to 5 \right),\text{ }\left( B\text{ }\to \text{ }4 \right),\text{ }\left( C\text{ }\to \text{ }1 \right),\text{ }\left( D~\to 2 \right)\] ...
= = = = =
Let I = ……….(i) Putting I = = = = = = = = =
Solution: \[\left( b \right)\text{ }\left( A\text{ }\to 4 \right),\text{ }\left( B\text{ }\to \text{ }3 \right),\text{ }\left( C\text{ }\to \text{ }2 \right),\text{ }\left( D\text{ }\to \text{ }1...
Putting = = = = = = =
Let I = ……….(i) Putting From eq. (i), I = = = = = = = Putting and = =
Solution: Ambidendate ligands are those having diverse two restricting destinations. Example: Isothiocyanato Thiocyanato and Nitrite-N Nitrito-O The kind of isomerism when ambidentate ligands are...
Solution: In CuSO4.5H2O, there are water particles that go about as ligands. The electrons will invigorate to higher d orbital and show tone. While, in CuSO4, there are no water particles to go...
Let I = = = = = = = =
Solution: \[{{[CO{{F}_{\mathbf{6}}}]}^{\mathbf{3}-}}:\text{ }C{{o}^{\mathbf{3}+}}~({{d}^{\mathbf{6}}})\]
Let I = Putting I = = = = = = = = =
Solution: (a, c) NO2 and SCN are ambidentate ligands subsequently, show linkage isomerism.
Solution: hence, option a, b and c are correct
= = = = = =
Solution:
= = = = = =
= = = = = =
= [Applying product rule] = = = = = = = =
= = = = =
= = = = =
equals: (A) (B) (C) (D) SOLUTION: Let I = = = …(i) Putting Putting this value in eq. (i), I = = I = = = = = = Therefore,The correct option is option...
equals: (A) (B) (C) (D) SOLUTION: Let …….(i) = = = Therefore,The correct option is option (B).
Let I = …….(i) Putting From eq. (i), I = = = = = = = = = =
Let I = = = …(i) Putting Putting this value in eq. (i), I = = I = = = = = =
Let I = …….(i) Putting From eq. (i), I = = = = = = = = =
…….(i) Putting = …….(ii) = ….(iii) Let …….(iv) = = = = =
Let I = …….(i) Putting From eq. (i), I = = = = = = = = =
Let I = Multiplying both numerator and denominator by , I = = ……..(i) Putting From eq. (i), I = = = = = = = =
= Putting , = …..(i) Comparing the coefficients of A + B = 0 ……(ii) Comparing constants A – B = 1 …….(iii) On solving the eq. (ii) and (iii), we get A = B = Putting the values of A, B...
Let I = …….(i) Putting Putting this value in eq. (i), I = = = = = = = = =
= …….(i) = = = = = = = =
Option (i) is correct During polymerisation, extensive cross linking results in the creation of a three-dimensional network that is rigid, insoluble, and infusible.
Option (v) is correct Natural rubber is made up of isoprene molecules, while neoprene, a synthetic rubber, is made up of chloroprene polymers.
Option (ii) is correct Polyamides, such as nylon, are the most often used fibres. Because of the strong intermolecular hydrogen connection, they have a high tensile strength.
Option (iv) is correct Enzymatic hydrolysis and environmental oxidation do not destroy the majority of synthetic polymers. Toxic characteristics are not produced via polymerization.
(i) is d (ii) is a (iii) is b (iv) is e (v) is c
(i) is d (ii) is a (iii) is b
(i) is d (ii) is e (iii) is a (iv) is f (v) is b (vi) is c
(i) is b (ii) is c (iii) is a (iv) is e (v) is d
(i) is c (ii) is a (iii) is b (iv) is e (v) is d
Intermolecular H –bonding exists between the two terminal groups –C=O and –NH, allowing one Nylon molecule to join another. Because intermolecular H –bonding is relatively strong, it can result in...
HDP stands for high-density polymer with a linear structure, whereas LDP stands for low-density polymer with a branching structure. HDP has a greater melting point and is chemically inert, whereas...
The polymer may be stretched by applying force due to the presence of these weak forces. When the external force is released, the polymer recovers to its original condition, showing elastic...
The two starting monomers for this Resin intermediate are melamine and formaldehyde. Melamine polymer is the result of their condensed polymerization.
The enzyme functions as a catalyst, speeding up biological processes. Proteins make up their structure. Proteins are polymers made up of monomeric units called amino acids. As a result, enzymes are...
Rubber may be stretched by applying force and then returned to its original shape once the force is removed. As a result, they are referred to as elastomers.
This is a network or cross-linked polymer. Cross-linking connects two linear polymers.
In this illustration, a chain-growth polymerization process combines two monomers to produce a polymer. Only monomers react with each other to keep the chain expanding.
Solution: (b, d) In complexes, [Fe(NH3)4Cl2]+ and [CO(NH3)4Cl2], metal is bonded to more than one sort of ligands subsequently, they are heteroleptic.
Solution: (a, c) In complexes [Co(NH3)6]3+ and [Ni(CN)4]2-, both Co and Ni are connected to one sort of ligands just subsequently, they are homoleptic. hence, option a and c are correct
Solution: (b, c) Aqueous pink arrangement of cobalt (II) chloride is because of electronic progress of electron from t2g to eg energy level of [Co(H2O)6]2+ complex. At the point when overabundance...
Option (i) and (iv) are the answers. Nylon and polyester are thread-forming fibres with a high melting point and tensile strength. Intermolecular forces such as hydrogen bonding are strong.
Solution: hence, option a and c are correct for [Fe(CN)6]3- complex
Solution: so, option a and c are the correct answer
Option (i) and (iii) are the answers. Polychloroprene and Buna-N are examples of a synthetic rubber
Solution: hence , a and c are the correct answer
Solution: (c) [Pt(NH3)2Cl(NO2)] is diamminechloridonitrito-N-platinum (II)
Solution: (b) The given mixtures have diverse number of water atoms inside and outside the organize circle.
Option (iii) and (iv) are the answers. Because of strong intermolecular interactions such as H-bonding, nylon and terylene are employed as fibres, resulting in tight chain packing and therefore...
Option (i) and (iv) are the answers. Teflon and polystyrene are thermoplastics because they can be melted and moulded anew.
Option (i) and (iii) are the answers. Thermosetting polymers have a lot of branching cross links. They can't be used again since they don't melt when heated and can't be remoulded.
Solution: (b) Ligand should give a pair of electrons or inexactly held electron pair to metal and shape a M – L bond,
Solution: (a) Thiosulphate or S2O3–isn't a chelating specialist since it is a monodentate ligand.
Solution: (d) [Co(SO4)(NH3)5]Br and [CO(SO4)(NH3)5]Cl address no isomerism since they are different compounds.
Solution: (a) The ligands having two distinctive holding locales are known as ambident ligands e.g., NCS, NO2, and so forth Here, NCS has two restricting locales at N and S. Subsequently, NCS...
Solution:
Solution:
Solution: (a) [Pt(NH3)2Cl2] is diamminedichloridoplatinum (II) .
Solution: (d) 3 mol of AgCl implies 3Cl are given in the arrangement henceforth, the equation of the complex will be [Cr(H2O)6]Cl3.
Solution: (b) One mole of AgNO3 accelerates one mole of chloride particle. In the above response, when 0.1 mole COCl3(NH3)5 is treated with abundance of AgNO3, 0.2 mole of AgCl are obtained hence,...
Solution:
Solution: (b) The greater the value of log K, the more prominent will be strength of complicated compound shaped. For response, For this response, log K has most elevated worth among the...
SOLUTION: = …….(i) [On dividing numerator by denominator] Let = = …….(ii) Comparing coefficients of : A + B = 2 …….(iii) Comparing constants: –A + B = 1 …….(iv) On solving eq. (iii) and...
SOLUTION: = = …….(i) Comparing coefficients of : A + +B + C = 0 …….(ii) Comparing coefficients of : –B + 3C= 5 …….(iii) Comparing constants: –4A – 2B + 2C = 0 …….(iv) On solving eq. (i),...
SOLUTION: = = …….(i) Comparing coefficients of : 2A + 2B + C = 0 …….(ii) Comparing coefficients of : 5A + B = 2 …….(iii) Comparing constants: 3A – 3B – C = –3 …….(iv) On solving eq. (i),...
SOLUTION: = = = = = …….(i) Comparing coefficients of : A + C = 0 …….(ii) Comparing coefficients of : B – 2C= 3 …….(iii) Comparing constants: –2A + B + C = 5 …….(iv) On solving eq. (i),...
SOLUTION: = …….(i) Comparing coefficients of : A + C = 0 …….(ii) Comparing coefficients of : A + B – 2C= 1 …….(iii) Comparing constants: –2A + 2B + C = 0 …….(iv) On solving eq. (i), (ii) and...
SOLUTION: …….(i) Comparing coefficients of A + C = 0 …….(ii) Comparing coefficients of , –A + B = 1 …….(iii) Comparing constant terms, –B + C = 0 …….(iv) Solving eq. (ii), (iii) and (iv), we...
SOLUTION: = = = (Dividing numerator by denominator) = …….(i) Now = …….(ii) Comparing coefficients of on both sides –2A + B = …..(iii) Comparing constants A = 1 …….(iv) Solving eq....
SOLUTION: = = = = ….(i) Comparing coefficients of on both sides A + B = 2 …….(ii) Comparing constants 2A + B = 0 …..(iii) Solving eq. (ii) and (iii), we get A = and B = 4 Putting these...
SOLUTION: = …….(i) Comparing coefficients of : A + B + C = 0 …….(ii) Comparing coefficients of : –5A – 4B – 3C= 1 5A + 4B + 3C = –1 …….(iii) Comparing constants: 6A + 3B + 2C = 0 …….(iv) On...
SOLUTION: = …….(i) Comparing coefficients of : A + B + C = 0 …….(ii) Comparing coefficients of : –5A – 4B – 3C= 3 5A + 4B + 3C = –3 …….(iii) Comparing constants: 6A + 3B + 2C = –1 …….(iv) On...
SOLUTION: = = =
SOLUTION: …….(i) Comparing coefficients of on both sides A + B = 1 …..(ii) Comparing constants 2A + B = 0 …..(iii) Solving eq. (ii) and (iii), we get A = and B = 2 Putting these values of A...