Write the formulas for the following coordination compounds:
Question: Write the formulas for the following coordination compounds: (i)Tetraamminediaquacobalt(III) chloride (ii)Potassium tetracyanonickelate(II) (iii)Tris(ethane1,2diamine) chromium(III) chloride (iv)Amminebromidochloridonitrito-N-platinate(II) (v)Dichloridobis(ethane1,2diamine)platinum(IV) nitrate (vi)Iron(III) hexacyanoferrate(II) Solution: (i) $\left[\mathrm{CO}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{3}$ (ii) $\mathrm{K}_{2}\left[\ma...
Read More →Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, ∠A = 700 PR = 9 cm
Question: Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, A = 700PR = 9 cm,P = 700and PQ = 4.5 cm. Show that△ABC △PQR and state similarity theorem. Solution: Now, In △ABC and △PQRA = P = 700 (Given) $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \quad\left[\because \frac{3}{4.5}=\frac{6}{9} \Rightarrow \frac{1}{1.5}=\frac{1}{1.5}\right]$ By SAS similarity criterion, △ABC △PQR...
Read More →A tent of height 77 dm is in the form a right circular cylinder of diameter 36 m and height 44 dm
Question: A tent of height $77 \mathrm{dm}$ is in the form a right circular cylinder of diameter $36 \mathrm{~m}$ and height $44 \mathrm{dm}$ surmounted by a right circular cone. Find the cost of the canvas at Rs $3.50$ per $\mathrm{m}^{2}$. [Use $\pi=22 / 7$ ] Solution: The height of the tent is 77dm = 7.7m. The height of the upper portion of the tent is $44 \mathrm{dm}=4.4 \mathrm{~m}$. Therefore, the height of the cylindrical part is $77-44=33 \mathrm{dm}=3.3 \mathrm{~m}$. The radius of the c...
Read More →If D, E and F are respectively the midpoints of sides AB, BC and CA of
Question: If D, E and F are respectively the midpoints of sides AB, BC and CA of △ABC then what is the ratio of the areas of △DEF and △ABC? Solution: By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side. $\therefore \mathrm{DF} \| \mathrm{BC}$ and $\mathrm{DF}=\frac{1}{2} \mathrm{BC}$ $\Rightarrow \mathrm{DF}=\mathrm{BE}$ Since, the opposite sides of the quadrilateral are para...
Read More →State the converse of Pythagoras' theorem.
Question: State the converse of Pythagoras' theorem. Solution: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle...
Read More →What can be inferred from the magnetic moment values of the following complex species?
Question: What can be inferred from the magnetic moment values of the following complex species? Example Magnetic Moment (BM) K4[Mn(CN)6] 2.2 [Fe(H2O)6]2+5.3 K2[MnCl4] 5.9 Solution: Magnetic moment $(\mu)$ is given as $\mu=\sqrt{n(n+2)}$. For value $n=1, \mu=\sqrt{1(1+2)}=\sqrt{3}=1.732$. For value $n=2, \mu=\sqrt{2(2+2)}=\sqrt{8}=2.83$. For value $n=3, \mu=\sqrt{3(3+2)}=\sqrt{15}=3.87$. For value $n=4, \mu=\sqrt{4(4+2)}=\sqrt{24}=4.899$. For value $n=5, \mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$. (i)K4[...
Read More →State the Pythagoras' theorem.
Question: State the Pythagoras' theorem. Solution: The square of the hypotenuse is equal to the sum of the squares of the other two sides.Here, the hypotenuse is the longest side and it's always opposite the right angle....
Read More →An iron spherical ball has been melted and recast into smaller balls of equal size.
Question: An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball. Solution: Let the radius of the big metallic ball is 4r. Therefore, the volume of the big metallic ball is $V=\frac{4}{3} \pi \times(4 r)^{3}$ The metallic sphere is melted to produce small...
Read More →State the SAS-similarity criterion.
Question: State the SAS-similarity criterion. Solution: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar....
Read More →State the SSS-similarity criterion.
Question: State the SSS-similarity criterion. Solution: If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar....
Read More →An iron s$ rac{S_{2}}{S}= rac{64 imes 4 pi imes(r)^{2}}{4 pi imes(4 r)^{2}}=4$pherical ball has been melted and recast into smaller balls of equal size.
Question: An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball. Solution: Let the radius of the big metallic ball is 4r. Therefore, the volume of the big metallic ball is $V=\frac{4}{3} \pi \times(4 r)^{3}$ The metallic sphere is melted to produce small...
Read More →State the AA-similarity criterion.
Question: State the AA-similarity criterion. Solution: If two angles of a triangle are correspondingly equal to the two angles of another triangle, then the two triangles are similar....
Read More →Comment on the statement that elements of
Question: Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. Solution: The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways. (i)The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2ndand 3rdtransition series). However, the atomic sizes of the elements in th...
Read More →State the AAA-similarity criterion.
Question: State the AAA-similarity criterion. Solution: If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and hence the triangles are similar....
Read More →State the mid point theorem.
Question: State the mid point theorem. Solution: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side...
Read More →Write down the number of 3d electrons in each of the following ions:
Question: Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, CO2+, Ni2+and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). Solution:...
Read More →State the converse of Thale's theorem.
Question: State the converse of Thale's theorem. Solution: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side....
Read More →The diameter of a metallic sphere is equal to 9 cm.
Question: The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire. Solution: The radius of the metallic sphere is $\frac{9}{2}=4.5 \mathrm{~cm}=45 \mathrm{~mm}$. Therefore, the volume of the metallic sphere is $V=\frac{4}{3} \pi \times(45)^{3}$ Cubic $\mathrm{mm}$ The metallic sphere is melted to produce a long wire of uniform cross section of radius $\frac{2}{2}=1 \mathrm{~mm}$. Let th...
Read More →State the basic proportionality theorem.
Question: State the basic proportionality theorem. Solution: If a line is drawn parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio....
Read More →Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns.
Question: Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i)electronic configurations, (ii)oxidation states, (iii)ionisation enthalpies, and (iv)atomic sizes. Solution: (i) In the $1^{\text {st }}, 2^{\text {nd }}$ and $3^{\text {rd }}$ transition series, the $3 d, 4 d$ and $5 d$ orbitals are respectively filled. We know that elemen...
Read More →State the two properties which are necessary
Question: State the two properties which are necessary for given two triangles to be similar. Solution: The two triangles are similar if and only if1. The corresponding sides are in proportion.2. The corresponding angles are equal...
Read More →Naman is doing fly-fishing in a stream. The tip of his fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m
Question: Naman is doing fly-fishing in a stream. The tip of his fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from him and 2.4 m from the point directly under the tip of the road. Assuming that the string(from the top of his road to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from him...
Read More →Prove that the surface area of a sphere is equal
Question: Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder. Solution: We have the following figure to visualize the situation Let the radius of the sphere isr. Therefore, the surface area of the sphere is $S=4 \pi \times r^{2}$ $=4 \pi r^{2}$ The circumscribed cylinder of the sphere must have radiusrcm and height 2rcm. Therefore, the curved surface area of the cylinder is $S_{1}=2 \pi r \times 2 r$ $=4 \pi r^{2}$ Hence,SandS1are same. Thus...
Read More →Solve this
Question: In △ABC, AD is a median and AL BC.Prove that (i) $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$ (ii) $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DL}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$ (iii) $\mathrm{AB}^{2}+\mathrm{AC}^{2}=2 \mathrm{AD}^{2}-\frac{1}{4} \mathrm{BC}^{2}$ Solution: (a)In right triangle ALDUsing Pythagoras theorem, we haveAL2= AD2 DL2 .....(1)Again, In right triangle ACLUsing Pythagoras theorem, we...
Read More →Write the electronic configurations of the elements with the atomic numbers
Question: Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. Solution:...
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