Give two examples to show the anomalous behaviour of fluorine.
Question: Give two examples to show the anomalous behaviour of fluorine. Solution: Anomalous behaviour of fluorine (i)It forms only one oxoacid as compared to other halogens that form a number of oxoacids. (ii)Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expecte...
Read More →Considering the parameters such as bond dissociation enthalpy,
Question: Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2and Cl2. Solution: Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors. 1.Bond dissociation energy 2.Electron gain enthalpy 3.Hydration enthalpy The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than t...
Read More →Find the circumference and area of a circle of radius 4.2 cm.
Question: Find the circumference and area of a circle of radius 4.2 cm. Solution: We know that the circumference $C$ and area $A$ of a circle of radius $r$ are given by $C=2 \pi r$ and $A=\pi r^{2}$ respectively. Here, $r=4.2 \mathrm{~cm}$ So substituting the value ofrin above formulas, Circumference of the circle $C=2 \times \pi \times 4.2 \mathrm{~cm}$ $=2 \times \frac{22}{7} \times 4.2 \mathrm{~cm}$ $=26.4 \mathrm{~cm}$ Area of the circle $A=\pi \times 4.2 \times 4.2 \mathrm{~cm}^{2}$ $=\frac...
Read More →Solve the following
Question: Why is $K_{\mathrm{a}_{2}} \ll K_{\mathrm{a}_{1}}$ for $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water? Solution: $\mathrm{H}_{2} \mathrm{SO}_{4(\text { (at })}+\mathrm{H}_{2} \mathrm{O}_{(\eta)} \longrightarrow \mathrm{H}_{3} \mathrm{O}_{(\text {(a) })}^{+}+\mathrm{HSO}_{4(\text { aq })}^{-} ; \quad K_{a_{1}}10$ $\mathrm{HSO}_{4(a q)}^{-}+\mathrm{H}_{2} \mathrm{O}_{(\ell)} \longrightarrow \mathrm{H}_{3} \mathrm{O}_{(a q)}^{+}+\mathrm{SO}_{4(a q)}^{-} ; \quad K_{a_{2}}=1.2 \times 10^{-2}$ It...
Read More →∆ ABC ∼ ∆ DEF and their areas are respectively 64 cm2
Question: $\triangle A B C \sim \triangle D E F$ and their areas are respectively $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$. If $E F=15.4 \mathrm{~cm}$, find $B C$. Solution: It is given that $\triangle A B C \sim \triangle D E F$. Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. $\frac{a r(\triangle A B C)}{a r(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}$ Let $B C$ be $x \mathrm{~cm}$. $\Rightarrow \frac{64}{121}=\frac{x^{...
Read More →Solve the following
Question: Why is $K_{\mathrm{a}_{2}} \ll K_{\mathrm{a}_{1}}$ for $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water? Solution: It can benoticed that This is becausea neutral H2SO4has a much higher tendency to lose a proton than the negatively charged. Thus, the former is a much stronger acid than the latter....
Read More →If π is taken as
Question: If $\pi$ is taken as $\frac{22}{7}$, the distance (in metres) covered by a wheel of diameter $35 \mathrm{~cm}$, in one revolution, is (a) 2.2(b) 1.1(c) 9.625(d) 96.25 Solution: The disstance covered by the wheel in one revolution is equal to the circumference of the wheel. Circumference $=\pi r^{2}$ $=\frac{22}{7}\left(\frac{35}{2}\right)^{2}$ $=\frac{22}{7}\left(\frac{35}{2}\right)^{2}$ $=962.5 \mathrm{~cm}$ The distance covered by the wheel in one revolution (in m) is given by $\frac...
Read More →Write the conditions to maximize the yield of
Question: Write the conditions to maximize the yield of H2SO4by Contact process. Solution: Manufacture of sulphuric acid by Contact process involves three steps. 1.Burning of ores to form SO2 2.Conversion of SO2to SO3by the reaction of the former with O2 (V2O5is used in this process as a catalyst.) 3.Absorption of SO3in H2SO4to give oleum (H2S2O7) The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this ...
Read More →If AD and PM are medians of ΔABC and ΔPQR respectively, where ΔABC ~ ΔPQR; prove that
Question: If $A D$ and $P M$ are medians of $\triangle A B C$ and $\triangle P Q R$ respectively, where $\triangle A B C \sim \triangle P Q R ;$ prove that $\frac{A B}{P Q}=\frac{A D}{P M}$. Solution: Since,ADandPMare the medians of ΔABCand ΔPQR respectively Therefore, $B D=D C=\frac{B C}{2}$ and $Q M=M R=\frac{Q R}{2}$ ...(1) Now,ΔABC~ ΔPQRAs we know, corresponding sides of similar triangles are proportional. Thus, $\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$ ...(2) Also, $\angle A=\angle ...
Read More →If the area of a circle is equal to the sum of the
Question: If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the large circle (in cm) is(a) 34(b) 26(c) 17(d) 14 Solution: Let the diameter of the larger circle bed Now, Area of larger circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=\pi\left(\frac{10}{2}\right)^{2}+\pi\left(\frac{24}{2}\right)^{2}$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=(5)^{2}+(...
Read More →Mention three areas in which
Question: Mention three areas in which H2SO4plays an important role. Solution: Sulphuric acid is an important industrial chemicaland is used for a lot of purposes. Some important uses of sulphuric acid are given below. (i)It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate. (ii)It is used in the manufacture of pigments, paints, and detergents. (iii)It is used in the manufacture of storage batteries....
Read More →How is the presence of
Question: How is the presence of SO2detected? Solution: SO2is a colourless and pungent smelling gas. It can be detected with the help of potassium permanganate solution. When SO2is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces...
Read More →ABCD is a square of side 4 cm.
Question: ABCDis a square of side 4 cm. IfEis a point in the interior of the square such that ΔCEDis equilateral, then area of ΔACEis (a) $2 \sqrt{3}-1 \mathrm{~cm}^{2}$ (b) $4 \sqrt{3}-1 \mathrm{~cm}^{2}$ (c) $6 \sqrt{3}-1 \mathrm{~cm}^{2}$ (d) $8 \sqrt{3}-1 \mathrm{~cm}^{2}$ Solution: We have the following diagram. Since $\triangle$ CED is equilateral, Therefore, $\mathrm{EC}=\mathrm{CD}=\mathrm{DE}=4 \mathrm{~cm}$ Now, $\angle \mathrm{ECD}=60^{\circ}$ Since AC is diagonal of sqr.ABCD Therefor...
Read More →Comment on the nature of two S−O bonds formed in
Question: Comment on the nature of two SO bonds formed in SO2molecule. Are the two SO bonds in this molecule equal? Solution: The electronic configuration of S is 1s22s22p63s23p4. During the formation ofSO2, one electron from 3porbital goes to the 3dorbital and S undergoessp2hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair.p-orbital andd-orbital contain an unpaired electron each. One of these electrons formsp-pbond with one oxygen ato...
Read More →What happens when sulphur dioxide is passed through an aqueous solution of
Question: What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt? Solution: SO2acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions....
Read More →In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC.
Question: In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BDAC. If DPAB and DQBC thenprove that(a) DQ2= DP.QC(b) DP2= DQ.AP Solution: We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.(a) Now using the same property in In △BDC, we get $\triangle \mathrm{CQD} \sim \triangle \mathrm{DQB}$ ...
Read More →Solve the following
Question: How is O3estimated quantitatively? Solution: Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below....
Read More →Solve the following
Question: Why does O3act as a powerful oxidising agent? Solution: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive. Therefore, ozone acts as a powerful oxidising agent...
Read More →Complete the following reactions:
Question: Complete the following reactions: (i)C2H4+O2 (ii)4Al +3O2 Solution: (i) (ii)...
Read More →The hour hand of a clock is 6 cm long.
Question: The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is (a) $2.75 \mathrm{~cm}^{2}$ (b) $5.5 \mathrm{~cm}^{2}$ (c) $11 \mathrm{~cm}^{2}$ (d) $10 \mathrm{~cm}^{2}$ Solution: Hour hand moves $\left(\frac{1}{2}^{\circ}\right)$ in one minute. So, area, $=\frac{1}{2}\left(r^{2}\right)\left(\frac{\theta}{180} \pi\right)$ $=\frac{1}{2}(36)\left(\frac{35}{2(180)} \pi\right)$ $=5.5 \mathrm{~cm}^{2}$ So the answer is (b)...
Read More →Which of the following does not react with oxygen directly?
Question: Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe Solution: Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly....
Read More →Solve the following
Question: Why is H2O a liquid and H2S a gas? Solution: H2O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H2O, which is absent in H2S. Molecules of H2S are held together only by weak van der Waals forces of attraction. Hence, H2O exists as a liquid while H2S as a gas....
Read More →Write the order of thermal stability of the hydrides of Group 16 elements.
Question: Write the order of thermal stability of the hydrides of Group 16 elements. Solution: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy(HE) of hydrides on moving down the group. Therefore,...
Read More →If area of a circle inscribed in an equilateral triangle
Question: If area of a circle inscribed in an equilateral triangle is 48 square units, then perimeter of the triangle is (a) $17 \sqrt{3}$ units (b) 36 units (c) 72 units (d) $48 \sqrt{3}$ units Solution: Let the circle of radiusrbe inscribed in an equilateral triangle of sidea. Area of the circle is given as 48. $\Rightarrow \pi r^{2}=48 \pi$ $\Rightarrow r^{2}=48$ $\Rightarrow r=4 \sqrt{3}$ Now, it is clear that ONBC. So, ON is the height ofΔOBC corresponding to BC. Area ofΔABC = Area ofΔOBC +...
Read More →List the important sources of sulphur.
Question: List the important sources of sulphur. Solution: Sulphur mainly exists in combined form in the earths crust primarily as sulphates [gypsum (CaSO4.2H2O), Epsom salt (MgSO4.7H2O), baryte (BaSO4)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS2)]....
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