Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\frac{x-1}{2 x^{2}}\right) d x$ Solution: $\int e^{x}\left(\frac{x-1}{2 x^{2}}\right) d x$ Let $I=\int e^{x} \frac{1}{2 x} d x-\int e^{x} \frac{1}{2 x^{2}} d x$ Integrating by parts, $=\frac{\mathrm{e}^{\mathrm{x}}}{2 \mathrm{x}}-\int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \mathrm{x}}\right)\right) \mathrm{dx}-\int \frac{\mathrm{e}^{\mathrm{x}}}{2 \mathrm{x}^{2}} \mathrm{dx}$ $=\frac{\mathrm{e}^{\ma...
Read More →The electronic configuration of Cu(II)
Question: The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct? (i) Cu(II) is more stable (ii) Cu(II) is less stable (iii) Cu(I) and Cu(II) are equally stable (iv) Stability of Cu(I) and Cu(II) depends on the nature of copper salts Solution: Option (i)Cu(II) is more stableis the answer....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x$ Solution: Let $\mathrm{I}=\int \mathrm{e}^{\mathrm{x}}\left(\cot \mathrm{x}-\operatorname{cosec}^{2} \mathrm{x}\right) \mathrm{dx}$ $=\int e^{x} \cot x d x-\int e^{x} \operatorname{cosec}^{2} x d x$ Integrating by parts, $=\cot \int e^{x} d x-\int \frac{d}{d x} \cot \int e^{x} d x-\int e^{x} \operatorname{cosec}^{2} x d x$ $=\cot x e^{x}+\int e^{x} \operatorname{cosec}^{2} x d x-\int e^{x} ...
Read More →Electronic configuration of a transition element
Question: Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number? (i) 25 (ii) 26 (iii) 27 (iv) 24 Solution: Option (ii) 26 is the answer....
Read More →Find the value
Question: Let $f: R \rightarrow R: f(x)=\left(x^{2}+3 x+1\right)$ and $g: R \rightarrow R: g(x)=(2 x-3)$. Write down the formulae for (i) g o f (ii) f o g (iii) g o g Solution: (i) $g \circ f$ To find: g o f Formula used: g o f = g(f(x)) Given: (i) $f: R \rightarrow R: f(x)=\left(x^{2}+3 x+1\right)$ (ii) g: R R : g(x) = (2x - 3) Solution: We have, $g \circ f=g(f(x))=g\left(x^{2}+3 x+1\right)=\left[2\left(x^{2}+3 x+1\right)-3\right]$ $\Rightarrow 2 x^{2}+6 x+2-3$ $\Rightarrow 2 x^{2}+6 x-1$ Ans)....
Read More →Phosphorus forms a number of oxoacids.
Question: Phosphorus forms a number of oxoacids. Out of these oxoacidsphosphinicacid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour. Solution: 4AgNO3 + H3PO2 + 2H2O 4Ag + H3PO4 + 4HNO3 2HgCl2 + H3PO2 + 2H2O 2Hg + H3PO4 + 4HCl...
Read More →PCl5 reacts with finely divided silver on heating
Question: PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution. Write the reactions involved to explain what happens. Solution: PCl5 reacts with silver to form white silver salt(AgCl). Which than dissolves on adding excess aqueous NH3 to form a soluble complex PCl5 + 2Ag 2AgCl(white ppt.) + PCl3 AgCl + 2NH3(aq) [Ag(NH3)]+Cl-...
Read More →Give an example to show the effect
Question: Give an example to show the effect of concentration of nitric acid on the formation of oxidation product. Solution: 4Zn +10HNO3(dil) 4Zn(NO3)2 + NH4NO3 + 3H2O 4Zn +10HNO3(dil) 4Zn(NO3)2 + N2O + 5H2O 4Zn +4HNO3(conc.) Zn(NO3)2 + NO2 + 2H2O...
Read More →Phosphorus has three allotropic forms —
Question: Phosphorus has three allotropic forms (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity. Solution: 1. White phosphorus (i) It is soft waxy solid having garlic order (ii) It is poisonous (iii) Has low melting point and boiling point because P4 molecules are held together by weak van der Waal forces of attraction 2. Red Phosphorus (i) It is hard, crystalline, odourless ...
Read More →Name three oxoacids of nitrogen.
Question: Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state. Solution: There are three oxoacids of nitrogen. Nitric acid(HNO3), Nitrous acid (HNO2), Hyponitrous acid (H2N2O1) + 3 oxidation is shown by HNO2, Therefore it undergoes disproportion reaction, To calculate the oxidation state, consider the oxidation state of N is x. 3HNO2 HNO3 + 2NO +H2O The oxidation state of nitrogen in HNO2 is +3 X + (+1) +2*...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$ Solution: Let I $=\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$ We know that, $\sin ^{2} x+\cos ^{2} x=1$ and $\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}$ $=e^{x}\left(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)$ $=\frac{e^{x}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}{2 \cos ^{2} \frac{x}{2}}$ $=\frac{1}{2} ...
Read More →White phosphorus reacts with chlorine and
Question: White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water. Solution: When white phosphorous reacts with chlorine: P4 + 6Cl2 4PCl3 PCl3 + 3H2O H3PO3 + 4HCl ]*4 P4 +6Cl2 +12H2O 4H3PO3 +12HCl Moles of white P= 62/124 = 0.5mol 1mol of white P4 produces HCl = 12mol 0.5 mol of white P4 will produce HCl =...
Read More →P4O6 reacts with water according to equation
Question: P4O6 reacts with water according to equation P4O6 + 6H2O 4H3PO3. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P4O6 in H2O. Solution: P4O6 + 6H2O 4H3PO3 4H3PO3 + 8NaOH Na2HPO3 + 8H20 Overall reaction : P4O6 + 8NaOH 4Na2HPO3 + 2H2O Moles of P4O6 = 1.1/220 = 0.5 Acid formed by one mole of P4o6 requires = 8 mol Acid formed by 0.005 mol of NaOH is present in 100mL solution 0.04 of NaOH is present in solution = 1000/0.1 *0.04 = 400...
Read More →Explain why ozone is thermodynamically
Question: Explain why ozone is thermodynamically less stable than oxygen. Solution: Ozone is thermodynamically less stable because it decomposes into oxygen and this decomposition leads result in the liberation of heat, so its entropy is positive and free energy is negative....
Read More →Explain why the stability of oxoacids of
Question: Explain why the stability of oxoacids of chlorine increases in the order given below: HClO HClO2 HClO3 HClO4 Solution: As the electronegativity of halogen decreases, the tendency of XO3 group ( X = halogens) to withdraw electrons of the O-H bond towards itself decreases and hence the acid strength of the perhalic acid decreases....
Read More →In the ring test of NO3– ion,
Question: In the ring test of NO3 ion, Fe2+ ion reduces nitrate ion to nitric oxide, which combines with Fe2+ (aq) ion to form the brown complex. Write the reactions involved in the formation of a brown ring. Solution: NO3+ 3Fe2++ 4H+ NO + 3Fe2+ 2H2O [Fe(H2O)6]SO4 + NO [Fe(H2O)5NO]SO4 Thus the complex formed is brown....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x$ Solution: Let I $=\int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x$ $=\int e^{x} x^{-2} d x-2 \int e^{x} x^{-3} d x$ Integrating by parts $=x^{-2} \int e^{x} d x-\int \frac{d}{d x} x^{-2} \int e^{x} d x-2 \int e^{x} x^{-3} d x$ We know that, $\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}$ $=e^{x} x^{-2}+2 \int e^{x} x^{-3} d x-2 \int e^{x} x...
Read More →On reaction with Cl2,
Question: On reaction with Cl2, phosphorus forms two types of halides A and B. Halide A is a yellowish-white powder but halide B is a colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products. Solution: Halide A is PCl5 because it is a yellowish-white powder Halide B is PCl3 because it is a colourless oily liquid. PCl3 + 3H2O H3PO3 + 3HCl PCl5 undergoes a violent hydrolysis PCl5 + H2O POCl3 + 2HCl PCl5 + 4H2O H3PO4 + 5HCl...
Read More →SF6 is known but SCl6 is not.
Question: SF6 is known but SCl6 is not. Why? Solution: Fluorine is the strongest oxidizing agent and it can oxidize sulphur to its maximum oxidation state +6 to form SF6. Chlorine is not a good oxidizing agent, it cannot oxidize sulphur to its maximum oxidation state. Chlorine can oxidize sulphur to only +4 oxidation state. Hence it can form SCl4 but not SCl6....
Read More →Out of H2O and H2S,
Question: Out of H2O and H2S, which one has a higher bond angle and why? Solution: H2O has higher bond angle than H2S because as we move from oxygen to sulphur the size of the central atom increases and electronegativity decreases due to which bond pair goes away from the central atom which results in a decrease in bond pair repulsion and hence bond angle decreases....
Read More →Give the reason to explain
Question: Give the reason to explain why ClF3 exists but FCl3 does not exist. Solution: Chlorine has vacant d orbitals hence it can show an oxidation state of +3. Fluorine has no d orbitals, it cannot show a positive oxidation state. Fluorine shows only -1 oxidation state. Therefore FCl3 does not exists....
Read More →Why is nitric oxide paramagnetic in gaseous
Question: Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic? Solution: Nitric acid in the gaseous state exists in monomer form. It consists of only one unpaired electron therefore it is paramagnetic. In solid-state, it exists as a dimer (N2O2). There is no unpaired electron in its dimer form, therefore it is diamagnetic....
Read More →In PCl5, phosphorus is in sp3d hybridised
Question: In PCl5, phosphorus is in sp3d hybridised state but all its five bonds are not equivalent. Justify your answer with reason. Solution: The size of axial bonds is greater than the size of equatorial bonds to overcome repulsion because the three equatorial bonds cause more repulsion. Therefore two axial P-Cl bonds are longer and different from equatorial bonds....
Read More →PH3 forms bubbles when passed slowly
Question: PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why? Solution: PH3 does not form hydrogen bonding with water. Therefore it is not soluble in water and it escapes as a gas and forms bubbles whereas ammonia forms hydrogen bonding and soluble in water....
Read More →Write a balanced chemical equation
Question: Write a balanced chemical equation for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen. Solution: NH3 + O2 (Pt gauze) 4NO + 6H2O Pt gauze reacts as a catalyst, which is added to increase the rate of reaction....
Read More →