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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
Give reason for the following observations : (i) When Silver nitrate solution is added to Potassium iodide soltuion , a negatively charge colloidal solution is formed. (ii) Finely divided substance is more effective as an adsorbent. (iii) Lyophilic colloid are also called reversible sols. |
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Answer» (i) If silver nitrate solution is added to postassium iodide solution, the precipitated Agl will adsorb iodide ions formt the dispersion medium and a negatively charged colloidal solution will form. (ii) Finely divided substance is more effective as an adsorbent, more effective is the adsorption, more effective is the adsorption. (iii) Lyophillic colloids are reversible sols because the residue left on evaporation can be readily transferrred back into a colloidal state simply by adding the solvent. |
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| 202. |
Complete the following chemical equtions : (i) `Ca_(3)P_(2)+H_(2)O to` (ii) `Cu+ H_(2)SO (conc) to` |
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Answer» (i) `Ca_(3)P_(2)+ 6H_(2)O to3Ca(OH)_(2) + 2PH_(3)` (ii) `Cu+ H_(2)SO (conc) toCuSO_(4) + SO_(2) + 2H_(2)O` |
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| 203. |
Complete the following chemical equations : (i) `Ca_(3)P_(2)(s) + H_(2)O(l) rarr.........` (ii) `Cu^(2+)(aq) + underset(("excess"))(NH_(3)(aq))rarr.........` (iii) `F_(2)(g) + H_(2)O(l) rarr ..........` |
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Answer» (i) `Ca_(3)P_(2)(s) + H_(2)O(l)rarr3Ca(OH)_(2)(aq)+2PH_(3)(g)` OR `Ca_(3)P_(2)+ 6H_(2)Orarr3Ca(OH)_(2)+2PH_(3)` (ii) `underset("blue ")(Cu^(2+))(aq)+4NH_(3)(aq)iffunderset("deep blue")([Cu(NH_(3))_(4)])^(2+)(aq)` (iii) `2F_(2)(g) + 2H_(2)O(l) rarr 4H^(+)(aq) + 4F^(-)(aq) + O_(2)(g)` |
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| 204. |
Write chemical equations for the following processes : (i) Chlorine reacts with a hot concentrated solution of sodium hydroxide (ii) Orthophosphorous acid is heated (iii) `PtF_(6)` and xenon are mixed together |
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Answer» (i) `3Cl_(2)+6NaOH rarr 5 NaCl + NaClO_(3) + 3H_(2)O` (ii) `4H_(3) PO_(3) overset("heat")rarr3H_(3)PO_(4)+PH_(3)` (iii) `PtF_(6) + Xe rarrunderset("red colour compound")(Xe^(+)PtF_(6)^(-))` |
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| 205. |
(a) Write balanced equations for the following reactions : (i) Chlorine reacts with dry slacked lime. (ii) Carbon reacts with concentrated `H_2SO_4`. (b) Describe the contact process for the manufacture of sulphuric acid with special referecen to the reaction conditions, catalysts used and the yields in the process. |
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Answer» (a) (i) `2Cl_2+2Ca(OH)_2 to CaCl_2+Ca(OCL_2)+2H_2O` (ii) `C+2H_2SO_4to CO_2+2SO_22H_2O` (b) In the contact process the sulphur dioxide is mixed with air and the mixture passed over a catalyst of vanadium (v) oxide `V_2O_5` at a relatively high temperature of about `450^@C` and at a pressure of between 1-2 atm. it is an exothermic oxidation . `(1) 2SO_2(g)+O_2(g) hArr 2SO_3(g)" "(DeltaH=-196kJmol^(-))` `K_(P)=(PSO_(3)^(2))/(PSO_(2)^(2)PO_2)` (2) The sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming sulphuric acid (oleum). `SO_((3)(g))+H_2SO_((4)(l)) hArrH_2SO_((4)(l))` (3) Water is then carefully added ot the oleum to produce concentrated sulphuric acid (`98%H_2SO_4`) `H_2S_2O_(7(l))+H_2O_(l)hArr2H_2SO_(4(l)` If the sulphur trioxide is added directly to water an acid mist from which is difficult to contain because the reaction to form sulphuric acid solution is very exothermic with a big K value. |
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| 206. |
What is the oxidation number of phosphorus in `H_(3)PO_(2)` molecule ? |
| Answer» Correct Answer - `+ 3 or +1` | |
| 207. |
What is the coordination number of each type of ions in a rock-salt type crystal structure? |
| Answer» Rock salt type crystal structure has 6:6 coordination number of each type of ion. Example : In NaCl, coordination no. of `Na^(+)=6` and coordination no. of `Cl^(-)=6`. | |
| 208. |
why is Frenkel defects not found in pure alkali metal halides ? |
| Answer» Frenkel defects is not found in alkali metal halides because the ions cannot get into the interstitial sites due to their larger size. | |
| 209. |
What is point defects. Describe two types of point defects. |
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Answer» Point defects : Irregulatrities from ideal arrangement around a point in a crystalline sybstance. (A) Stoichiometric defects : The point defects that do not disturb the stoichiometric defects of cation and anion as per the chemical formula of the solid called stoichiometric defects. (B) Non-stoichiometric defects : These defects arise when stoichiometry of a substance is disturbed. |
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| 210. |
Define modulation index. Why is it generally kept less than one ? |
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Answer» Modulation index is the ratio of amplitude of modulating signal to amplitude of carrier waves. `mu=(A_(m))/(A_(c ))` It is kept low to decrease or noise. A band pass filter rejects low and high frequencies and allows a band of frequencies to pass. |
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| 211. |
Calculate the de-Brogile wavelength of the elctron orbiting in the `n=2` state of hydrogen atom. |
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Answer» Energy in orbit for which `n=2`. `E=(-13.6)/(2^(2))eV=-3.4eV` Now, `3.4xx1.6x10^(-19)=(hc)/(lambda)implies=(6.6xx10^(-34)xx3xx10^(8))/(3.4xx1.6xx10^(-19))=3.639xx10^(-7)=3639Å`. |
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| 212. |
Write the name of monomers used for getting the following polymers : (i)Telflon (ii) Buna-N |
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Answer» (i) `CF_(2)=CF_(2)`(Tetrafuoro ethylene) (ii) `CH_(2)=CH-CH-CH_(2)+CH=overset(CN)overset(|)CH` |
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| 213. |
(i) Write the type of magnetism observed when the magnetism observed when the magnetic moments are aligned in parallel and anti parallel direction in unequal number . ltbnrgt (ii) Which stoichiometric defect decreases the density of the crytal? |
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Answer» (i) Antioferromagnetism (ii) The densitiy of fernkel defect does not change |
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| 214. |
(a) Define the following terms : (i) Limiting molar conductivity, (ii) Fuel cell (b) Resistance of a conductivity cell filled with `0.1 molL^(-1)KCl` solution is 100 Ohm. If the resistance of the same cell when filled with `0.2 mol L^(-1)KCl` solution is 520 Ohm , calculate the conductivity and molar conductivity of `0.2 mol L^(-1) KCl` solution. The conductivity of `0.1 mol L^(-1) KCl` solution is ` 1.29xx10^(-1) Scm^(-1)` . |
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Answer» (i) When concentration of an electrolyte approaches zero or the molar comductiveity of a solution at infinite dilution. Then its molar conductivity is known as limiting molar conductivity. (ii) Fuel cells are the galvanic cells in which the energy of combustion of the fuels like hydrogen, methanol. etc is directly converted into electrical energy. (b) Given that : Concentration of the KCl solution = `0.1 mol L^(-1)` Resistance of cell filled with `0.1 mol L^(-1)` KCl solution = 1000 ohm Cell constant = G =conductivity `xx resistence 1.29 xx10^(-2)ohm^(-1) cm^(-1)xx100` ohm `1.29 cm^(-1)` Cell constant for a particular conductivity cell ia a consant. Conductivity of `0..2 mol L-1 KCl` solution `k= (G^(ast))/(R)= (1.29)/(520) = 2.48xx10^(-3)` Now, Molar conductivity = `wedge_(m)` `(kxx100)/(M)=(2.48xx10^(-3)xx1000)/(0.02) = 124 S cm^(2) mol^(-1) ` . |
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| 215. |
Define the following terms : (i) Fuel cell , (ii)Limiting molar conductivity |
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Answer» (i) A device in which the heat produced as a combustion of a fuel like `H_(2)` in the presence of `O_(2)` is converted in to electricity is called fuel cell (ii) when concentratio approaches zero molar conductivity reaches to limiting value |
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| 216. |
What are the dispersed phase and dispersion medium in milk |
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Answer» Dispersed phase = fat Dispersed Medium : water |
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| 217. |
Define the following terms: (i) Glycosidic linkage (ii) Invert sugar (iii) Oligosaccharides |
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Answer» (i) Two mono saccharides joined together by oxide linkage formed by loss of water melecule (ii) Invert sugar : Hydrolysis of sucrose brings a change in sign of ratation from dextro to leavo hence it is known as invert (iii) oligosaccharides : carbohydrates which on hydrolysis give 2- 10 units of mono saccharides |
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| 218. |
How does an algal bloom cause eutrophication of a water body? Name the weed that can grow in such a eutrophic lake. |
| Answer» Algal bloom in the lake or any other water body forms a scum. The scum depletes the oxygen in the water leading to foul smelling of the water body. The oxygen depletion affects the aquatic life adversely resulting in the death of fish and ultimately the eutrophic lake itself dies. Eicchornia crassipes/water hycinth grows in a eutrophic lake. | |
| 219. |
Why a signal transmitted from a TV tower cannot received beyound a certain distance ? Write the expression for the optimum separation separation between the receiving and the transmitting antenna. |
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Answer» A signal transmitted from a TV tower cannot be received beyound a centarin distance due to attenuation and curvature of earth. Every antenna has the range. Rahge is the distance between a source and a destination up to which a signal is received with sufficient strength. Optimum distance between the receiving and transmitting antenna. `d_(M)=sqrt(2Rh_(T))+sqrt(2Rh_(R))` ` "where R = Radius of earth "` `h_(T)="Height of transmitting antenna"` `h_(R)="Height of receiving antenna"` |
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| 220. |
Given below is the representation of amino acid composition of the relevant translated portion of `beta`-chain of haemoglobin, related to the shape of human red blood cells. (a) Is this representation indicating a normal human or a sufferer from certain related genetic disease? Give reason in support of your answer. (b) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene? (c) Who are likely to suffer more from the defect related to the gene represented the males, the females or both males and females equally? And why? |
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Answer» (a) This representation indicates a normal human because in the respective amino acid chain, Glutamic and is present at the 6th position. (b) In the sufferer who exhibits sickle cell trait, defect is caused by the substitution of Glutamic acid (Glu) by valine (Val) at the 6th position of Beta globin chain of the haemoglobin. (c) Both the males and females suffer equally because sickle cell anaemia is not a sex linked disease. It is an autosomal disease and sickle shaped RBC will cause equal deficiency of oxygen in both males and females. |
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| 221. |
Mention the important considerations required while fabricating a p-n junction diode to be used as light emitting diode (LED). What should be the order of the band gap of an LED if it is required to emit light in the visible region. |
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Answer» (1) The reverse breakdown voltages of LEDs are very low, typically arround 5V. So, care should be taken while fabricating a pn- junction diode so that the high reverse voltages do not appear across them. (2) There is very little resistance to limit the current in LED. Therefore, a resistor must be used in series with the LED to avoid any damage to it. (3) The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8eV. |
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| 222. |
Describe the process of decomposition of detritus under the following heads : Fragmentational leaching, catabolism, humification and mineralization. |
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Answer» Decomposition is the process in which the complex organic matter is broken down into simpler organic substances and ultimately into inorganic compounds. Steps of Decomposition. (i) Fragmentation : It is the process of breaking of the detritus into smaller particles by detritivores like earthworm. (ii) Leaching : It is the process in which water soluble inorganic substances run down mto soil horizon and get precipitated as unavailable salts. (iii) Catabolism: The enzymatic conversion of the detritus into simple organic compounds and then into inorganic compounds is called catabolism. (iv) Humification : Humification during decomposition leads to the accumulation of a dark cloured amorphous substance called humus. (v) Mineralisation : It is the process in which the humus is degraded by certain microbes and the inorganic nutrients are relased. The products of decomposition are simple organic moleculers and inorganic compounds. |
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| 223. |
Account for the following : (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction. (b) `pK_(a)` value of 4-nitrobenzoic acid is lower than that of benzoic acid . |
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Answer» (a) Aromatic carboxylic acids do not undergo Friedal-Crafts reaction because the carboxylic group is strongly deactivating and the catalyst `AlCl_(3)` (lewis acid ) gets bonded to the carboxylic group strongly . (b) Lower the value `pK_(a)` - greater will be the acidic character , 4-nitrobenzoic acid is stronger base than that of benzoic acid . Nitro group is electron withdrawing group . It has powerful electron-withdrawing resonance effects as well as electron-withdrawing inductive effect due to that it is easy to release the hydrogen . Hence , to make 4-nitrobenzoic acid stronger base. |
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| 224. |
Which compound in each of the following pairs will react faster in `S_(N)2` reaction with -OH ? Why? (i) `CH_(3)Br or CH_(3)I` (ii) `(CH_(3))_(3)C Cl or CH_(3)Cl` |
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Answer» (i) Since `I^(-)` ion is a better leaving group than `Br^(-)` ion, therefore, `CH_(3)I` reacts faster than `CH_(3)` Br in `S_(N)2` reaction with `OH^(-)` ion. (ii) On steric grounds, `1^(@)` alkyl halides are more reactive than tertalkyl halides in `S_(N)2` reactions. Therefore, `CH_(3)I`. will react at a faster rate than `(CH_(3))_(3) C Cl` in a `S_(N)2` reaction with `OH^(-)` ion. |
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| 225. |
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy `2//3^(rd)` of tetrahedral voids ? |
| Answer» The number of tetrahderal voids formed is equal to twice the number of atoms of element B and only `(2)/(3)`rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is `2xx(2//3) : 1 or 4 : 3` and the formula of the compound is `A_(4)B_(3).` | |
| 226. |
Which would undergo `S_(N)1` reaction faster in the following pair : `CH_(3)-CH_(2)-CH_(2)-Br and CH_(3)-underset(Br)underset(|)CH-CH_(3)` |
| Answer» A tertiary alkyl tends to undergo the `S_(N)1` mechanism because it can form a tertiary carbocation, which is stablized by the three alkyl groups attached to it. As alkyl group are electron donating, they allow the positive charge in the carbocation to be delocalised by the induction effect. Hence out of the given pair `(CH_(3))_(3)` CBr would undergo `S_(N)1` reaction faster than `CH_(3)-CH_(2)-Br` | |
| 227. |
Name the type of association that genus Glomus exhibits with higher plants. |
| Answer» Mycorrhizae association. | |
| 228. |
List the symptoms of Ascariasis. How does a healthy person acquire this infection ? |
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Answer» Ascaris, the common roundworm and wucheria, the filarial worm, are some of the helminths which are known to be pathogenic to Man Ascaris, an intestinal Parasite causes ascariasis. Symptoms : In this disease include internal bleeding, muscular pain, fever, anemia and blockage of the intestinal passage. Mode of transmission of disease : The egges of the parasite are excreted along with the faeces of infected person which contaminate soil, water, plants etc. A healthy person acquires has this infection through contaminated water, vegetable, fruits etc. |
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| 229. |
Explain the significant role of the genus Nucleopoyhedrovirus in ab ecological sensitive area. |
| Answer» The genus Nucleopolyhedrovirus provides viruses that are being used for species-specific narrow spectrum insecticidal application. They have been shown to have no negative impact on plants, birds, animals, fish or even non-targetinsects. They have been shown to have no negative impact on plants, birds, animals fish or even non-targetinsects. They are useful in overall integrated pest managemant or ecologically sensitive areas. | |
| 230. |
Name the type of interaction seen in each of the following examples (i) Ascaris worms living in the intestine of human (ii) Wasp pollinating fig inflorescence. (iii) Clown fish living among the tentacles of sea-anemone (iv) Mycorrhizae living on the roots of higher plants (v) OPCh-id growing on a branch of a mango tree (vi) Disappearance of smaller barnacles when Balanus dominated in the Coast of Scotland. |
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Answer» (i) Parasitism (ii) Mlltualism (iii) Commensalism ltbRrgt (iv). Mutualism - (v) Commensalism (vi) Competition |
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| 231. |
Name the interaction in each of the following : (a) Ascaris worms living in the intestine of human (b) Sucker fish attached to the shark (c) Smallerbarnacles disappeared when Balanus dominated in the coast of Scotland |
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Answer» a = Parasitism b = Commensaliqm c = Competition |
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| 232. |
Name the interaction that exists between sucker fish and shark. |
| Answer» Correct Answer - Commensalism. | |
| 233. |
Explain the causes, inheritance pattern and symptoms of any two Mendelian genetic disorders . |
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Answer» Haemophilia (i) It is a sex-linked recessive disorder. (ii) Patient continues to bleed even on a minor cut as the patient does not possess natural phenomenon of blood clotting. (iii) The gene for haemophilia is located on X- chromosome. (iv) So, more males suffer from haemophilia than females because in males single gene for the defect is able to express (v) The defective alleles produce non-functional cascade of protein involved in blood clotting (vi) Females suffer from this disease only in homozygous condition, i.e.,` X^(C)X^(C)`. Sickle-cell Anaemia (i) It is an autosomal linked recessive trait (ii) The disease is controlled by a single pair of allele `H_(b)^(A) and H_(b)^(S)` (iii) Only the homozygous individuals for `H_(b)^(S)`,ie., `H_(b)^(S)H_(b)^(S)` show the diseased phenotype. (iv) The heterozygous individual are carriers `(H_(b)^(A)H_(b)^(S))` (v) Due to point mutation Glutamic acid (Glu) is replaced by Valine (Val) at sixth positions of beta globin chain of haemoglobin molecule. (vi) `H_(b)^(S)` behaves as normal haemoglobin except under oxygen stress where erythrocytes lose their circular shape and become sickle-shaped. As a result, the cells cannot pass through narrow capillaries are clogged and thus, affect blood supply to different organs. |
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| 234. |
Mention the site where syngamy occurs in amphibians and reptiles respectively. |
| Answer» In amphibians, external fertilisation occurs hence, syngamy occurs in the medium of water. In reptiles, internal fertilization occurs hence, syngamy occurs within the body. | |
| 235. |
(a) Why is bithional added to soap ? (b) What is tincture of iodine ? Write its one use. (c) Among the following , which one acts as a food preservative ? Aspartame , Aspirin , Sodium Benzoate , Paracetamol |
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Answer» (a) Bithional is a disinfectant and used as an antiseptic in medicated soap . (b) Tincture of iodine is 2-3% solution of iodine in alcohol and water . It is a powerful antiseptic and applied on wounds . (c) Sodium Benzoate acts as a food preservative . |
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| 236. |
Write the therapeutic action of following on human body and mention the class of drugs to which each of these belong : (i) Ranitidine (ii) Morphine (iii) Aspirin |
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Answer» (i) Ranitidine : It hinders the interaction of histamine with the receptors present in stomach wall due to that less HCl is released and hyperacidity problem is solved. It belongs to Antacid drugs. (ii) Morphine : They are used for the relief of post operative pain, cardiac pain, pains of terminal cancer and in child birth. They belong to Narcotic analgesics (addictive). (iii) Aspirin : It inhibits the synthesis of prostaglandin which stimulate inflammation in tissues and cause pain. It is effective in relieving skeletal pain due to arthritis, prevents blood platelet coagulation. It belongs to non-narcotic analgesics (non-addictive). |
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| 237. |
Mention one use each of the following drugs : (i) Ranitidine (ii) Paracetamol (iii) Tincture of iodine |
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Answer» (i) Rantidine is used as antacid. (ii) Paractamol is used to bring down the body temperature during high fever. (iii) Tincture of iodine is used as an antiseptic, Its, is `2-3%` soltution of idodide in alchool and water. |
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| 238. |
Study the given aquatic food chain and answer the questions that follow: (i) Give reasons why there is a continuous increase in the DDT content in different trophic levels of the chain. (ii) Name the phenomenon responsible for the increase in DDT content. |
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Answer» (i) When DDT enters the food chain it reaches water. Since, it not degradable it gets accumulated progressively at each trophic level. AB it reaches higher trophic levels, the concentration of DDT also increases. (ii) Biological magnification. |
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| 239. |
What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V` |
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Answer» `2Cr(s) + 3Fe^(2+)(0.1M) rarr2Cr^(3+)(0.01M) + 3Fe(s)` `E_("cell")=E_("cell")^(@)-(0.0591)/(n) log. ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` Cell reaction : `underset(("s"))(Cr) " | "underset(("0.01M")) (Cr^(3+))" || " underset(("0.01M"))(Fe^(2+))" | "Fe(s)` `E_("cell")^(@)=E_("cathode")^(@)-E_("amode")^(@)` `=E^(@)(Fe^(2+)|Fe) - E^(@) (Cr^(3+)|Cr)` = `- 0.44 - (-0.74)` `E_("cell")^(@) = - 0.44 + 0.74 = 0.30V` `n = 6e^(-)` `E_("cell")^(@) = 0.30 - (0.0591)/(6)log.((0.01)^(2))/((0.1)^(3)) = 0.30 - (0.0591)/(6) log. 10^(-1)` `0.30 - (0.0591)/(6) xx (-1) = 0.30 + (0.0591)/(6)` =0.30 + 0.00985 = 0.309V |
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| 240. |
Express the relation between.conductivity and molar conductivity of a solution. |
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Answer» Molar conductivity `(Lambda_M)` is related to conductivity (K) as : `Lambda_m=(Kxx1000)/M` Where M be the molarity of the solution. |
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| 241. |
(a) Express the relationship amongst cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solute related to conductivity of its solution ? (b) Calculate the equilibrium constant for the reaction `Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd (s)` (Given : `E_(Cd^(2+)|Cd)^(@) = 0.40 V , E_(Fe^(2+) |Fe)^(@) = -0.44 V`). |
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Answer» (b) `E_("cell")^(@) = E_("cathod")^(@) - E_("anode")` `= -0.40 - (-0.44)` = `-0.40 + 0.44 = 0.04V` Since `" " E_("cell")^(@) = (0.059)/(n) "log" Kc` `0.04 = (0.059)/(2) "log" Kc` log Kc = `(2xx 0.04)/(0.059) = 1.356` Kc = anti-log `(1.356) = 22.70` |
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| 242. |
Describe the following : (i) Tyndall effect (ii) Shape-selective catalysis |
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Answer» (i) Tyndall effect : The phenomenon of scattering of light by colloidal particles as a result of which the path of beam becomes visible is called Tyndall effect. (ii) Shape-selective catalysis : The catalytic reaction which depend upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. |
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| 243. |
Given that the standard electrode` (E^(@))` of metals are : `K^(+)//K=- 2.93 V, Ag^(+)//Ag = 0.80 V, Cu^(2+)//Cu = 0. 34 V, ` ` Mg^(2+)//Mg =- 2.37 V, Cr^(3+)// Cr =- 0.74 V, Fe^(2+)//Fe =- 0.44 V`. Arrange these metals in an increasing order of their reducing power. Or Two half -reactions of an electrochemical cell are given below : `MnO_4^(-)`(aq) + 8 `H^(+)` (aq) + 5 `e^(-)` `rarr``Mn^(2+)` (aq) + 4 `H_(2) O` (l) , `E^(@)` = + 1.51 V, ` Sn^(2+) (aq) to Sn^(4+) (aq) + 2e^(-) , E^(@) = + 0.15^(V)` construct redox equation and predict if the reaction is reactant or product favoured. |
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Answer» Metals in an increasing order of their reducing power . `Ag^(+)//Ag, Cu^2//Cu, Fe^(2+)//Fe,Cr^(3+)//Cr, Mg^(2+)//Mg, K^(+) //K` ` i.e." " Ag lt Cu lt Fe lt Cr lt Mg lt K` Or Cathode : `[MnObar(4) (aq) + 8 H^(=) (aq) + 5 e^(-) to Mn^(2+) (aq) + 4 H_(2) O (l) ]xx 2` Anode : `[Sn^(2+) (aq) to Sn^(4+) (aq) + 2 e^(-) ] xx 5` Cell reaction : `2MnO^(-)._(4)(aq)+ 16H^(+)(aq)+5 Sn^(2+)(aq) to 2Mn^(2)+5 Sn^(4+)(aq)+8 H_(2)O` `E^(@)._("cell") = E_("Cathode-")E _ ("anode") = 1.51 - (-0.15) = 1.66 V` `:. ` EMF of the cell is positive. Hence the reaction favoured product. |
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| 244. |
A circular coil of closely wound N terns and radius r carriers a current I. Write the expressions for the following : (i) the magnetic field at its centre (ii) the magnetic moment of this coil |
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Answer» (i) At centre `B_(C)=(mu_(0)Ni)/(2r).` Here, current in the coil is anticlockwise and the direction of magnetic field is perpendicular to the plane of coil upward, but if the current in the coil is clockwise, then the direction of magnetic field will be perpendicular to the plane of coil downward. (ii) The magnetic field due to a circular coil of radius r, carrying current I at its centre is `B=(mu_(0) I)/(2r)` The direction of magnetic field is perpendicular to plane of coil, directed outward if current is anticlock wise and inward if current is clockwise. |
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| 245. |
A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why ? |
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Answer» An electron has the larger wavelength. Reason : de-Broglie wavelength in terms of kinetic energy is `lamda=(h)/(sqrt(2mE_(k)))alpha=(1)/(sqrt(m))` for the same kinetic energy. As an electron has a smaller mass than a proton, an electron an electron has larger de-Broglie wavelength than a proton for the same kinetic energy. |
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| 246. |
Mention two characteristic properties of the material suitable for making core of a transformer. |
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Answer» (i) An output voltage increases, the output current automatically decreases to keep the power, same. Thus, there is no violation of conservation of energy in step-up-transformer. (ii) The source of energy loss in a transformer are eddy current losses and flux leakage losses. |
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| 247. |
List four causes of biodiversity loss. |
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Answer» (i) Habitat loss and Fragmentation (ii) Over-exploitation (iii) Alien species invasion (iv) Co-extinction. |
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| 248. |
Name two metals used in a catalytic converter. How do they help in keeping the environment clean ? |
| Answer» Platinum-palladuim and Rhodium. The catalyst converts unburnt hydrocarbons into water and carbondioxide and carbon monoxide is charged to carbondioxide and nitric oxide to nitrogen gas. | |
| 249. |
Write the IUPAC Name of the compound. |
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Answer» The IUPAC Name of the given organic compound is 3-Hydroxybutan-I-oic acid. `CH_(3)-underset(OH)underset(|)CH-CH_(2)-COOH` |
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| 250. |
Which of the following complexes is more stable and why ? `[Co(en)_3]^(3+)` and `[Co(NH_3)_6]^(3+)` |
| Answer» Chelating ligands from more stable complexes complexes compared to non-chelating ligands. Since ethylene diammine is a bidentate ligand and froms stable chelate, `[Co(en)_(3)]^(3+)` | |