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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.
| 10401. |
The common method of extraction of metals from oxide ores isA. reduction of AlB. reduction with HC. reduction with CD. None of these |
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Answer» Correct Answer - C The common method of extraction of metals from oxide involves reduction with carbon. |
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| 10402. |
What is the role of graphite rod in the electrometallurgy of aluminium?A. It acts as anodeB. It acts as cathodeC. It acts as oxidising agentD. Both (a) and (c) |
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Answer» Correct Answer - A Graphite rods act as anode during the electrolytic reduction of alumina (Hall-Heroult process). At anode , `O_(2)` gas is produced which reacts with the carbon of anode (rods ) to produce `CO_(2)` gas , so these graphite rods are consumed slowly and need to be replaced from time to time. `Al_(2)O_(3) iff 2Al^(3+) +3O^(2-)` At cathode `Al^(3+) +3e^(-) to Al ` At anode `3O^(2-) to 3[O]+6e^(-) xx 2 ` `or 3[O]+3[O] to3O_(2)` `C(s)+O_(2)(g) to CO_(2)(g)` |
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| 10403. |
Which of the following metals are extracted by electrolytic reduction ?A. CuB. AlC. MgD. Ag. |
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Answer» Correct Answer - B::C Both Mg and Al are extracted by electrolytic reduction process. |
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| 10404. |
Which metals are generally extracted by the electrolytic processes ? What positions these metals generally occupy in the periodic table? |
| Answer» Electrolytic process is used for the extraction of active metals like `Na,Ca,Mg,Al` etc. The cases in which all other methods do not apply. Except Al and few other metals, these metals belong to s-block elements. | |
| 10405. |
Out of C and CO, which is better reducing agent for ZnO? |
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Answer» The two reduction reactions are : `ZnO(s)+C(s) to Zn(s)+CO(g)` `ZnO(s)+CO(g)to Zn(s)+CO_(2)(g)` In the first case, there is increase in the magnitude of `DeltaS^(@)` while in the second case, it almost remains the same. In other words, `DeltaG^(@)` will have more negative value in the first case when C(s) is a better reducing agent. |
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| 10406. |
Why is zinc not extracted from zinc oxide through reduction using CO? |
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Answer» The chemical reaction involving the reduction of ZnO by CO is : `ZnO(s)+CO(g) to Zn(s)+CO_(2)(g)` The process is thermodynamically not feasible because there is hardly any change in entropy as a result of the reaction. This is quite evident from the physical states of the reactants and products involved in the reaction |
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| 10407. |
The value of`Delta _(f) G^(Ө)` for formation of `Cr_(2) O_(3)` is `– 540 kJmol^(−1) ` and that of `Al_(2) O_(3)` is `– 827 kJmol^(−1)`. Is the reduction of `Cr_(2) O_(3)` possible with `Al` ? |
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Answer» The two thermochemical equations may be written as follows : `{:(4//3 Cr(s)+O_(2)(g)to 2//3Cr_(2)O_(3)(s), Delta_(f)G^(@)=-540 kJ),(4//3Al(s)+O_(2)(g)to 2//3Al_(2)O_(3)(s), Delta_(f)G^(@)=-827 kJ):}/(2//3Cr_(2)O_(3)(s)+4//3Al(s)to 2//3Al_(2)O_(3)(s)+4//3Cr(s), Delta_(f)G^(@)=-287 kJ)` |
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| 10408. |
Assertion : `Delta_(f)G^(@)` for the formation of `Al_(2)O_(3) " and " Cr_(2)O_(3)` involving one mole of oxygen are `-827 " kJ mol"^(-1) " and " -540 " kJ mol"^(-1)` respectively Reason : Al can reduce `Cr_(2)O_(3) " since " Delta_(f)G^(2)` for its formation is less.A. Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.B. Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.C. Assertion is correct, but reason is wrong statement.D. Assertion is wrong, but reason is correct statement. |
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Answer» Correct Answer - A Reason is the correct explanation for Assertion. |
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| 10409. |
Is it true that under certain conditions, `Mg` can reduce `SiO_(2)`, and `Si` can reduce `MgO` ? |
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Answer» (i) `SiO_(2) + 2Mg rarr Si + 2MgO` Well-powdered quartz is mixted with magnesium and heated in a fire clay crucible along with some quantify of calcined magnesia. (ii) `2MgO overset(2000^@C)rarr SiO_(2) + 2Mg`. |
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| 10410. |
The extraction of `Au` by leaching with `NaCN` both oxidation and reduction. Justify giving equations. |
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Answer» During leaching process, `Au` is first oxidised to `Au^(olpus)` by `O_(2)` of the air, which then combines with `CN^(Ө)` ions to form the soluble complex, sodium dicyaniodaurate `(I)` `underset("Gold(impure)") (4Au_((s))) + 8NaCN_((aq)) + 2H_(2) O_((1)) + O_(2(g)) rarr underset(("Soluble complex"))underset("Sodium dicyaniodaurate(I)")(4Na[Au(CN)_(2)]_((aq)))+ 4NaOH_((aq))` Gold is then extracted from this complex by displacement method using a more electropositive zinc metal. In this reaction, `Zn` acts as a reducing agent. It reduces `Au^(oplus)` to `Au` while it itself gets oxidised to `Zn^(2+)`, which combines with `CN^(Ө)` ions to form a soluble complex, sodium tetracyanidozincate `(II)` `2Na[Au(CN)_(2)]_((aq)) + Zn_((s)) rarr underset("Gold (pure)") (2 Au_((s))) + underset("Sodium tetracyanidozincate (II)") (Na_(2)[Zn(CN)_(4)]_((aq)))` Thus, extraction of `Au` by leaching with `NaCN` involves both oxidation and reduction. |
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| 10411. |
Indicate the temperature at which carbon can be used as a reducing agent for `FeO`. |
| Answer» Correct Answer - Above `1123 K`. | |
| 10412. |
(a) Indicate the temperature at which carbon can be used as reducing agent for `FeO`. (b) Define flux. ( c) Metal usually do not occur in nature as nitrates. Why ? (d) Metal such as `Cu, Ag, Zn` etc. occur in nature as sulphide rather than oxides. Why ? |
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Answer» (a) At `1073 K` or above, the standard free energy of formation of `CO` from `C` is much below the standard free energy of formation of `FeO`. Hence, above `1073 K`, carbon can reduce `FeO` to `Fe`. (b) Flux is a substance that combines with gauge, which mat still be present in the roasted or calcinaed ore to form an easily fusible material called slag. ( c) Nitrates of all metals are soluble in water. Hence, if metal nitrates are present in the crust of earth, these would be slowly and gradually washed by rain water into the sea. Hence, metals usually do not occur in nature as nitrates. (d) The cations of `Cu, Ag` and `Zn`, i.e. `Cu^(oplus), Ag^(oplus)` and `Zn^(2+)` [pseudo inert gas configuration, `(ns^(2) p^(6) d^(10))`] have high polarising power and hence easily polarises the bigger sulphide `(S^(2-))` ion than the small oxide `(O^(2-))` ion. Hence, sulphides of these metals are more stable than the oxides and these metals occur in nature as sulphides rather than oxides. |
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| 10413. |
Interpret the following Ellingham diagram. . |
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Answer» (1) `Si_((s)) + O_(2(g)) rarr 2 Si O_(2 (s))` As temperature increases, `Delta_(f) G^(Ө)` becomes less negative , hence `SiO_(2(s))` becomes less stable, i.e., at `500 K, SiO_(2)` is more stable than `2500 K`. (2) `2Mg_((s)) + O_(2(g)) rarr 2MgO_((s))` Abrupt changes in the plot at `~ 800 K` and `~ 1400 K` denote phase change//transition from `s rarr 1` and `1 rarr g`, respectively and represent melting point `(~800 K)` and boiling point `(~ 1400 K)`. `MgO_((s))` becomes less stable with increase in temperature. At `T_(1)`, lines of `(1)` and `(2)` intersect. One can conclude the following : (a) At `T_(1)`, both `MgO` and `SiO_(2)` are equally stable. (b) At temperature `gt T_(1), MgO` is less stable than `SiO_(2)` Hence, `Si` will reduce `MgO` to `Mg`. `2MgO + Si rarr SiO_(2) + 2Mg` ( c) At temperature `lt T_(1), SiO_(2)` is less stable than `MgO`. Hence, `Mg` will reduce `SiO_(2)` to `Si` `SiO_(2) + 2Mg rarr 2MgO + Si` Relative stability of `MgO` and `SiO_(2)` at given temperature can be predicted with the help of Ellingham diagram. (3) `2Zn_((s)) + O_(2(g)) rarr 2ZnO_((s))` `Delta_(f) G^(Ө) (ZnO) gt Delta_(f) G^(Ө) (MgO)` `Delta_(f) G^(Ө) (ZnO) gt Delta_(f)^(Ө) (SiO_(2))` Hence, both `Mg_((s))` and `Si_((s))` can be used as reducing agent for reducing `ZnO_((s))` to `Zn_((s))`. |
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| 10414. |
Why is it advantageous to roast a sulphide ore to the oxide before reduction ? |
| Answer» The free energies of formation `(Delta_(f) G^(Ө))` of most sulphides are greater than those for `CS_(2)` and `H_(2) S`. Carbon disulphide is, in fact, an endothermic compound. So neither carbon nor hydrogen is a suitable reudcing agent for metal sulphides. Moreover, the roasting of a sulphide to the oxide is quite advantageous thermodynamically. Hence, the common practice is to roast sulphide ore to the oxide prior to reduction. | |
| 10415. |
(a) Suggest a condition under which magnesium could reduce alumina. (b) Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ? ( c) What is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction ? (d) At a site, low grade copper ores oare available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why ? |
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Answer» (a) The two equations are : (i) `(4)/(3) Al + O_(2) rarr (2)/(3) Al_(2) O_(3)` (ii) `2Mg + O_(2) rarr 2MgO` All the point of intersection of the `Al_(2) O_(3)` and `MgO` curves, `Delta_(r) G^(Ө)` becomes zer for the reaction : `(2)/(3) Al_(2) O_(3) + 2Mg rarr 2 MgO + (4)/(3) Al` Below that point magnesium can reduce alumina. (b) Temperatures below the point of intersection of `Al_(2) O_(3)` and `MgO` curves, magnesium can reduce alumina. But the process will be uneconomical as magnesium is much costlier than alumina. ( c) The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change `(Delta_(r) G^(Ө))` of the reduction process is more on positive side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus, the value of `Delta_(r) G^(Ө)` becomes more on negative side and the reduction becomes easier. (d) Zinc being above iron in the electrochemical series (more reactive metal is zinc), the reduction will be faster in case zinc scraps are used, But zinc is costlier metal than iron, so using iron scraps will be advisable and advantageous. |
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| 10416. |
Product of reaction is:A. B. C. D. |
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Answer» Correct Answer - C `OH^(-)` converts phenol to phenoxide ion which is strongly activating so attack occurs at ortho position w.r.t. hydroxyl group |
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| 10417. |
The correct order of matching of complex compound in column I with the properties in column II: `{:(,"Column I",,"Column II",),((A),[Cr(NH_(3))_(6)]^(3+),(P),"Tetrahedral and paaramagnetic",),((B),[Co(CN)_(6)]^(3-),(Q),"Octahedral and diamagnetic",),((C ),[Ni(CN)_(4)]^(2-),(R ),"Octahedral and paramagnetic",),((D),[Ni(Cl)_(4)]^(2-),(S),"Square planar and diamagnetic",):}`A. `A-R, B-Q, C-P, d-S`B. `A-Q, B-R, C-P, D-S`C. `A-R, B-Q, C-S, D-P`D. `A-Q, B-R, C-S, D-P` |
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Answer» Correct Answer - C `[Cr(NH_(3))_(6)]^(3+) - (d^(3)) - d^(2)sp^(3)` hybridised & paramagnetic `[C0(CN)_(6)]^(3-) - (d^(6)) - d^(2)sp^(3)` hybridised (`CN^(+)` is a strong field ligand) |
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| 10418. |
Which of the following is non-reducing sugar a)isomaltose b) maltose c) lactose d) trehalose |
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Answer» Trehalose is a non-reducing sugar.It is made up to 2 molecules of glucose. |
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| 10419. |
The function not performed by villi is :(a) To increase the surface area for absorption(b) To ensure rich supply of blood vessels(c) Absorption of food(d) Egestion of food |
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Answer» Correct answer is (d) Egestion of food Villi is a part of small intestive responsible for absorption of food by in hearing the surface are. It is richly supplied by blood vessels so that energy is transported to another part of body. |
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| 10420. |
Four test tubes A, B, C and D are taken. In test tube A iron nail is dipped in copper sulphate solution. In test tube B copper wire is dipped in ferrous sulphate solution. In test tube C zinc metal is dipped in ferrous sulphate solution and in test tube D iron nail is dipped in zinc sulphate solution. The reactivity order has been found to be zn > Fe > Cu. In which test tubes was the colour change observed ?(a) A and C(b) A and B(c) B and C(d) B and D |
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Answer» Correct answer is (a) A and C A more reactive metal can replace a less reactive metal, but a less reactive one cannot replace a more reactive metal. ∵ The reactivity order has been found to be zn > fe > cu Therefore, In test tube A \(\longrightarrow\) Iron can replace Cu metal from CuSO4 solution. \(\underset{\text{blue color}}{CuSO_4(aq)} + Fe(s)\) \(\longrightarrow\) \(Cu(s) + \underset{\text{pale green}}{FeSO_4(aq)}\) In test tube B - Copper can not replace Fe metal from FeSO4 solution. There will no reaction between copper metal and FeSO4 solution. Therefore, no change in colour. In test tube C - Zinc can replace, Fe metal from FeSO4 solution. \(\underset{\text{Pale green}}{FeSO_4(aq)} + Zn(s)\) \(\longrightarrow\) \(\underset{\text{colorless}}{ZnSO_4(aq)} + Fe(s)\) In test tube D - Iron cannot replace, Zn metal from ZnSO4 solution. There will no reaction between Iron nail and ZnSO4 solution. Therefore, no change in colour. |
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| 10421. |
Assertion (A) : The rate of breathing in aquatic organisms is much slower than that seen in terrestrial organisms.Reason (R) : The amount of oxygen dissolved in water is very low as compared to the amount of oxygen in air.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. |
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Answer» Correct answer is (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). |
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| 10422. |
Match the metal (Column I) with its reaction with oxygen (Column II)Column IColumn IIA. Potassium(i) Does not react event at high temperaturesB. Zinc(ii) Gets coated with black-coloured layer of oxideC. Copper(iii) Does not burn at ordinary temperatureD. Silver(iv) Burns vigorously(a) A-(iv), B-(iii), C-(ii), D-(i)(b) A-(iv), B-(ii), C-(i), D-(iii)(c) A-(iii), B-(ii), C-(i), D-(iv)(d) A-(iv), B-(ii), C-(iii), D-(i) |
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Answer» Correct answer is (a) A-(iv), B-(iii), C-(ii), D-(i) (A → iv) Potassium is highly reactive metal. so it reacts vigorously with oxygen and burn. (B → iii) Zinc does not burn at ordinary temperature because zinc is less reactive so we need to apply heat to make it able to react. (C → ii) Copper metal get coated with black colored layer of oxide - \(2Cu + O_2\) \(\longrightarrow\) \(\overset{2CuO}{\text{black in color}}\) (D → i) Silver metal does not react even at high temperature because silver metal is very less reactive like (Au, pt) so that it does not react even at high temperature. |
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| 10423. |
Assertion (A) : Kerosene having higher refractive index is optically denser than water, although its mass density is less than that of water.Reason (R) : The speed of light decides whether a medium is optically rarer or optically denser. An optically denser medium may not possess greater mass density.(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true. |
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Answer» Correct answer is (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). |
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| 10424. |
In Living organisms During respiration which of the following Products are not formed if oxygen is not available? (a) Carbon dioxide + Water (b) Carbon dioxide + Alcohol (c) Lactic acid + Alcohol (d) Carbon dioxide + Lactic Acid |
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Answer» Correct option is (b) Carbon dioxide + Alcohol |
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| 10425. |
The Correct Statements with reference to single called organisms are (i) Complex substances are not broken down in to simpler substances. (ii) Simple diffusion is sufficient to meet the requirement of exchange of gases. (iii) Specialized tissues perform different functions in the organism. (iv) Entire surface of the organism is in contact with the environment for taking in food. (a) (i) and (iii) (b) (ii) and (iii) (c) (ii) and (iv) (d) (i) and (iv) |
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Answer» Correct option is (c) (ii) and (iv) |
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| 10426. |
Write the name of an organism which increase the fertility of soil. |
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Answer» The microorganisms such fungi and bacteria Earthworms increases it Some bacteria like rhizobium and blue green algae are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called biological nitrogen fixers. |
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| 10427. |
The pair(s) which will show displacement reaction is/are (i) NaCl solution and copper metal (ii) AgNO3 solution and copper metal (iii) Al2(SO4)3 solution and magnesium metal (iv) ZnSO4 solution and iron metal(a) (ii) only (b) (ii) and (iii) (c) (iii) and (iv) (d) (i) and (ii) |
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Answer» Correct option is (c) (iii) and (iv) |
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| 10428. |
Which one among the following is not removed as a waste product from the body of a plant? (a) Resins and Gums (b) Urea (c) Dry Leaves (d) Excess Water |
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Answer» Correct option is (b) Urea |
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| 10429. |
In which of the following is a concave mirror used? (a) A solar cooker (b) A rear view mirror in vehicles (c) A safety mirror in shopping malls (d) In Viewing full size image of distant tall buildings. |
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Answer» Correct option is (a) A solar cooker |
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| 10430. |
which of the following statements is not true for scattering of light? (a) Colour of the scattered tight depends on the size of particles of the atmosphere. (b) Red light is least scattered in the atmosphere. (c) Scattering of light Lakes place as various colours of white light travel with different speed in fir. (d) The fine particles in the atmospheric air scatter the blue, light more strongly than red. So the scattered blue light enters our eyes. |
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Answer» (c) Scattering of light Lakes place as various colours of white light travel with different speed in fir. |
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| 10431. |
In kingdom plantae, alternation of generation is not associated with(1) Sporophytic and gametophytic phase (2) Length of haploid & diploid phases (3) Number of haploid and diploid phases (4) Freeliving or dependent nature of haploid & diploid phases |
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Answer» Correct option is (3) Number of haploid and diploid phases |
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| 10432. |
A ray of light starting from air passes through medium A of refractive index 1.50, enters medium B of refractive index 1.33 and finally enters medium C of refractive index 2.42. If this ray emerges out in air from C, then for which of the following pairs of media the bending of light least? (a) air-A (b) A-B (c) B-C (d) C-air |
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Answer» Correct option is (b) A-B |
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| 10433. |
The recommendation of Narashimham Committee Report was admitted in the year ………(a) 1990 (b) 1991 (c) 1995 (d) 2000 |
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Answer» The recommendation of Narashimham Committee Report was admitted in the year 1991. |
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| 10434. |
The selection of the right controller for the application depends on 1. The degree of control required by the application. 2. The individual characteristics of the plants. 3. The desirable performance level including required response, steady-state deviation and stability. Which of the above statements are correct ? (a) 1 and 2 only(b) 1 and 3 only (c) 2 and 3 only (d) 1, 2 and 3 |
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Answer» (d) 1, 2 and 3 |
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| 10435. |
Greenhouse effect was first described by.......(a) Yues Chauvin(b) Einstein(c) Fourier(d) Newton |
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Answer» Answer: (c) Fourier |
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| 10436. |
The rate law for a reaction between the substances A and B is given by rate `= K[A]^(n) [B]^(m)`. On doubling the concentration of A and having the concentration of B, the ratio of the new rate to the earlier rate of the reactio will be:A. `1//2^(m + n)`B. `2^(m - n)`C. `1//2^(n - m)`D. `2^(n - m)` |
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Answer» Correct Answer - D `r_(1) = k[A]^(n) [B]^(m)` `r_(2) = k [2A]^(n) [(B)/(2)] = 2^(n-m) k [A]^(n) [B]^(m)` `(r_(1))/(r_(2)) = 2^(n - m)` |
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| 10437. |
The following electronic transition occurs when lithium atoms are sprayed into a hot flame, `2s overset(I)rarr 2p overset(II)rarr 3d overset(III)rarr 3p overset(IV)rarr 4s overset(V)rarr 3p`. Which of the transition would result in the emission of light?A. I, II and IVB. III and VC. III, IV and VD. All of these steps |
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Answer» Correct Answer - B Subshell with lower `(n + l)` value is lower in eenergy and energy is emitted in transition from a subshell of higher energy to lower energy subshell. |
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| 10438. |
Which of the following reaction will not proceed via carbanion intermediate ?A. Perkin reactionB. cannizaro reactionC. Claisen condensaationD. Aldol condensation |
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Answer» Correct Answer - B Cannizaro reaction involves attack of `OH^(-)` on carbonyl carbon and `H^(-)` shift on another carbonyl carbon, no carbanion intermediate. |
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| 10439. |
State “Law of definite proportions”. |
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Answer» The given compound always contains exactly the same proportion of elements by weight. |
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| 10440. |
Correct statement about classical smog (I) and photochemical smog (II) is:A. `I=` reducing smog, `II=` oxidizing smogB. `I=` oxidizing smog, `II=` reducing smogC. Both I and II are oxidizing smogD. Bothe I and II are reducing smog |
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Answer» Correct Answer - A Classical smog is a reducing mixture, while photochemical smog has high concentration of oxidizing agents. |
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| 10441. |
Describe the following in brief: (i) Ozone depletion over Antarctica (ii) BOD and COD (iii) Eutrophication |
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Answer» (i) Ozone depletion over Antarctica: In summer season, nitrogen dioxide and methane react with chlorine monoxide preventing ozone depletion whereas in winter, special type of clouds called polar stratospheric could s are formed over Antarctica. These clouds provide surface on which chlorine nitrate molecules formed gets hydrolysed to form HOCl. It also reacts with HCls to give Cl2. When sunlight returns to Antarctica again ozone depletion starts by free radicals. (ii) BOD: It is amount of oxygen required by bacteria to decompose organize wastes present in water. COD: It is the amount of oxygen (in ppm) required to oxidize the contaminants. COD is determined by using chemical oxidizing agent K2Cr2O7. (iii) Eutrophication: The process in which nutrient-enriched water bodies support a dense plant population which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication. |
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| 10442. |
Why is dipole movement of ammonia more than NF3 |
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Answer» The dipole moment of ammonia (1.47D) is higher than the dipole moment of NF3 (0.24D). The molecular geometry is pyramidal for both the molecules. In each molecule, N atom has one lone pair. F is more electronegative than H and N−F bond is more polar than N−H bond. Hence, NF3 is expected to have much larger dipole moment than NH3. However reverse is true as in case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of NF3 it is opposite. Thus, in ammonia molecule, individual dipole moment vectors add whereas in NF3, they cancel each other. |
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| 10443. |
Find out the standard Gibbs free energy and the equilibrium constant of the cell reaction given below which has standard electrode potential of 0.236 V at 298 K.(Take F=96487 C, Antilog 0.981=9.37)2 Fe3+(aq) + 2I-(aq) → 2 Fe+2(aq) + I2(s) |
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Answer» We have given, E0 = 0.236 v at 298 k f = 96487 c 2 Fe3+(aq) + 2I-(aq) → 2 Fe+2(aq) + I2(s) We have, \(\triangle G^{\circ}=-nFE^{\circ}\) Here, n = 2 = No. of transfered electrons. So, \(\triangle G^{\circ}=-2 \times 96487 c\times0.236 v\) \(\triangle G^{\circ}=-45541.864 J\) At equilibrium \(\triangle G^{\circ}=-RTlnK_{eq}\) -45541.864 = -8.314 x 298 ln Keq or 45541.864 = 8.314 x 298 x 2.303 x log Keq ⇒ Keq = 9.6 x 107 Hence, for given reaction at 298k, standard cribbs free energy (\(\Delta G^{\circ}\)) = -4454.864 J and equilibrium constant = 9.6 x 107 |
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| 10444. |
मात्रा विशुदू एल्कोहाल की मिला दो जाती फेआन: (freons) किसे कहते है |
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Answer» मेथेन व एथेन से बने पोली क्लोरो और फ्लोरो व्युत्पन्न फ्रियान कहलाते है। इन्हे CFC (क्लोरो फ्लोरो कार्बन ) भी कहते है। |
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| 10445. |
explanation of formula mass |
| Answer» The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. | |
| 10446. |
Which of the following statements is incorrect ? Standard Gibbs free energy change is always zero at equilibriumA. Addtion of solid does not affect equlibriumB. Addition of solid does not affect equlibriumC. On addition of catalyst the value of equilibrium constants is not affectedD. Equilibrium constant for a reaction with negative `DeltaH` value decreases as the temperature increases |
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Answer» Correct Answer - 1 `DeltaG^(@)` is not always zero at equilibrium |
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| 10447. |
A particle is moving along a straight line. Its displacement-time graph is shown `{:(,,"Column-I",,"Column-II"),(,(A),"Work done on particle from 0 to "t_(1),(P),"Positive"),(,(B),"Work done on particle from " t_(1) " to " t_(2),(Q),"Negative"),(,(C),"Work done on particle from " t_(2) " to " t_(3),(R),"zero"),(,(D),"Work done on particle from " t_(3) " to " t_(4),(S),"Unpredectible"):}`A. `AtoR,BtoR,CtoP,DtoQ`B. `AtoS,BtoR,CtoP,DtoQ`C. `AtoR,BtoS,CtoP,DtoQ`D. `AtoR,BtoR,CtoQ,DtoP` |
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Answer» Correct Answer - 1 For 0`gttgtt_(0)to`slope=constnt W=0 For `t_(1)lttltt_(2)to`slope=0 W=0 For `t_(2)lttltt_(3)to` slope is increasing W=+ve For `t_(3)lttltt_(4)to` slope is decreasing W=-ve |
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| 10448. |
Three blocks of A, B and C of equal mass m are placed one over the other on a smooth horizontal ground as shown in figure. Cofficient of friction between any two blocks of A,B and C is 1/2. The maximum value of mass of block D so that the blocks A,B and C move without slipping over each other is- A. 6 mB. 5 mC. 3 mD. 4 m |
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Answer» Correct Answer - 3 Block B & C move together but For A `malemu(mg)` So `a_(max)=mu_(s)g` So `a_(max)=(Mg)/(M+3m)` M=3m |
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| 10449. |
A block `A` of mass `4 kg` is kept on ground. The coefficient of friction between the block and the ground is `0.8`. The external force of magnitude `30 N` is applied parallel to the ground. The resultant force exerted by the ground on the block is `(g = 10 m//s^(2))`A. `40N`B. `30N`C. `0N`D. `50N` |
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Answer» Correct Answer - D `N=mg=40` `(f_(s))_(max)=muN=(0.8)(40)=32` `f_(s)=ext. ` force `=30` `R^(2)=N^(2)+f_(s)^(2)=(50)^(2):.R=50N`. |
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| 10450. |
Figure shows variation of acceleration due to gravity with distance from centre of a uniform spherical planet, Radius of planet is R. What is `r_(1) - r_(1)` A. `(R)/(4)`B. `(7R)/(4)`C. `(4R)/(3)`D. `2 R` |
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Answer» Correct Answer - B Inside planet `(9_(0))/(4)g_(0)(1-(d)/(R))rArr d=(3R)/(4)` Outside planet `g = g_(0)(R^(2))/((r+h)^(2))rArr h=R` . |
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