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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.
| 89151. |
A hot silver foil is tarnished when exposed to ozone. Why ? |
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Answer» Solution :Silver METAL is oxidised by ozone to silver oxide. This oxide is REDUCES BACK to finely divided silver metal, which is black. `2AG + O_3 rarr Ag_2O + O_2` (oxidation) `Ag_2O + O_3 rarr 2 Ag + 2O_2` (reduction) These are called MUTUAL reduction reactions. |
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| 89152. |
A homopolymer is obtained by polymerisatation of: |
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Answer» One TYPE of MONOMER units |
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| 89153. |
A horizonal piston-cylinder arrangement is placed in a constant temperature bath.The piston slides in the cylinder with negligible friction, and an external force holds it in place against an initial gas pressure of 14 bar.The initial gas volume is 0.03 m^3 (a)The external force on the piston is reduced gradually, allowing the gas to expand until its volume doubles.Calculate the work done by the gas in moving the external force. (b)How much work would be done if the same expansion is carried out by removing a part of the external force suddenly.Also calculate % efficiency of this process as compared with the reversible process (%of reversible work). Take ln2=0.7 Report your answer as (% efficiency xx 0.7). |
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Answer» `W=-P_1V_1` In `V_2/V_1=-14xx0.03"In"0.06/0.03"bar"m^3= -14xx0.7xx0.03= -0.294 "bar" m^3` (B)`P_1V_1=P_2V_2` `:. P_2=(P_1V_1)/V_2=(14xx0.03)/(0.06)=7` bar `:. W=-P_(EXT)(V_2-V_1)=-7(0.06-0.03)=-7xx0.03= -0.21 "bar"m^3` Efficiency `=0.21/0.294=71.43%` |
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| 89154. |
A lab assistant prepared a solution by adding a calculated quantity of HCI gas 25^@ C to get a solution with [H_3O^+]=4xx10^(-5)M. Is the solution neutral (or) acidic (or) basic. |
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Answer» SOLUTION :`[H_3O^+]=4XX10^(-5)M` `pH=-logp[H_3 O^+]` `=-LOG(4xx10^(-5))` `pH=4.398` The solution is BASIC. |
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| 89155. |
(A): HNO_(3) is a stronger acid than HNO_(2). (R): There are two nitrogen-oxygen bonds in HNO_(3), whereas in HNO_(2) there is only one such bond. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 89156. |
(A) HNO_(2) disproportionate to give HNO_(3) and NO (R ) In basic media HNO_(2) is less stable |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 89158. |
(a) HIO_(4) decomposes 1,2-glycols but not 1,3- or higher glycols. (R ) Only 1,2-glycols form cyclic esters which subsequently undergo cleavage to form carbonyl compounds. |
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Answer» If both (A) and (R ) are CORRECT and (R ) is correct EXPLANATION of (A ). |
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| 89159. |
A knocking sound is produced more in the engine when the fuel contains mainly: |
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Answer» n-alkanes |
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| 89160. |
(A) Highest atomicity of an interhalogen compound is 8 (R ) Highest valency exhibited by a halogen , other than fluorine is 7 |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 89161. |
A: Higher the charge density on the central ion, greater will be stability of the complex R: Hard acid show a greater tendency for forming complexes with hard ligands such as F |
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Answer» If both Assertion & REASON are true and the reason is the correct EXPLANATION of the assertion, then mark (1) |
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| 89162. |
A KMnO_(4) solution can be standarised by titration against As_(2)O_(3(s)). A 0.1156 g sample of As_(2)O_(3) requires 27.06mL of the KMnO_(4(aq.)) for its titration. What is the molarity of the KMnO_(4(aq.)) [As =75]? 5As_(2)O_(3)+4MnO_(4)^(-)+9H_(2)O+12H^(+)rarr10H_(3)AsO_(4)+4Mn^(2+) |
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Answer» 0.0172 M |
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| 89163. |
(A): High purity gallium for semiconducting purpose can be obtained by zone refining (R): Zone refining works on the principle of fractional crystallisation, leading to high purity |
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Answer» Both A & R are TRUE, R is the correct EXPLANATION of A |
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| 89164. |
A high molecular weight molecular which does not contain repeating structural units is called a |
| Answer» Answer :B | |
| 89165. |
A kettle which becomes furred-up in use has inside it a deposit composed mainly of : |
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Answer» CALCIUM carbonate |
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| 89166. |
A hexapeptide has composition "Ala-Gly-Phe-Val"_(3). Both N - teriminal & C - terminal units are val. The number of val - val peptide bonds present, half of total 1^(@) - structures hexapeptide passible which satidfy these conditions are ________. |
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| 89167. |
A heterogeneous catalyst system follows |
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Answer» order kinetics in beginning and after some time it BECOMES a zero order REACTION |
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| 89168. |
A hexa-coordinated complex of formula CoCl_(3).6H_(2)O undergoes 100% ioisation is aqueous solutin to give van't Hoff factor equal to 3. Which of the following is/are correct regarding the given complex? |
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Answer» Effective atomic number of the complex is 36. `EAN=27-3+12=36` Anhydrous `CaCl_(2)` is dehydrating agent which absorbs the water of crystallization outside coordination sphere. |
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| 89169. |
A ketones reacted with C_2H_5MgBr reagent followed by hydrolysis gave a product which on dehydration gives an alkene.The alkane on ozonolysis gave diethyl ketone and acetaldehyde.The ketone is: |
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Answer» DIMETHYL ketone |
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| 89170. |
(A) Helium is used in modem diving apparatus (R) Helium is chemically inert |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 89171. |
(A) : Helium was discovered in chromosphere of the sum ( R ): D_3 line is observed along with D_1 and D_21 in the solar spectrum The correct answer is |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 89172. |
(A) Helium is used in gas cooled nuclear reactors(R) Helium has high thermal conductivity |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 89173. |
(A) : Ketones have higher boiling point than aldehydes. (R) : Ketones are more polar than aldehydes. |
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Answer» Both A & R are TRUE, R is the CORRECT explanation of A |
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| 89174. |
(A) Helium is regarded as the least volatile of all liquids.(R) The boiling point of liquid helium is 4.2 K which is the highest of all known liquids. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 89175. |
(A) Helium is present in radio active minerals like pitchblende, monazite and clevite(R) The main commercial source of helium is natural gas |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 89176. |
A ketone on reduction gives : |
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Answer» PRIMARY alcohol |
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| 89177. |
(A) Helium is preferred than hydrogen in filling ballons for meterological observations(R) Helium is lighter than hydrogen and it is non-combustible in nature |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 89178. |
(A): The name helium is made after sun (R): Helium is abundant gas in sun's chromosphere |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 89179. |
A ketone C_(8)H_(16)O (A) reacts with ethylmagnesium bromide to give a product which on hydrolysis yields an alcohol (B). Compound (B) undergoes dehydration with concentrated sulphuric acid to form (C) which on ozonolysis gives 3-pentanone as the only product. (A) is |
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Answer» 4-ethyl-3-hexanone (B) = 3, 4-Diethyl-3-hexanol (C) = 3, 4-Diethyl-3-hexene |
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| 89180. |
(A): Helium is found in radioactive minerals (R): During alpha-day, helium is formed and is occluded in radioactive minerals |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 89181. |
A helium atom on loosing an electron becomes: |
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Answer» ALPHA -particle |
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| 89182. |
A ketone A which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono-ozonide D. The compound D on hydrolysis in presence of zinc dust gives only acetaldehyde. Write the structures and IUPAC names of A, B and C. Write down the reactions involved. |
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Answer» Solution :COMPOUND D on hydrolysis in the presence of zinc DUST gives acetaldehyde. Therefore, the compound C must be `underset(C" (But-2-ene)")(CH_(3)-CH=CH-CH_(3))` The compound C is obtained from B, therefore, B should be `underset("B(Butan-2-ol)")(CH_(3)-underset(OH)underset(|)(CH)-CH_(2)-CH_(3))` B is obtained by the reductionof A, therefore A should be `underset("A(Butan-2-one)")(CH_(3)-underset(O)underset(||)C-CH_(2)-CH_(3))` Compounds containing `CH_(3)CO` - group give haloform reaction. The REACTIONS can be explained as under : 
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| 89183. |
A helium atom on losing an electron becomes |
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Answer» `alpha -` particle |
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| 89184. |
A heat pump is used to draw water from a well. The temperature of the water is 15^(@)C and that of the atmosphere is 40^(@)C . If the amount of water drawn is 10 kg from the depth of 12 m, Calculate the amount of heat supplied to the well. |
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| 89185. |
A ketone A(C_(4)H_(8)O) which undergoes a haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives a compound C whch forms mono-ozonide D. D on hydrolysis in presence of zinc dust gives only acetaldehyde E. Identify A,B,C,D and E. Write the reaction involved. |
Answer» SOLUTION :
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| 89186. |
(a) Heat of adsorption is greater for chemisorption than physisorption. Why? (b) What is colloidion? (c) Differentiate between peptization and coagulation. |
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Answer» Solution :(a) Due to the formation of chemical bond between adsorbate and ADSORBENT. (b) 4% solution of nitro CELLULOSE in a MIXTURE of alcohol and ether. (c) Peptisation is the process of converting a precipitate into colloidal sol by adding an electrolyte. But coagulation is the SETTLING of colloidal particles. |
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| 89187. |
A heat engine is a system that converts heat into mechanical work. A heat "source" generates thermal energy that brings a workingsubstanceto a high temperature. The working substance then generates work in the engine while transferring heat to a sink at a lower temperature . The working of a heat engine is shownn figure 1. Heat engines can be modelled using themodynamic cycles. The heat engine given in Figure -2 is of a working substance which is 1.00 mol of a monoatomic ideal gas . The thermodynamic cycle begins at the point designated as '1' and goes clockwise and the values of P and l or Vat each point is as given belowFig-2 (P_(1)=1.00 atm "and" V_(1) = 24.6 L , P_(2)=2.00 atm , V_(3) =49.2L, P_(4)=1.00 atm) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible , what is the temperature T_(3)? |
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| 89188. |
a. Ketone (A) which undergoes haloform reaction gives compound (B) on reduction. (B) on heating with sulphuric acid gives compounds (C ) which forms monozonide (D). (D) on hydrolysis in the presence of Zn dust gives only acetaldehyde. Identify (A), (B), (C ). b. C_(6)H_(14)O (A) on heating with KOH and I_(2) gives yellow precipitate (A) on dehydration using Al_(2)O_(3) gives (B) which on catalytic hydration gives (C ) which gives Lucas test readily. Identify (A), (B), and (C ). c. Compound (A) C_(5)H_(10)O forms a phenylhydrazone and gives a negative Tollens and negative iodoform tests. (A) on reduction gives n-pentane. Identify A. d. One gram mixture of CH_(3)OH and CH_(3)CHO reacts with Bendict's reagent to give a red precipitate. Themass of the red precipitate obtained is 1/43 gm. Calculate the % of CH_(3)CHO in the mixture. |
Answer» Solution :a. `{:(Aimplies,"2-Butan-2-ONE",,Bimplies,"Butan-2-ol"),(Cimplies,"But-2-ene",,Dimplies,"But-2-enozonide"):}`  c. `Aimplies 3`-Pentan-3-one b. Benedict's regent reacts only with `CH_(2)CHO`. One mole of `CH_(3)CHO` gives 1 MOL `Cu_(2)O. (Mw of CH_(3)CHO=44 gm), (Mw of Cu_(2)O=143 gm) 143 gm` of `Cu_(2)O` is obtained from `44 gm` of `CH_(3)CHO` `1.43 gm` of `Cu_(2)O` is obtained from `IMPLIES (44xx1.43)/143` `implies 0.44 G. of CH_(3)CHO` `%` of `CH_(3)CHO=0.44/1xx100 =44%` |
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| 89189. |
A heat engine is a system that converts heat into mechanical work. A heat "source" generates thermal energy that brings a workingsubstanceto a high temperature. The working substance then generates work in the engine while transferring heat to a sink at a lower temperature . The working of a heat engine is shownn figure 1. Heat engines can be modelled using themodynamic cycles. The heat engine given in Figure -2 is of a working substance which is 1.00 mol of a monoatomic ideal gas . The thermodynamic cycle begins at the point designated as '1' and goes clockwise and the values of P and l or Vat each point is as given belowFig-2 (P_(1)=1.00 atm "and" V_(1) = 24.6 L , P_(2)=2.00 atm , V_(3) =49.2L, P_(4)=1.00 atm) Calculate the efficiency of the cycle given in figure 2. Another system completes a cycle consistings of six quasi-static steps, during which the total work done by the system is 100J During step 1 the system absorbs 300J of heat form a reservior at 300K , during step 3 the system absorbs 200Jof heat from a reservior at 400K, and during step5 it abosrbs heat from a reservior at temperature T_(3) steps 2,4,6 are adiabatic such that the temperature of one reservoir changes to that of next. |
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| 89190. |
A ketone (A) which undergoes haloform reaction gives compound (B) on reduction. (B) on heating with H_(2)SO_(4) gives compound (C ), which forms monozonide (D). (D) on hydrolysis in the presence of Zn dust gives only acetaldehyde. Identify (A), (B) and (C ) |
| Answer» Solution :(A) `CH_(3)COCH_(2)CH_(3)` (B) `CH_(3) UNDERSET(underset(OH)(|))(C )HCH_(2)CH_(3)` ( C) `CH_(3)CH=CH_(2)CH_(3)` | |
| 89191. |
A heat engine operating between 227^(@)Cand27^(@)C absorbs 2 Kcal of heat from the 227^(@)C reservoir reversibly per cycle. The amount of work done in one cycle is : |
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Answer» 0.4 Kcal |
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| 89192. |
A ketone A (C_(4)H_(8)O), which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives a compound C which forms nono-ozonide D.D on hydrolysis in presence of zinc dust gives only acetaldehyde E. Identify A,B,C,D and Egt write the reactions involved. |
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Answer» Solution :(i) Since ketone A `(C_(4)H_(8)O)` undergoes haloform reaction, it must CONTAIN the GROUPING, B must be 2-butanol ketone `A(C_(4)H_(8)O)` must be butanone `(CH_(3)COCH_(2)CH_(3))`. (ii) Since ketone A, i.e., butanone gives compound (B) on readuction, therefore, B must be 2-butanol `(CH_(3)CHOHCH_(2)CH_(3))`. (iii) Since B, i.e., 2-butanol on heating with `H_(2)SO_(4)` gives compound C which forms a mono-ozonide, therefore, compound C must be an alkene. (iv) Since alkene C forms a mono-ozonide D which on hydrolysis in presence of ZINC dust (i.e., reductive ozonolysis) gives only ACETALDEHYDE E, therefore, C must be a symmetrical alkene, i.e., 2-Butene. (v) All the reactions involved in the problem can now be explained as follows: 
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| 89193. |
A heat engine is a system that converts heat into mechanical work. A heat "source" generates thermal energy that brings a workingsubstanceto a high temperature. The working substance then generates work in the engine while transferring heat to a sink at a lower temperature . The working of a heat engine is shownn figure 1. Heat engines can be modelled using themodynamic cycles. The heat engine given in Figure -2 is of a working substance which is 1.00 mol of a monoatomic ideal gas . The thermodynamic cycle begins at the point designated as '1' and goes clockwise and the values of P and l or Vat each point is as given belowFig-2 (P_(1)=1.00 atm "and" V_(1) = 24.6 L , P_(2)=2.00 atm , V_(3) =49.2L, P_(4)=1.00 atm) Calculate DeltaE for the paths (i) 2 rarr 2"" (ii) 2 rarr 3""(iii) 3rarr 4 "" (iv) 4 rarr 1 The heal engine depicted in this problem is a 'camot heat engine' and the themodynamic cycleof operationsof this engine is known as a 'camot cycle' named after Saudi camot , an engine cum thermodynamist . The efficiency of a camot engine as (1-T_(1)//T_(2)) where T_(1) and T_(2) are the temperature of the sink and the source. |
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| 89194. |
A heat engine is a system that converts heat into mechanical work. A heat "source" generates thermal energy that brings a workingsubstanceto a high temperature. The working substance then generates work in the engine while transferring heat to a sink at a lower temperature . The working of a heat engine is shownn figure 1. Heat engines can be modelled using themodynamic cycles. The heat engine given in Figure -2 is of a working substance which is 1.00 mol of a monoatomic ideal gas . The thermodynamic cycle begins at the point designated as '1' and goes clockwise and the values of P and l or Vat each point is as given belowFig-2 (P_(1)=1.00 atm " and " V_(1) = 24.6 L , P_(2)=2.00 atm , V_(3) =49.2L, P_(4)=1.00 atm) Calculate T_(1),T_(2),T_(3) "and" T_(4) |
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| 89195. |
A+KBrtoYellowppt.(B)A+conc.H_(2)SO_(4)overset(Delta)toBrown vapour intensified with Cu-turnings.Dissolves in hgyp forming a soluble complex(C ) What are (A),(B) and (C ) and explain their reraction? |
| Answer» Solution :(A)`AgNO_(3)`(B)AgBr (C )`Na_(3)[Ag(S_(2)O_(3))_(2)]` | |
| 89196. |
A heat engine absorbs heat Q_(1) at temperature T_(1) and Q_(2) at temperature T_(2) . Work done by the engine is (Q_(1)+Q_(2)) . This data: |
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Answer» Violates `I^(ST)` law of thermodynamics |
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| 89197. |
A heat engine absorbs 760 kJ heat from a source at 380 K. It rejects (i) 650 kJ, (ii) 560 kJ, (iii) 504 kJ of heat to sink at 280 K. State which of these represent a reversible, an irreversible and an impossible cycle. |
| Answer» SOLUTION :(i) irreversible, (II) reversible, (III) IMPOSSIBLE | |
| 89198. |
(A): K_(2)CrO_(4) has yellow colour due to charge transfer (R) : CrO_(4)^(2-)ion is tetrahedral in shape. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 89199. |
(A) HClO is stronger acid than HBrO (R ) Greater is the electronegativity of the halogen , greater will be attraction of electron pair towards it and hence more easily the H^(+) ion will be released. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 89200. |
(a) K_(2)Cr_(2)O_(7)is treated with SnCL_(2), in acidic media. (b) Pb_(3)O_(4) is trerated with conc. HNO_(3) (c) Pb(CH_(3)COO)_(2) is treated with bleaching powder. (d) PbO_(2), is treated with NaOH (e) Ferrous sulphate is ignited (F) FeCl_(3) is treated with sodium carbonate. The resulting precipitate is further heated. |
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Answer» Solution :(a) ` K_(2)Cr_(2)O_(7) + 14HCl + 3SnCl_(2) rarr 3SnCl_(4) + 2KCl + 2CrCl_(3) + 7H_(2)O` (b) ` Pb_(3)O_(4) + 4HNO_(3) rarr 2PB(NO_(3))_(2) + PbO_(2) + H_(2)O` (c ) `Pb(CH_(3)COO)_(2) + Ca(OCL)Cl + H_(2)O rarr PbO_(2) + CaCl_(2) + 2CH_(3)COOH` (d) ` PbO_(2) + 2NaOH rarr Na_(2)PbO_(3) + H_(2)O` (e) `2FeS_(4) rarr Fe_(2)O_(3) + SO_(2) + SO_(3)` (f) `2FeCl_(3) + 2Na_(2)CO_(3) + 3H_(2)O rarr 2Fe(OH)_(3) 6NaCl + 3CO_(2)` `2Fe(OH)_(3) rarr Fe_(2)O_(3) + 3H_(2)O` |
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