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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.
| 30501. |
The greater reactivity of fluroine is due to: |
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Answer» LOWER energy of F-F BOND |
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| 30502. |
The greater reactivity of fluorine is due to |
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Answer» LOW ENERGY of the F-F bound |
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| 30503. |
The graphs representing distribution of molecular speeds at 300 K for gases CI_(2) and N_(2) are as shown below (Atomic mass N = 14, CI = 35.5): |
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Answer» I graph is for `N_(2)` and II is for `CI_(2)` |
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| 30505. |
The graph obtained by taking vapour pressure (P) of a liquid on y-axis and temperature (T) on x-axis will be |
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Answer»
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| 30506. |
The graph between. The log K verrus (1)/(T) is a straight line. The slope of the line is |
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Answer» `(-2.303R)/(EA)` |
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| 30507. |
The graph between log t_(1//2) and log a at a given temperature is Rate of this reaction will …….with passage of time |
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Answer» INCREASE |
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| 30508. |
The graph between log k and 1/T[K is rate constant (sec^(-1)) and T the temperature (K)] is a straight line with OX=5 and theta=tan^(-1)(-1/2.303).Calculate the value of E_a is cal. ? |
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Answer» THUS, `-E_a/(2.303R)=TAN theta=-1/2.303` `:. E_a=R=2` CAL. 
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| 30509. |
The graph alogside showsthe heating curve for a pure substance. The temperature rises with time as the substance is heated : What is the physical state of the substance at the point A,B,C and D? What is the melting point of the substance ? What is its boiling point ? What happens to the temperature while the substance is chaging state ? The substace is not wter. how can you judge from the graph ? |
Answer» SOLUTION : (a)At point A : The substancve is in the solid STATE. At point B : The SUBSTANCE has started has started melting. It exists both in the solid and liquid states. At point C : The substace is in the liquid state. At point D : The substance has started boiling. It exists both in the liquid and gaseous states. (b) The melting point of the substance is `15^(@)C`. (c) The melting point of the substance is `110^(@)C`. (d) The temperature remains the same during the chage of state. (e) Had substance been wagter, its melting point should have been `0^(@)C` and boiling point `100^(@)C`. It is therefore,not water. |
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| 30510. |
The gram molar volume of a gas is the volume occupied at S.T.P. by |
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Answer» ONE gram of the GAS |
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| 30511. |
The gradual decrease in size (actionoid contraction) is from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction). |
| Answer» SOLUTION :Actinoid contraction is greater than LANTHANOID contraction. This is DUE to the POOR shielding by 5f-electrons than by 4f electrons. | |
| 30512. |
The gradiual addition of KJ solutionto Bi(NO_(3))_(3) solution identifyproduces a darkprecipitatewhichdissslovesin excess of KJ to give a yellowsolution.Writethechemicalequationfor the above reactions. |
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Answer» Solution :The gradual additionof K! solution of` Bi(NO_(3))_(2)` solutioninitialyproduces darkbrownprecipitate which dissolve in EXCESS of Ki to give a YELLOWSOLUTION IODIDE isoxidesed to iodine whichdissolve in KI to give a yellow solution of `KI_(3)` `Bi^(3+) +H_(2)O hArr Bi(OH)^(2+) +H^(o+)` `[NO_(3)^(Theta) + 4H^(o+) + 3e^(Theta) NO + 2H_(2)O] xx2` `[21^(Theta) rarr I_(2) + 2e^(Theta)] xx 3 ` `ulbar (2NO_(3)^(Theta) + 8H^(o+) + 61^(Theta) rarr 2NO + 4H_(2)O + 31_(2))` `KI +I_(2) rarr KI_(3)` |
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| 30513. |
The good method for converting benzene into n- propyl benzene is: |
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Answer» `C_(6)H_(6) + CH_(3)CH_(2)CH_(3)Cl +` Anhyd.`AlCl_(3)` |
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| 30514. |
The gold numbers of three protective colloids A, B and Care 0.05, 0.01 and 0.5 respectively. The decreasing order of their protective power is: |
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Answer» `B. C, A` |
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| 30515. |
The gold numbers of protective colloids A,B,D and D are 0.04,0.004,10 and 40 respectively. The protective powers of A,B,C, and D are in the order : |
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Answer» `AgtBgtCgtD` |
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| 30516. |
The gold numbers of lyophilic colloids A, B ,C and D are 125, 10, 0.1 and 0.005 respectively. Their protective power will be in the order- |
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Answer» `B GT D gt C gt A` |
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| 30517. |
The gold numbers of a few protective colloids are: A Starch 25, (B) Gelatin - 0.005 -0.01. (C) Haemoglobin = 0.03. (D) Dextrin = 6 - 123 Which is the best protective colloid? |
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Answer» A |
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| 30518. |
The gold number of A, B, C and D are 0.04, 0.002, 10 and 25 respectively. The protective powers of A, B, C and D are in order |
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Answer» `A GT B gt C gt D` |
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| 30519. |
The gold number of protective colloids A,B,C and D are 0.02, 0.002, 10 and 30 respectively. then the protective powers of A,B,C and D are in the order |
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Answer» `DgtCgtAgtB` |
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| 30520. |
The Glycosidic linkage present in sucrose is between |
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Answer» C - l of `ALPHA`-glucose and C - 2 0f `beta`-fructose 
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| 30521. |
The glycosidic linkage involved in linking the glucose units in amylose part of starch is |
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Answer» `C_1 -C_4 beta` - LINKAGE |
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| 30522. |
The glycosidic linkage in carbohydrates is |
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Answer» Link between two carbon atoms in a carbohydrate by a covalent bond |
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| 30523. |
The glycolipids abundantly found in |
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Answer» oils |
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| 30524. |
The glucose gets oxidized to saccharic acid in presence of HNO_(3). This indicates the presence of |
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Answer» PRIMARY alcohol groups in GLUCOSE |
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| 30525. |
Explain the classification of carbohydrates. |
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Answer» Link between two CARBON ATOMS in a CARBOHYDRATE by a covalent bond |
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| 30526. |
The globular protein is present in ………………….. |
| Answer» SOLUTION :Blood | |
| 30527. |
The glass having smallest coefficient of thermal expansion is : |
| Answer» Answer :D | |
| 30528. |
The given structure of alpha-amino acid will exist at which pH? oveset(o+)NH_3- underset( R) underset(|) overset(COOH)overset(|)C -H |
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Answer» 7 |
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| 30529. |
The given reaction is an example of : C_2H_5Br + KCN(aq.) rarr C_2H_5CN +KBr |
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Answer» Elmination |
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| 30530. |
The given reaction is an example of (AAK_MCP_37_NEET_CHE_E37_018_Q01) |
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Answer» STEPHEN reaction |
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| 30531. |
The given reaction FeCl_(3)+SnCl_(2) to 2FeCl_(2)+SnCl_(4) is an example of |
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Answer» FIRST ORDER concentration |
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| 30532. |
The given reaction is an example of(AAK_MCP_35_NEET_CHE_E35_015_Q01) |
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Answer» WURTZ reaction |
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| 30533. |
The given reaction 2FeCl_3+SnCl_2rarrSnCl_4+2FeCl_2 is an example of : |
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Answer» FIRST ORDER reaction |
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| 30534. |
The given reaction 2FeCl_3 + Sn Cl_2 to 2FeCl + Sn Cl_4 is an example of |
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Answer» FIRST order REACTION |
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| 30535. |
The given reaction2 NO + O_(2) to 2NO_(2) is an example of |
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Answer» FIRST ORDER REACTION Rate = `k [NO]^(2) [O_(2)]^(1) therefore O.R = 2 + 1 = 3`. |
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| 30536. |
Which of the following octahedral complex does not show geometrical isomerism (A and B are monodentate ligands) ? |
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Answer» `[MA_5B]` |
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| 30537. |
The given monosaccharidecan exist as aldohexose in its D-configuration. What is the total number of stereoisomers which can exist in pyranose form ? |
Answer» Solution :The pyranose form of the stereoisomers having D-configuration is shown :  There are three chiral `(C^(**))` ATOMS. THEREFORE, chiral centres = 3. Total number of stereoisomers `= 2^(n)=2^(3)=8`. |
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| 30538. |
The given graph shows the vapour pressure- temperature curves for some liquids. Liquids A,B,C, and D respectively are |
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Answer» diethyl ETHER, acetone, ETHYL alcohol, whater |
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| 30539. |
The given graph represents the variation of Z (compressibility factor =(PV)/(nRT)) versus P, for three real gases A,B and C. Identify the only incorrect statement. |
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Answer» For the gas A,a=0 and its dependence on P is linear at all pressures `(P+a/(V^(2)))(V-b)=RT` When `a=0,P(V-b)=RT` or `PV=RT+Pb` i.e. `Z=1+(Pb)/(RT)`. Thus `Zgt1` and INCREASES with INCREASE of pressure. Hence a is correct. When `b=0(P+a/(V^(2)))V=RT` or `PV+a/V=RT` i.e. `Z=1-a/(RTV)`. Thus `Zlt1` and decreases with increase of pressure.Hence b is correct. Statement d is also correct. |
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| 30540. |
The given graph/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice (s) about I, II, III and IV is (are) correct. |
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Answer» I is PHYSISORPTION and II is CHEMISORPTION |
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| 30541. |
The given graph/data I, II, III and IV represent general trends for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice (s) about I, II, III and IV is (are) correct |
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Answer» I is PHYSISORPTION and II is CHEMISORPTION 
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| 30542. |
The given figure shows a change of state A to state C by two paths. ABC and AC for an ideal gas. Calculate the , (a) Path along which work done is least. (b) Internal energy at C if the internal energy of gas at A is 10 J and amount of heat supplied to change its state to C through the path AC is 200 J. ( c) Amount of heat supplied to the gas to gd from A to B, if internal energy of gas at state B is 10 J. (ii) In an insulated contained 1 mole of a liquid, molar volume 100 mL. is at 1 bar liquid is steeply taken to 100 bar, where volume of liquid decreases by 1 mL. Find DeltaE and DeltaH for the process? (iii) 14 g of oxygen at 0^(@) C and 10 atm are subjected to reversible adiabatic expansion to a pressure of 1 atm. Calculate the work done in (i) Litre atm. (ii) Calorie (Given, C_(p)/C_(v) = 1.4). |
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Answer» SOLUTION :(i) `+3262.88` cal (ii) 9900 BAR mL (iii) `W_("rev") = -287.88` cal. |
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| 30543. |
The given compound (X) has: |
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Answer» chirality |
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| 30544. |
The gibbs free energy for a reversible reaction at equilibrium is |
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Answer» positive `DeltaG=DeltaH-TDeltaS` For REVERSIBLE REACTION at EQUILIBRIUM, `DeltaG=0` |
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| 30545. |
The Gibb's energy for the decomposition of Al_(2)O_(3) at 500^(@)C is as follows (2)/(3)Al_(2)O_(3)to(4)/(3)Al+O_(2),Delta_(r)G=+966kJmol^(-1) The potential difference needed for electrolytic reduction of Al_(2)O_(3) at 500^(@)C is at least |
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Answer» 5.0V `E^(0)=-(966xx1000)/(4xx96500)=-2.5V` |
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| 30546. |
The Gibbs energy for the decomposition of Al_(2)O_(3) at 500^(@)C is as follows: (2)/(3)Al_(2)O_(3)to(4)/(3)Al+O_(2),Delta_(r)G=+966 kJ mol^(-1) The potential difference needed for electrolytic reduction of Al_(2)O_(3) at 500^(@)C is at least. |
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Answer» 2.5 V Thus, `(2)/(3)XX(3O^(2-)), i.e., 2O^(2-)toO_(2),n=4` `DeltaG=-nFE` `therefore966xx10^(3)=-4xx96500xxE` or `E=-2.50V` ltBrgt Thus, minimum potential difference REQUIRED =2.50V. |
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| 30547. |
The Gibb's energy for the decomposition of Al_(2)O_(3) at 500^(@) C is as follows: 2/3Al_(2)O_(3) to 4/3 Al + O_(2) , DeltaG = +966 kJ mol^(-1) The potential difference needed for electrolytic reduction of Al_(2)O_(3)at 500^(@) C is at least |
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Answer» 5.0 V |
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| 30548. |
The Gibbs energy for decomposition of Al_(2)O_(3) at 500^(@)C is as follows: 2/3Al_(2)O_(3)to4/3Al+O_(2),""DeltaG=+966kJ The potential difference needed for electrolytic reduction of Al_(2)O_(3) at 500^(@)C is at least |
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Answer» 4.5 V |
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| 30549. |
The geometry with respect to the central atom of the following molecules are : N(SiH_(3))_(3):Me_(3)N: (SiH_(3))_(3)P |
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Answer» PLANAR, PYRAMIDAL, planar 
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| 30550. |
The geometry possible in [FeF_(6)]^(4-) and [CoF_(6)]^(4-) is ………….. |
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Answer» TRIGONAL bipyramidal |
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