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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 49601. |
A reaction S_((g))hArr 4S_(2(g))is s carried out by taking 2 moles of S_(8(g))and 0.2 mole of S_(2(g))in a reaction vessel of l lit. Which one is not correct. If K_(c) = 6.3 xx 10^(-6). |
| Answer» <html><body><p> Reaction quotient is `8 xx 1O^(-4)`<br/>Reaction proceeds in backward direction<br/>Reaction proceeds in forward direction<br/>`K_p = 2.55 "atm"^3`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`S_(8) harr 4S_(2)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>=((0.2)^(4))/(2)=8 xx 10-^(-4) lt K` <br/> `:.` reaction proceds backward `KP = KC (RT)^(Delta n_(g))`, T not given</body></html> | |
| 49602. |
A reaction rate contant doubles between 300 and 310 k. By which of the following factors does the rate constant increases between 400-410 K. |
| Answer» <html><body><p>3.39<br/>2<br/>1.48<br/>1.1</p>Solution :For any reaction , the value of <a href="https://interviewquestions.tuteehub.com/tag/rate-1177476" style="font-weight:bold;" target="_blank" title="Click to know more about RATE">RATE</a> <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> <a href="https://interviewquestions.tuteehub.com/tag/doubles-958969" style="font-weight:bold;" target="_blank" title="Click to know more about DOUBLES">DOUBLES</a> between 300 and 310 . For any other temp the rate constant decrease <br/> `"log"("log"(k_(2))/(k_(1)))/("log"(k_(2))/(k_(1)))=(10)/(400xx410)*(300xx310)/(10)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>'_(2)` and `k'_(1)` are rate constant at 410 and 400 K respectively. `k_(2)` and `k_(1)` are rate constant at 310 and 300 K respectively .Here `k_(2).k_(1)=2` <br/> `therefore "log"(k_(2)')/(k_(1)')=(3xx31)/(4xx41)xx"log"2` <br/> `=(3xx31xx0.3010)/(4xx41)=0.1707` <br/> `(k_(2))/(k_(1))`=antilog 0.1707 =1.48</body></html> | |
| 49603. |
(A) : Reaction of NaOH with chlorine is a disproportionation reaction . (R) : All redox reactions are disproportionation reactions . |
| Answer» <html><body><p><br/></p>Solution :All <a href="https://interviewquestions.tuteehub.com/tag/redox-2246707" style="font-weight:bold;" target="_blank" title="Click to know more about REDOX">REDOX</a> reactionsare not disproportianation reaction. So , statement A is <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> but reason is false.</body></html> | |
| 49604. |
(A) : Reaction of 1-butene with HBr gives 1-bromobutane as major product (R ) : Addition of hydrogen halides to alkenes proceeds according to Markovnikov's rule |
| Answer» <html><body><p>A and <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a>, R <a href="https://interviewquestions.tuteehub.com/tag/explains-2627999" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLAINS">EXPLAINS</a> A<br/>A and R are true, R does not explain A<br/>A is true, but R is false <br/>A is false, but R is true'</p>Answer :D</body></html> | |
| 49605. |
A reaction occursspontaneously if |
| Answer» <html><body><p>`T Delta S <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> Delta H ` and `Delta H` is`+ve` and`Delta S` is -ve<br/>`T <a href="https://interviewquestions.tuteehub.com/tag/deltas-947743" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAS">DELTAS</a> =Delta H ` and both `Delta H` and `DeltaS` are `+ve`<br/>`T DeltaS ltDeltaH` and both `DeltaH ` and `DeltaS` are `+ve`<br/>`T DeltaS gt Delta H` and both `DeltaH ` and `DeltaS` are `+ve`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :d</body></html> | |
| 49606. |
A reaction occurs spontaneously if |
| Answer» <html><body><p>`T DeltaS lt DeltaH` and both `DeltaH` and `DeltaS` are +ve<br/>`T DeltaS <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> DeltaH` and`DeltaH` is +ve and `DeltaS` are -ve<br/>`T DeltaS gt DeltaH` and both `DeltaH` and `DeltaS` are +ve<br/>`T DeltaS = DeltaH` and both `DeltaH` and `DeltaS` are +ve</p>Solution :For a <a href="https://interviewquestions.tuteehub.com/tag/spontaneous-632196" style="font-weight:bold;" target="_blank" title="Click to know more about SPONTANEOUS">SPONTANEOUS</a> <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>, <br/> `<a href="https://interviewquestions.tuteehub.com/tag/deltag-2053246" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAG">DELTAG</a> = (-ve), DeltaS = + ve and DeltaH = + ve and T DeltaS gt DeltaH`</body></html> | |
| 49607. |
A reaction N_(2)+ 3H_(2) hArr 2NH_(3) + 92k.j is at equilibrium. If the concentration of N_(2)is increased the temperature of the system |
| Answer» <html><body><p>decreases<br/><a href="https://interviewquestions.tuteehub.com/tag/increases-1040626" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASES">INCREASES</a><br/>remains constant<br/>becomes half</p>Solution :on adding `N_(2)` <a href="https://interviewquestions.tuteehub.com/tag/eqm-446398" style="font-weight:bold;" target="_blank" title="Click to know more about EQM">EQM</a> <a href="https://interviewquestions.tuteehub.com/tag/shifts-1205400" style="font-weight:bold;" target="_blank" title="Click to know more about SHIFTS">SHIFTS</a> forward, i.e. more heat released <br/> `:.` T of system increases</body></html> | |
| 49608. |
A reaction is carried out aniline as a reactant as well as a solvent. How will you remove unreacted aniline ? |
| Answer» <html><body><p></p>Solution :The b.p. of <a href="https://interviewquestions.tuteehub.com/tag/aniline-378475" style="font-weight:bold;" target="_blank" title="Click to know more about ANILINE">ANILINE</a> (457 K) is very high. If aniline (large excess) is <a href="https://interviewquestions.tuteehub.com/tag/distilled-7316446" style="font-weight:bold;" target="_blank" title="Click to know more about DISTILLED">DISTILLED</a> froma small amount of the product by <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> distillation, it may cause <a href="https://interviewquestions.tuteehub.com/tag/decomposition-946042" style="font-weight:bold;" target="_blank" title="Click to know more about DECOMPOSITION">DECOMPOSITION</a> of the product. Therefore, to avoid decomposition of the product, aniline is <a href="https://interviewquestions.tuteehub.com/tag/removed-2986295" style="font-weight:bold;" target="_blank" title="Click to know more about REMOVED">REMOVED</a> either by vacuum distillation or by steam distillation.</body></html> | |
| 49609. |
A reaction having equal energies of activation for toward and reverse reaction has |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> =<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`<br/>`Delta H = Delta <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> =Delta S =0`<br/>`Delta S =0`<br/>`Delta G =0` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 49610. |
A reaction has both DeltaH and DeltaS - ve.The rate of reaction |
| Answer» <html><body><p>cannot be <a href="https://interviewquestions.tuteehub.com/tag/predicted-7271267" style="font-weight:bold;" target="_blank" title="Click to know more about PREDICTED">PREDICTED</a> for change in temperature<br/>increases with <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> in temperature<br/>increases with <a href="https://interviewquestions.tuteehub.com/tag/decrease-946104" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASE">DECREASE</a> in temperature<br/>remains unaffected by change in temperature.</p>Answer :C</body></html> | |
| 49611. |
A reaction between ammonia and boron trifluoride is given below: NH_3 + BF_3 to H_3N :BF_3 Identify the acid and base in the given reaction. Which theory explains it? What is the hybridisation of B and N in the reactants ? |
| Answer» <html><body><p></p>Solution :As `BF_3` does not have a <a href="https://interviewquestions.tuteehub.com/tag/proton-1170983" style="font-weight:bold;" target="_blank" title="Click to know more about PROTON">PROTON</a> but acts as <a href="https://interviewquestions.tuteehub.com/tag/lewis-541347" style="font-weight:bold;" target="_blank" title="Click to know more about LEWIS">LEWIS</a> acid as it is an electron deficient molecule. It reacts with `NH_3` by accepting the lone pair of electrons from `NH_3` and <a href="https://interviewquestions.tuteehub.com/tag/complete-423576" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETE">COMPLETE</a> its octet. The reaction is <br/> `BF+:NH_3 to BF_3 larr NH_3` <br/> Lewis theory of acids and bases can <a href="https://interviewquestions.tuteehub.com/tag/explain-447165" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLAIN">EXPLAIN</a> it. <a href="https://interviewquestions.tuteehub.com/tag/boron-400820" style="font-weight:bold;" target="_blank" title="Click to know more about BORON">BORON</a> in `BF_3` is `sp^2` hybridised where N in `NH_3` is `sp^3` hybridised.</body></html> | |
| 49612. |
A reaction A (g) + B (g) hArr 2 C (g)is in equilibrium at a certain temperature. Can we increase the amount of products by (i) adding catalyst (ii) increasing pressure ? |
| Answer» <html><body><p></p>Solution :(i) No, because <a href="https://interviewquestions.tuteehub.com/tag/catalyst-21033" style="font-weight:bold;" target="_blank" title="Click to know more about CATALYST">CATALYST</a> does not <a href="https://interviewquestions.tuteehub.com/tag/disturb-957156" style="font-weight:bold;" target="_blank" title="Click to know more about DISTURB">DISTURB</a> the <a href="https://interviewquestions.tuteehub.com/tag/state-21805" style="font-weight:bold;" target="_blank" title="Click to know more about STATE">STATE</a> of equilibrium (ii) No, because `n_(p) = n_(r)`</body></html> | |
| 49613. |
A reaction, A+B to C+D+q is found to have a positive entropy change. The reaction will be |
| Answer» <html><body><p>possible at <a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a>.<br/>possible only at low temperature.<br/>not possible at any temperature.<br/>possible at any temperature.</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/deltah-2053275" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAH">DELTAH</a>= -ve and <a href="https://interviewquestions.tuteehub.com/tag/deltas-947743" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAS">DELTAS</a> =+ve` <br/> Reaction is spontaneous so `<a href="https://interviewquestions.tuteehub.com/tag/deltag-2053246" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAG">DELTAG</a>= -ve` So, Reaction will be spontaneous at any temperature.</body></html> | |
| 49614. |
A reaction, A+B rarrC + D+ q is found to havea positive entropy change. The reaction will be (i) possible at high temperature(ii) possible only at low temperature (iii) not possible at any temperature(iv)possible at any temperature |
| Answer» <html><body><p></p>Solution :Here, `DeltaH =-ve` and `DeltaS = + ve,Delta <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> = DeltaH - T DeltaS`. For the reaction to be <a href="https://interviewquestions.tuteehub.com/tag/spontaneous-632196" style="font-weight:bold;" target="_blank" title="Click to know more about SPONTANEOUS">SPONTANEOUS</a>, `DeltaG` shoulbebe `-ve` which will <a href="https://interviewquestions.tuteehub.com/tag/beso-8286785" style="font-weight:bold;" target="_blank" title="Click to know more about BESO">BESO</a> at any <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a>, i.e., option (iv) is correct.</body></html> | |
| 49615. |
A reaction has both DH and DS negative. The rate of reaction |
| Answer» <html><body><p>Increase with increase in <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> <br/>Cannot be <a href="https://interviewquestions.tuteehub.com/tag/predicted-7271267" style="font-weight:bold;" target="_blank" title="Click to know more about PREDICTED">PREDICTED</a> for <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a> in temperature <br/>Increase with decrease in temperature <br/>Remains unaffected by change in temperature </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :C</body></html> | |
| 49616. |
Rate of diffusion of gases is not influenced by |
| Answer» <html><body><p>Both A and R are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a>. R is the correct explanation of A. <br/>Both A and R are correct. R is not the correct explanation of A.<br/>A is true, but R is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a><br/> A is false, but R is true </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 49617. |
A rarr B, Graph between long_(10) P and (1)/(T) is a straight line of slope (1)/(4.606). Hence, Delta H is |
| Answer» <html><body><p>1 cal<br/>`-4.18` J<br/>4 cal<br/>`-1` cal</p>Solution :`Delta G^(0) = - <a href="https://interviewquestions.tuteehub.com/tag/rt-615359" style="font-weight:bold;" target="_blank" title="Click to know more about RT">RT</a> ln K_(P) = - RT ln P= - 2.303RT log P` <br/> log P `= (-Delta G^(0))/(2.303R).(1)/(T) = (-Delta <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(0))/(2.303RT) + (Delta S^(0))/(2.303R)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>` <a href="https://interviewquestions.tuteehub.com/tag/slope-15649" style="font-weight:bold;" target="_blank" title="Click to know more about SLOPE">SLOPE</a> `= (-Delta H^(0))/(2.303RT) = (1)/(4.606)` <br/> `rArr Delta H^(0) = - 1 "cal" = -4.18J`</body></html> | |
| 49618. |
A radioisotope will not emit |
| Answer» <html><body><p>alpha and <a href="https://interviewquestions.tuteehub.com/tag/beta-2557" style="font-weight:bold;" target="_blank" title="Click to know more about BETA">BETA</a> <a href="https://interviewquestions.tuteehub.com/tag/rays-1178020" style="font-weight:bold;" target="_blank" title="Click to know more about RAYS">RAYS</a> simultaneously<br/>beta and gamma rays simultaneously<br/>gamma and alpha rays simultaneously<br/>gamma rays only</p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/radio-4998" style="font-weight:bold;" target="_blank" title="Click to know more about RADIO">RADIO</a> isotope first <a href="https://interviewquestions.tuteehub.com/tag/emits-450754" style="font-weight:bold;" target="_blank" title="Click to know more about EMITS">EMITS</a> `alpha` or `beta` particles , then it becomes <a href="https://interviewquestions.tuteehub.com/tag/unstable-717952" style="font-weight:bold;" target="_blank" title="Click to know more about UNSTABLE">UNSTABLE</a> and emits `gamma`-rays</body></html> | |
| 49619. |
A radioactive substance has a constant actity of 2000 disintegraion/mintue. The material is swparated into two fractions, one of which has an initlial activity of 100 disntegrationper secoundwhile the other fraction decays with t_(1//2) = 24 hour. The total activity in both samples after 48 hour of separation is: |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1500-275573" style="font-weight:bold;" target="_blank" title="Click to know more about 1500">1500</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>`<br/>`1250`<br/>`2000`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The problem refers that rate is constant.</body></html> | |
| 49620. |
A radioactive isotope decays as ._(Z)A^(m) rarr ._(Z-2)B^(m-4) rarr ._(Z-1)C^(m-4) The half lives of A and B are 6 months respectively. Assuming that initially only A was present, will it be possibel to achieeveradioactive equilibrium fo B? If so, what would be the ratio ofnuclei A and B? If so, what wouldhappen if the half-livesfor A and B were 10 months adn 6 months respectively? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`0.6`, No</body></html> | |
| 49621. |
A radioactive isotope decays as ._(Z)A^(m) rarr ._(Z-2)B^(m-4) rarr ._(Z-1)C^(m-4) The half lives of A and B are 6 months respectively. Assuming that initially only A was present, will it be possibel to achieeveradioactive equilibrium fo B? If so, what would be the ratio ot A and B? If so, what wouldhappen if the half-livesfor A and B were 10 months adn 6 months respectively? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`0.6, 1.66`</body></html> | |
| 49622. |
A radiation of 2000Å falls on the metal whose work function is 4.2 eV. Then the kinetic energy of the fastest photo eletron is |
| Answer» <html><body><p>`6.4xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)` <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a><br/>`16xx10^(-10)` J<br/>`1.6xx10^(-19)` J<br/>`3.2xx10^(-19)` J</p>Solution :Kinetic energy (<a href="https://interviewquestions.tuteehub.com/tag/ke-527890" style="font-weight:bold;" target="_blank" title="Click to know more about KE">KE</a>) `=hv-W` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/lambda-4.2eV(6.62xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/34-308171" style="font-weight:bold;" target="_blank" title="Click to know more about 34">34</a>)xx3xx10^(8))/(2000xx10^(-10))=-4.2xx1.6xx10^(-19)` <br/> `=9.9xx10^(-19)j-6.7xx10^(-19)J=3.2xx10^(-19)` J</body></html> | |
| 49623. |
A quantity of KMnO_(4) was boiled with HCl and the gas evolved was led into a solution of KI. When the reaction was complete, the I_(2) liberated was titrated with titrated with a solution of hypo containing 124 g of Na_(2)S_(2)O_(3). 5H_(2)O per litre. It was found that exactly 60 mL were required to decolourise the solution of I_(2). What weight of KMnO_(4) was used ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 49624. |
"A quantity of " PCl_(5) " was heated in a " 10 dm^(3) " vessel at " 250^(@)C : PCl _ (5) (g) hArr PCl_(3) (g) + Cl_(2) (g) . At equilibrium, the vessel contains 0*1 mole of PCl_(5)and 0*2mole of Cl_(2). The equilibrium constant of the reaction is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>*<a href="https://interviewquestions.tuteehub.com/tag/04-256232" style="font-weight:bold;" target="_blank" title="Click to know more about 04">04</a>`<br/>`0*025`<br/>`0*<a href="https://interviewquestions.tuteehub.com/tag/02-256158" style="font-weight:bold;" target="_blank" title="Click to know more about 02">02</a>`<br/>`0*05`</p>Answer :A</body></html> | |
| 49625. |
A quantity of 4.0 moles of an ideal gas at 20^(@)C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 10.0L. What is the entropy change of the system (in cals)? |
| Answer» <html><body><p><br/></p>Solution :Expansion against constant pressure means the process is <a href="https://interviewquestions.tuteehub.com/tag/irreversible-519610" style="font-weight:bold;" target="_blank" title="Click to know more about IRREVERSIBLE">IRREVERSIBLE</a> <br/> `Delta S = nR l <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> (V_(2))/(V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)) = 4 xx 2 l n (10)/(1) = 18.424` <a href="https://interviewquestions.tuteehub.com/tag/cal-412083" style="font-weight:bold;" target="_blank" title="Click to know more about CAL">CAL</a></body></html> | |
| 49626. |
A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O IF 0.02 M K_(2)Cr_(2)O_(7) is used instead of 0.02 M KMnO_(4) its volume required in these titrations are respectively |
| Answer» <html><body><p>25 mL, <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> mL<br/>25 mL, 15 mL<br/>20.8 mL, 33.3 mL<br/>10.4 mL, 16.7 mL</p>Solution :m.eqts of `Fe^(+2)` = mEq.ts of `KMnO_(4)` <br/> implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` <br/> In <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of Zn, <br/> m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` <br/> In pressure of Zn, <br/> me.eqtsof `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` <br/> m.eqts of `Fe^(+3)=40xx0.02xx5=4` <br/> Now in the second case `K_(2)Cr_(2)O_(7)` <br/> used instead of `KMnO_(4)`, <br/> m eq. `K_(2)Cr_(2)O_(7)` =m eq of `Fe^(+2)` <br/> `0.02xx6xxv=2.5` <br/> `implies V=(2.5)/(0.12)=20.8ml` <br/> In pressure of Zn <br/> m. eqts of `K_(2)Cr_(2)O_(7)` = m eq of `Fe^(+3)` <br/> `0.02xx6xxv=2` <br/> `impliesV=(4)/(0.12)=33.3ml`</body></html> | |
| 49627. |
A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O Zinc aded in the second titration wil |
| Answer» <html><body><p>oxidize `Fe^(2+)` to `Fe^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>+)` <br/><a href="https://interviewquestions.tuteehub.com/tag/reduce-1181332" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCE">REDUCE</a> `Fe^(3+)` to `Fe^(2+)`<br/>reduce `Fe^(3+)` to Fe<br/>reduce `Fe^(2+)` to Fe</p>Solution :In presence of <a href="https://interviewquestions.tuteehub.com/tag/zn-751409" style="font-weight:bold;" target="_blank" title="Click to know more about ZN">ZN</a> `Fe^(+3)` reduced to `Fe^(+2)`</body></html> | |
| 49628. |
A proton is moving with kinetic energy 5 xx 10^(-27) J. What is the wavelength of the de Broglie wave associated with it ? |
| Answer» <html><body><p><br/></p>Solution :Mass of proton `= (1.008 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/(6.02 xx 10^(23)) <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> = 1.67 xx 10^(-27) kg` <br/> Further proceed as in Solved <a href="https://interviewquestions.tuteehub.com/tag/problem-25530" style="font-weight:bold;" target="_blank" title="Click to know more about PROBLEM">PROBLEM</a> 3.</body></html> | |
| 49629. |
A proton is accelerated to a velocity of3xx 10^(7) m s^(-1). If the velocity can be measured with a precision of +- 0.5%, calculate the uncertainty in position of proton [h = 6.6 xx 10^(-34)Js, mass of proton = 1.66 xx 10^(-27)kg] |
| Answer» <html><body><p><br/></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> v = (0.5)/(100) xx (3 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)) ms^(-1) = 1.5 xx 10^(5) m s^(-1)` <br/> `Delta <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> = (h)/(4pi m Delta v) = (6.6 xx 10^(-34) <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> m^(2) s^(-1))/(4 xx 3.14 xx 1.66 xx 10^(-27) kg xx 1.5 xx 10^(5) ms^(-1)) = 2.11 xx 10^(-13) m`</body></html> | |
| 49630. |
A proton is accelerated to 1/10th of the velocity of light. If its velocity can be measured with a precision of +- 0.5%, what must be its uncertanity in position? |
| Answer» <html><body><p></p>Solution :Velocity of <a href="https://interviewquestions.tuteehub.com/tag/light-1073401" style="font-weight:bold;" target="_blank" title="Click to know more about LIGHT">LIGHT</a> `= 3 xx 10^(8) ms^(-1)` <br/> `:.` Velocity of proton `= (1)/(10) xx 3 xx 10^(8) = 3 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>) ms^(-1)` <br/> `Delta v = (0.5)/(100) xx 3 xx 10^(7) = 1.5 xx 10^(5) ms^(-1)` <br/> Applying Heisenberg's uncertanity principle. <br/> `Delta x. m Delta v = (h)/(4pi)` <br/> We get `Delta x = (h)/(4pi xx m Delta v) = (6.626 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/34-308171" style="font-weight:bold;" target="_blank" title="Click to know more about 34">34</a>) kg m^(2) s^(-1))/(4 xx 31.43 xx 1.66 xx 10^(-27) kg xx 1.5 xx 10^(5) ms^(-1)) = 2.11 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>) m`</body></html> | |
| 49631. |
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1k V, its kinetic energy will be |
| Answer» <html><body><p>1840 keV<br/>1/1840 ke V<br/>1 keV<br/>920 keV</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> acquired by <a href="https://interviewquestions.tuteehub.com/tag/electron-968715" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRON">ELECTRON</a> (as K.E.) on being <a href="https://interviewquestions.tuteehub.com/tag/accelerated-366473" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATED">ACCELERATED</a> by a potential difference of 1kV = <a href="https://interviewquestions.tuteehub.com/tag/1kev-1803474" style="font-weight:bold;" target="_blank" title="Click to know more about 1KEV">1KEV</a>. As proton has the same <a href="https://interviewquestions.tuteehub.com/tag/charge-914384" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGE">CHARGE</a> and as energy = potential difference `xx` charge, hence proton will have the same kinetic energy</body></html> | |
| 49632. |
Electrons are accelerated through a potential difference of 150V. Calculate thede Broglie wavelength. |
| Answer» <html><body><p>`lambda_3 = lambda_p`<br/>`lambda_e lt lambda_p`<br/>`lambda_e gt lambda_p`<br/>`n_3 to n_1`</p>Solution :`(<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a> <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>)/(lambda p)=<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((<a href="https://interviewquestions.tuteehub.com/tag/mp-565167" style="font-weight:bold;" target="_blank" title="Click to know more about MP">MP</a>)/(me)) = sqrt(1837)`</body></html> | |
| 49633. |
An alpha-particle and a proton are accelerated from rest by the same potential, thenthe ratio of their de-Broglie wavelength is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt2-3056932" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT2">SQRT2</a>`<br/>`1/sqrt2`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2sqrt2-1838926" style="font-weight:bold;" target="_blank" title="Click to know more about 2SQRT2">2SQRT2</a>`<br/>`2`</p>Solution :`lambda = h/(sqrt(2qVm)) , lambda_(alpha) = h/(sqrt(2xx2xxV <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> m_(alpha)))` <br/>`implieslambda_(alpha) = (h)/(sqrt(2xx2xx V xx <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))` <br/>`lambda _P =(h)/(sqrt(2xx1xxVxx m_p)) , lambda_p = (h)/(sqrt(2xx1xxV xx 1))` <br/>`(lambda_p)/(lambda_(alpha)) = sqrt((16)/(2)) implies(lambda_P)/(lambda_(alpha)) = 2sqrt2`</body></html> | |
| 49634. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.05 M NH_(4)OH mixed with 100 mL of 0.10 M HCl solution |
| Answer» <html><body><p>1.6<br/>2.6<br/>3.6<br/>4.6</p>Solution :100 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of 0.05 M `NH_(4)OH = 5`mmol of `NH_(4)OH` <br/> 100 mL of 0.<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> M <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> = 10 mmol of HCl <br/> 5 mmol of `NH_(4)OH ` will react with5 mmol of HCl to <a href="https://interviewquestions.tuteehub.com/tag/form-996208" style="font-weight:bold;" target="_blank" title="Click to know more about FORM">FORM</a> 5 mmol of `NH_(4)Cl` <br/> HCl left = 5 mmol . Volume of solution = 200 mL<br/> `:. [HCl]=(5)/(200) = 0.025M`, <br/> `[H^(+)] =0.025 M`, <br/> `pH = - <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> (0.025) = 1.602`</body></html> | |
| 49635. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.10 M NH_(4)OH mixed with 100 of 0.05M HClsolution |
| Answer» <html><body><p>6.25<br/>7.25<br/>8.25<br/>9.25</p>Solution :100 ml of 0.10 M `NH_(4)OH` = 10mmol of `NH_(4)OH` <br/> 100 mL of 0.05 M HCl = <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> mmol of HCl <br/> 5 mmol of HCl will react with 5 mmol of `NH_(4)OH` to form 5 mmol of `NH_(4)Cl` <br/> `NH_(4)OH ` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> = 5 mmol , `NH_(4)Cl` <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 5 mmol <br/> Thus, it is a basic buffer with `[NH_(4)OH]=[NH_(4)Cl]` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/poh-1145204" style="font-weight:bold;" target="_blank" title="Click to know more about POH">POH</a> = pK_(b) + log .(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Base"])` <br/> `=-log (1.8xx10^(-5))=4.75` <br/> `:. pH = 14 - 4.759.25`</body></html> | |
| 49636. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.10 M NaOH mixed with 100 mL 0.10 M CH_(3)CO OH solution |
| Answer» <html><body><p>5.72<br/>6.72<br/>7.72<br/>8.72</p>Solution :100 ml of 0.1 M NaOH = 10 mmol of NaOH<br/> 100 ml of 0.10 M `CH_(3)CO <a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>` <br/> = 10 mmol of `CH_(3)CO OH` <br/> They will react completely to form 10 mmol of `CH_(3)CO ON a` <br/> Volume of solution = 200 mL<br/> `:. [CH_(3)CO ON a]=(10)/(200)= 0.05 M` <br/> As `CH_(3)CO ON a` is a salt of <a href="https://interviewquestions.tuteehub.com/tag/weak-729638" style="font-weight:bold;" target="_blank" title="Click to know more about WEAK">WEAK</a> acid with strong <a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a>, <br/> `pH= 7 +(1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) pK_(a)+(1)/(2) log c` <br/> `= 7 + (1)/(2) (4.74)+(1)/(2) log 0.05` <br/> `=7+2.37 + (1)/(2) (-1.301)=8.72`</body></html> | |
| 49637. |
(A) : Propene reacts with HBr to give isopropyl bromide. (R ) : Addition of hydrogen halide to alkenes follows Markownikoff's rule |
| Answer» <html><body><p>A and <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a>, R explains A<br/>A and R are true, R does not <a href="https://interviewquestions.tuteehub.com/tag/explain-447165" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLAIN">EXPLAIN</a> A<br/>A is true, but R is false <br/>A is false, but R is true'</p>Answer :A</body></html> | |
| 49638. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mLof 0.10 MNaOH mixed with 100 ml of 0.05 M CH_(3)C O O H solution |
| Answer» <html><body><p>10.4<br/>11.4<br/>12.4<br/>13.4</p>Solution :100 ml of 0.1 M <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> = 10 mmol of NaOH <br/> 100 mL of 0.05 M `CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)CO <a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>` <br/> = 5 mmol of `CH_(3)CO OH` <br/> 5 mmol will <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with 5 mmol of NaOH to form <br/> 5 mmol of `CH_(3)CO Ona` <br/> NaOH <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> = 5 mmol. <br/> Volume of the solution = 200 ml <br/> `:. [NaOH]=(5)/(200) M = 0.025M` <br/> or `[OH^(-)]=0.025M` <br/> or `[OH^(-)]=0.025M` <br/> `:. [H^(+)]=10^(-14)//0.025=4.0xx10^(-13)`, <br/> pH= 13- 0.602 = 12.398 `~=` 12.4</body></html> | |
| 49639. |
A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.05 M NaOH mixed with 100 ml of 0.10 M CH_(3)CO OH solution |
| Answer» <html><body><p>3.75<br/>4.75<br/>5.75<br/>6.75</p>Solution :100 ml of 0.05 M <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> = 5 mmol of NaOH <br/> 100 mL of 0. 1 M `CH_(3)CO OH= <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ` mmol of `CH_(3)CO OH` <br/> 5 mmol of NaOH will react with 5 mmol of `CH_(3)CO OH` to form 5 mmol of `CH_(3)CO ON a` <br/> `CH_(3)CO OH ` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> = 5 mmol, <br/> `CH_(3)CO ON a` <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 5 mmol <br/>Total <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> = 200 mL <br/> `:. [CH_(3)CO OH]=[CH_(3)CO ON a]` <br/> `=(5)/(200) M = 0.025 M` <br/> As it is a buffer of weak acid and its salt with strong base, <br/> `pH = pK_(a)+log. (["Salt"])/(["Base"])` <br/> `=-log (1.8xx10^(-5))+log 1` <br/> `=5- (0.255)=4.745`</body></html> | |
| 49640. |
(A) , product (A) is: |
| Answer» <html><body><p>`H_(2)C=CH-CH=CH_(2)`<br/>`CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)-C-=C-CH_(3)`<br/>`CH_(3)-CH_(2)-C-=CH`<br/>`CH_(2)-CH=C=CH_(2)`</p>Solution :Formation of <a href="https://interviewquestions.tuteehub.com/tag/vicinal-3261816" style="font-weight:bold;" target="_blank" title="Click to know more about VICINAL">VICINAL</a> di-halide followed by two consective `E_(2)` <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>. (elimination biomolecular)</body></html> | |
| 49641. |
(A) Product (A) is |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_CHM_XII_V04_C03_E01_100_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_CHM_XII_V04_C03_E01_100_O02.png" width="30%"/><br/>Both <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>& 2<br/>None of these</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/aldol-1972435" style="font-weight:bold;" target="_blank" title="Click to know more about ALDOL">ALDOL</a> <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a></body></html> | |
| 49642. |
A process is taking place at constant temperature and pressure. Then |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> H = Delta E`<br/>`Delta H = T DeltaS`<br/>`Delta H = 0`<br/>`Delta S = 0`</p>Solution :At const.T & P,<a href="https://interviewquestions.tuteehub.com/tag/process-11618" style="font-weight:bold;" target="_blank" title="Click to know more about PROCESS">PROCESS</a> is in <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> . `DeltaG=0`.Hence,`DeltaH = T DeltaS`</body></html> | |
| 49643. |
A process is spontaneous at all temperatures when |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> =- ve, Delta S=-ve `<br/>`Delta H =+ ve, Delta S=-ve `<br/>`Delta H =-ve, Delta S=+ ve`<br/>`Delta H =+ ve, Delta S =+ ve ` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 49644. |
A process in which no heat change takes place is called |
| Answer» <html><body><p>An <a href="https://interviewquestions.tuteehub.com/tag/isothermal-520014" style="font-weight:bold;" target="_blank" title="Click to know more about ISOTHERMAL">ISOTHERMAL</a> process <br/>An <a href="https://interviewquestions.tuteehub.com/tag/adiabatic-849856" style="font-weight:bold;" target="_blank" title="Click to know more about ADIABATIC">ADIABATIC</a> process <br/>An <a href="https://interviewquestions.tuteehub.com/tag/isobaric-1052329" style="font-weight:bold;" target="_blank" title="Click to know more about ISOBARIC">ISOBARIC</a> process<br/>An isochoric process </p>Answer :B</body></html> | |
| 49645. |
A process in which heat can flow from system to surroundings or vice-versa is called "…......................" process. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/adiabatic-849856" style="font-weight:bold;" target="_blank" title="Click to know more about ADIABATIC">ADIABATIC</a></body></html> | |
| 49646. |
(A): Pressure of the gas increases with the decrease in volume at constant temperature (R): Number of molecules per unit volume increases with decrease in volume consequently number of collisions on the wall increases |
| Answer» <html><body><p>Both A and <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> and R is the correct <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of A. <br/>Both A and R are correct but R is not the correct explanation of A. <br/>A is true but R is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a>. <br/>A is false but R is true. </p>Answer :B</body></html> | |
| 49647. |
A pressure cooker reduces cooking time for food because. |
| Answer» <html><body><p>Cooking <a href="https://interviewquestions.tuteehub.com/tag/involves-519086" style="font-weight:bold;" target="_blank" title="Click to know more about INVOLVES">INVOLVES</a> chemical changes helped by a rise in temperature <br/>Heat is more evenly distributed in the cooking space <br/><a href="https://interviewquestions.tuteehub.com/tag/boiling-17187" style="font-weight:bold;" target="_blank" title="Click to know more about BOILING">BOILING</a> point of water <a href="https://interviewquestions.tuteehub.com/tag/involved-7257329" style="font-weight:bold;" target="_blank" title="Click to know more about INVOLVED">INVOLVED</a> in cooking is increased <br/>The higherpressure inside the <a href="https://interviewquestions.tuteehub.com/tag/cooker-425792" style="font-weight:bold;" target="_blank" title="Click to know more about COOKER">COOKER</a> crushes the food <a href="https://interviewquestions.tuteehub.com/tag/material-1089170" style="font-weight:bold;" target="_blank" title="Click to know more about MATERIAL">MATERIAL</a> </p>Answer :C</body></html> | |
| 49648. |
Apre-weighed vessel was filled with oxygen at NTP and weighed. It was thenevacuated, filled with SO_2 at the same temperature andpressure and again weighed. The weight of oxygen will be |
| Answer» <html><body><p>The same as that of `SO_2`<br/>1/2 that of `SO_2`<br/><a href="https://interviewquestions.tuteehub.com/tag/twice-714133" style="font-weight:bold;" target="_blank" title="Click to know more about TWICE">TWICE</a> that of `SO_2`<br/>1/4 that of `SO_2`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 49649. |
A porpous cup is filled with H_(2) gas at the atmospheric pressure and is connected to a thin glass tube a vertical position. The second end of the tube is immersed in water below it. After some time, water rises in the glass tube. Explain giving reasons. |
| Answer» <html><body><p></p>Solution :Pressure is <a href="https://interviewquestions.tuteehub.com/tag/exerted-7272976" style="font-weight:bold;" target="_blank" title="Click to know more about EXERTED">EXERTED</a> <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to bombardment of gaseous molecues with the <a href="https://interviewquestions.tuteehub.com/tag/wall-1448655" style="font-weight:bold;" target="_blank" title="Click to know more about WALL">WALL</a> of <a href="https://interviewquestions.tuteehub.com/tag/container-20566" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINER">CONTAINER</a>.</body></html> | |
| 49650. |
A porous tube containing a mixture of H_2 and O_2 is placed in a flask. After the diffusion for 25 seconds into the flask, what would be the composition of the gases in the flask? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`4:1`</body></html> | |