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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.
| 49551. |
A sample of an ideal gas with initial pressure 'P' and volume 'V' is taken through an isothermal process during which entropy change is found to be DS. The work done by the gas is |
| Answer» <html><body><p>`(PV Delta S)/(<a href="https://interviewquestions.tuteehub.com/tag/nr-581903" style="font-weight:bold;" target="_blank" title="Click to know more about NR">NR</a>)`<br/>`n R Delta S`<br/>PV<br/>`(P Delta S)/(<a href="https://interviewquestions.tuteehub.com/tag/nrv-571854" style="font-weight:bold;" target="_blank" title="Click to know more about NRV">NRV</a>)`</p>Solution :`Delta S = nR l n (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>)), W = nRT l n (V_(2))/(V_(1))` <br/> `rArr W = T. Delta S = Delta S xx (PV)/(nR)`</body></html> | |
| 49552. |
A sample of an ideal gas has a volume of 0.5 litres at 27^@C and 750mm pressure. The number of moles of gas are |
| Answer» <html><body><p>0.02<br/>0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <br/>2 <br/>0.002 </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> | |
| 49553. |
A sample of air contains Nitrogen, Oxygen and saturated with water vapour under a total pressure of 640 mm. If the vapour pressure of water at that temperature is 40 mm and the molecular ratio of N_(2): O_(2) is 3:1, the partial pressure of Nitrogen in the sample is |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>><a href="https://interviewquestions.tuteehub.com/tag/480-1879936" style="font-weight:bold;" target="_blank" title="Click to know more about 480">480</a> mm <br/><a href="https://interviewquestions.tuteehub.com/tag/600-329515" style="font-weight:bold;" target="_blank" title="Click to know more about 600">600</a> mm <br/>450 mm <br/>160 mm </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> :`D_(N_2) = (D_(N_2))/(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>) cdot P_("Total") implies P_(N_2) = 3/4 xx 600 = 450 mm`.</body></html> | |
| 49554. |
A sampleof air consisting of N_(2) and O_(2) was heated to 2500 K until the equilibrium N_(2) (g) + O_(2) (g) hArr 2 NO (g) was established the intial composition of air in mole fraction of N_(2) and O_(2). |
| Answer» <html><body><p></p>Solution :`{:(,N_(2)(g),+,O_(2),hArr,2NO,),("Intial ",a,,b,,0,),(,,,,,,a+b = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>"" ...(ii)),("At eqm",a-x,,b-x,,2x,):}` <br/> ` K_(c)= (2x)^(2) /((a-x)(b-x)) = (4x^(2))/((a-x)(b-x))` <br/> In the question , we are given `2x = 1*8 ("because total moles " =100)`<br/> or ` x= 0*9 and K_(c) = 2*1 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-<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/>` :. 2*1 xx 10^(-3) = (1*8)^(2)/((a-0*9 )(b-0*9)) ` <br/> ` ab - 0*9 a- 0*9 b + 0*81 = <a href="https://interviewquestions.tuteehub.com/tag/1620-1794901" style="font-weight:bold;" target="_blank" title="Click to know more about 1620">1620</a>`<br/> ` ab - 0*9 (a+b) + 0*81 = 1620 ` <br/> ` ab- 0*9 xx 100 + 0*81 = 1620 `<br/> ` or ab = 1909 * 19 = 1709` <br/> ` Now(a-b)^(2) = (a+b)^(2) - 4 ab = (100)^(2)- 4 xx 709 = 3164 ` <br/> or ` a-b = sqrt (3164)= 56*2 "" `...(ii) <br/> Solving (i) and (ii), `a= 78*1 "moles"` <br/> Moles fraction of` N_(2) = (78*1 )/100 = 0* 781 `<br/> Mole fraction of ` O_(2) = 1 - 0* 781 = 0* 219 `</body></html> | |
| 49555. |
A sample of AgCl was treated with 5.00 ml of 1.5 M Na_(2)CO_(3) solution to give Ag_(2)CO_(3). The remaining solution contained 0.0026 g of Cl^(-) per litre. Calculate the solubility product of AgCl (K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2)) |
| 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> :`1.5 M Na_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) ` gives `[CO_(3)^(2-)]=1.5M` <br/> `K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]` <br/> `:. [Ag^(+)]=<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M` <br/> `K_(sp) "for" AgCl=[Ag^(+)][<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)`</body></html> | |
| 49556. |
A sample of AgCl was treated with 5 mL of 1.5 M Na_2 CO_3 solution to give Ag_2 CO_3. The remaining solution contained 0.0026 g of chloride per litre. If K_(sp) of Ag_2 CO_3is 8.2 xx 10^(-12), what is K_(sp) of AgCl ? |
| 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> :`1.7 <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>^(-10)`</body></html> | |
| 49557. |
A sample of a hydrate of barium chloride weighing 61g was heated until all the water of hydration is removed. The dried sample weighted 52g. The formular of the hydrated salt is: (Atomic mass, Ba = 137 amu, Cl = 35.5 amu) |
| Answer» <html><body><p>`BaCl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>). 2H_(2)O`<br/>`BaCl_(2).4H_(2)O`<br/>`BaCl_(2).H_(2)O`<br/>`BaCl_(2).3H_(2)O`</p>Answer :A</body></html> | |
| 49558. |
A sample of a gas was heated from 30^(@)C to 60^(@)C at constant pressure. Which of the following statement(s) is true? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> is doubled<br/>Boyle's law will appply<br/>Volume of the ga will be doubled<br/>None of the above</p>Answer :d</body></html> | |
| 49559. |
A sample of a gas occupies 650.0 cm at 100^@C. At what temperature the gas will occupy a volume of 1050.0 cm^3if the pressure is kept constant ? |
| Answer» <html><body><p></p>Solution :In the present case, <br/> `V_1 = 650.0 cm^3 "" V_2 = 1050.0 cm^3` <br/>`T_1 = 100^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 100 + 273 = 373 K, "" T_2` = ? <br/> The <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> remains constant during expansion. Hence, <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to Charles. law <br/> `V_1/T_1 = V_2/T_2` <br/> or `T_2 = (V_2 T_1)/(V_1)=(1050.0 xx 373)/(650.0)= 602.5 K`<br/>Hence, the required <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> is `602.5 - 273 = 329.5^@C`.</body></html> | |
| 49560. |
A sample of a gas has a volume of 8.5dm^(3) at an unknown tempreture, When the sample is submerged in ice water at 0^(@)C, its volume gets reduce to 6.37dm^(3) .What is its intial tempreture? |
| Answer» <html><body><p></p>Solution :`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>)=8.5dm^(3)``V_(2)=6.37dm^(3)` <br/> `T_(1)=?``T_(2)=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>^(@)C=273K` <br/> `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `V_(1)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>((T_(2))/(V_(2)))=T_(1)` <br/>`T_(1)=8.5cancel(dm^(3))xx(<a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> K)/(6.37cancel(dm^(3)))``T_(1)=364.28K`</body></html> | |
| 49561. |
A sample of a gas is found to occupy a volume of 800 cm^(3) "at" 27^(@) celsius. Calculate the temperature at which it will occupy a volume of 400 cm^(3), provided the pressure is kept constant. |
| 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> :Here `v_(1)=<a href="https://interviewquestions.tuteehub.com/tag/800-338448" style="font-weight:bold;" target="_blank" title="Click to know more about 800">800</a> cm^(-2) "" v_(2)= 400 cm^(3)` <br/> `T_(1)=(27+273)K=300K "" T_(2)= ?` <br/> Applying Charle.s law, `(V_(1))/(T_(1))=(V_(2))/(T_(2))` <br/> `T_(2)=(V_(2)xxT_(1))/(V_(1))=(400cm^(3)xx300K)/(800 cm^(3))=150 K` <br/> `=150-273=-123^(@)C` <br/> `:. T_(2)=-123^(@)C`</body></html> | |
| 49562. |
A sample of .^(14)CO_(2) was mixed with ordinary.^(12)CO_(2) for stuydinga biologicaltracerexperiment. The10mL of ths mixture at STP possess the rate of 10^(4) disintegration per minute. How manymilli-curie of radioactive carbon is needed to prepare 60 litre of such a mixture? |
| 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> :`0.27 <a href="https://interviewquestions.tuteehub.com/tag/mci-547186" style="font-weight:bold;" target="_blank" title="Click to know more about MCI">MCI</a>`</body></html> | |
| 49563. |
A sample of ._(131)^(53) I is iodide ion was admistered to a patient in a carrier consisting of 0.10 mg of stable iiodide ion. After 4 day 67.7% of initial radioactivity was detected in the throidgland of the patient. What mass of stable iodide ion hadmigrated to thyroid glad? (t_(1//2) for iodide ion = 8 day) |
| Answer» <html><body><p></p>Solution :If `t = 4`day, `lambda = (0.693)/(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)`, then since `r_(0) prop N_(0)` and `r prop N` <br/> `:. (r_(0))/(r) = (N_(0))/(N)` <br/> `:. t = (2.303)/(lambda) log (r_(0))/(r)` <br/> `4 = (2.303xx8)/(0.693) log (1)/(0.693) log(r_(0))/(r)` <br/> `r = 0.707 r_(0)` <br/> Thus iodide <a href="https://interviewquestions.tuteehub.com/tag/ion-1051153" style="font-weight:bold;" target="_blank" title="Click to know more about ION">ION</a> left is `0.707` part of intially injected sample, <a href="https://interviewquestions.tuteehub.com/tag/however-1032379" style="font-weight:bold;" target="_blank" title="Click to know more about HOWEVER">HOWEVER</a> the <a href="https://interviewquestions.tuteehub.com/tag/rate-1177476" style="font-weight:bold;" target="_blank" title="Click to know more about RATE">RATE</a> decreases only `67.7%` or `0.688` in 4 days, thus <br/> If `0.707` is left then iodidie ion migrated to thryorid `=1` Thus `0.677`is left than iodide ion migrated to thyroid <br/> `= (1xx0677)/(0.707) = 0.958` or `95.8%` <br/> of the iodide ion is migrated to gland.</body></html> | |
| 49564. |
A sample of 12 M concentrated hydrochloric acid has a density 1.2g L ^(-1)Calculated the molality. |
| Answer» <html><body><p></p>Solution :Given: Molarity `= 12 MHCl` <br/> <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> of the solution `=1.2 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> L ^(-1)` <br/> In 12 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> solution, there are 12 moles of HCl in 1 litre of the solution. <br/> Molality `= ("No. of moles of solute")/("Mass of solvent (in kg)")` <br/> Caculate mass of water (solvent) <br/> Mass of 1 litre HCl solution = density `xx` <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> <br/> `=1.2 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> L ^(-1) xx 100 mL = 1200g` <br/> Mass of HCl = No of moles of HCl` xx` molar mass of HCl <br/> Mass of water= mass of HCl solution - mass of HCl <br/> Mass of water `=1200 - 438 =762 g` <br/> Molality `= (12)/(0.762) =15.75m`</body></html> | |
| 49565. |
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in figure. What will be the value of Delta H for the cycle as a whole ? |
| Answer» <html><body><p></p>Solution :The net <a href="https://interviewquestions.tuteehub.com/tag/enthalpy-15226" style="font-weight:bold;" target="_blank" title="Click to know more about ENTHALPY">ENTHALPY</a> <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a>, `<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` for a cyclic process is zero as enthalpy change is a state function, i.e., `Delta H` (<a href="https://interviewquestions.tuteehub.com/tag/cycle-942437" style="font-weight:bold;" target="_blank" title="Click to know more about CYCLE">CYCLE</a>) `= 0`</body></html> | |
| 49566. |
A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. |
| Answer» <html><body><p></p>Solution :Step 1. To determine the volume of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a>. <br/> Volume of acid taken = <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> mL of 0.5 M `H_(2)SO_(4) = 25 mL` of 1 M `H_(2)SO_(4)` <br/> Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1 M NaOH. <br/> Now 1 mole of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/neutralizes-7700100" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALIZES">NEUTRALIZES</a> 2 moles of NaOH (i.e. `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`) <br/> `:.` 30 mL of 1 M `NaOH -= 15 mL` of 1 M `H_(2)SO_(4)` <br/> `:.` Volume of acid used by ammonia = 25 - 15 = 10 mL <br/> Step 2. To determine percentage of nitrogen.<br/> Again 1 mole of `H_(2)SO_(4)` neutralizes 2 moles of `NH_(3) :. 10 mL` of 1 M `H_(2)SO_(4) -= 20 mL` of 1 M `NH_(3)` <br/> But 1000 mL of 1 M `NH_(3)` contain nitrogen = <a href="https://interviewquestions.tuteehub.com/tag/14g-273935" style="font-weight:bold;" target="_blank" title="Click to know more about 14G">14G</a> <br/> `:.` 20 mL of 1 M `NH_(3)` will contain nitrogen `= (14)/(1000) xx 20 g` <br/> But this much amount of nitrogen is present in 0.5 g of the organic compound. <br/> `:.` Percentage of nitrogen `= (14)/(1000) xx (20)/(0.5) xx 100 = 56.0`. <br/> Alternatively, % of N can be determined by applying the following equation,<br/> `% N = (1.4 xx "Molarity of the acid" xx "Basicity of the acid" xx "Vol. of the acid used")/("Mass of substance taken")` <br/> Substituting the values of all the items in the above equation, we have, `% N = (1.4 xx 1 xx 2 xx 10)/(0.5) = 56.0`</body></html> | |
| 49567. |
A sample of 0.50g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50ml of 0.5M H_(2)SO_(4). The residual acid required 60mL of 0.5M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound |
| 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> :Initially taken `H_(2)SO_(4)= 50mL 0.5M` <br/> used NaOH = 60 mL 0.5M <br/> `:.` unused `H_(2)SO_(4)= <a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a> mL 0.5M H_(2)SO_(4)` <br/> used `H_(2)SO_(4)= (50-30)= 20 mL 0.5 H_(2)SO_(4)` <br/> `=40 mL 0.5M NH_(3)` <br/> `1000 mL 1M NH_(3)= 17g NH_(3)=14g` nitrogen 40mL 0.5M `NH_(3)=` <br/> `=(40 mL <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.5M xx 14g)/(1000mL xx 1M)= 0.280` g nitrogen <br/> % N `= (0.280 xx 100)/(0.5)= <a href="https://interviewquestions.tuteehub.com/tag/56-325545" style="font-weight:bold;" target="_blank" title="Click to know more about 56">56</a>` <br/> `:.` % of nitrogen = 56%</body></html> | |
| 49568. |
A sample of 0.5g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5M H_(2)SO_(4) . The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ............. . |
| Answer» <html><body><p>0.14<br/>0.28<br/>0.42<br/>0.56</p>Answer :<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> | |
| 49569. |
A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. What would be the percentage composition of nitrogen in the compound? |
| Answer» <html><body><p>50<br/>60<br/>56<br/>44</p>Solution :Given: <br/> Mass of compound taken=0.50g <br/> <a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a>. of `H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)=50`mL <br/> Molarity of `H_(2)SO_(4)=0.5M` <br/> Vol. of NaOH required=60mL <br/> Molarity of NaOH requried=0.5M <br/> <a href="https://interviewquestions.tuteehub.com/tag/method-559432" style="font-weight:bold;" target="_blank" title="Click to know more about METHOD">METHOD</a> adopted Kjeldahl's method <br/> Formula used: % of N`=(1.4xxMxx2[<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>-(V_(1))/(2)])/(m)`. .(i) <br/> by substituting the vlaues in the formula, we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `%` of `N=(1.4xx0.5xx2(50-60//2))/(0.5)=56` <br/> `therefore` % of N in the given compound=56%</body></html> | |
| 49570. |
A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. Ammonia evolved was absorbed in 50 mL of 0.5 M H_2SO_4. The residual acid required 60 mL or 0.5 M NaOH solution for neutralisation. Find the percentage composition of nitrogen in the compound. |
| 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> :`50 mL of 0.5 M H_2SO_4 = 50 mL 1N H_2SO_4` <br/> `60 mL 0.5 M NaOH = 60 mL 0.5 N NaoH = 60 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.5 = 30 mL 1N NaOH` <br/> `30mL 1N NaOH solution = 30 mL 1N H_2SO_4` <br/> <a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a>. of <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> used up = 50-30 = 20 mL 1N `H_2SO_4` <br/> Percentage `N_2 = (14)/1000 xx (20)/0.5 xx 100 = 5.6%`</body></html> | |
| 49571. |
A sample of 0.5 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is |
| Answer» <html><body><p>14<br/>28<br/>42<br/>56</p>Solution :Volume of acid <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> = 50 mL of 0.5 `M H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)SO_(4)` <br/> Let the volume of the acid left unused = v mL of M/10 `H_(2)SO_(4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> molarity equation, <br/> `n_(a)M_(a)V_(a)` (acid) `= n_(b)M_(b)V_(b)` (base), we have, `2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.5 xx v = 1 xx 0.5 xx 80` or v = 40 mL <br/> `:.` Volume of the acid used = 50 -40 <br/> = 10 mL of 0.5 M `H_(2)SO_(4)` <br/> Now `%N = (1.4 xx n_(a)M_(a)V_(a))/("wt. of substance taken")` <br/> `= (1.4 xx 2 xx 0.5 xx 10)/(0.5) = 28`</body></html> | |
| 49572. |
A sample contains two radioactive nuclie x and y with half-lives2 hour and 1 hour respectively. The nucleus x-decays to y and y-decays into a stable nucelus z.At t = 0, the activates of the components in the samewere equal. Find the ratio of the number of the active , nuclei of y at t = 4 hoursto the numberat t = 0. |
| 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> :`0.25`</body></html> | |
| 49573. |
A sample containing 0.496 gm of (NH_(4))_(2) C_(2)O_(4) (MW = 124) and inert material was dissolved in water and made strongly alkaline with KOH which converts NH_(4)^(+) into NH_(3). The liberated NH_(3) was distilled into exactly 50ml of 0.05M H_(2)SO_(4). The excess H_(2)SO_(4) was back titrated with 10ml of 0.1MNaOH. The percentage of (NH_(4))_(2) C_(2)O_(4) with sample is |
| Answer» <html><body><p>`40%`<br/>`50%`<br/>`60%`<br/>`75%`</p>Solution :m. eqts of <a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a> `H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) = 10 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.1 = 1` <br/> m.eqts of `H_(2)SO_(4)` taken `= 50 xx 0.05 xx 2 = <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>` <br/> m.eqts of `H_(2)SO_(4)` reacted with <br/> `NH_(4) = 4=` m.eqts of `(NH_(4))_(2) C_(2)O_(4)`<br/> wt of `(NH_(4))_(2)C_(2)O_(4) = 4 xx 62 xx 10^(-3)` <br/> % purity `=(4 xx 0.062)/(0.496) xx 100 = 50%`</body></html> | |
| 49574. |
A sample containing 0.4775 of (NH_(4))_(2)C_(2)O_(4) and inert material was dissolved in water and made strongly alkaline with KOH which converted NH_(4)^(o+) to NH_(3) The liberated NH_(3) was distilled of H_(2)SO_(4) was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) % of (NH_(4))_(2)C_(2)O_(4)=124.10 And atomic weight of N=14.0078. |
| Answer» <html><body><p></p>Solution :`(NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))_(2)C_(2)O+(4)+2KOHtoK_(2)C_(2)O_(4)+2NH_(3)+2H_(2)O` <br/> Total `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/used-2318798" style="font-weight:bold;" target="_blank" title="Click to know more about USED">USED</a> `=50xx0.05035xx2NH_(2)SO_(4)` <br/> `=5.035m" Eq of "H_(2)SO_(4)` <br/> Excess of `H_(2)SO_(4)=11.3xx0.124M NaOH` <br/> `=1.372 m" Eq of " NaOH` <br/> `=1.372 m" Eq of "H_(2)SO_(4)` <br/> `H_(2)SO_(4)` used `=5.035-1.372` <br/> `=3.663 m" Eq of "H_(2)SO_(4)` <br/> `=3.663 <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a> NH_(3)` <br/> `=3.663 m" Eq of "(NH_(4))_(2)C_(2)O_(4)` <br/> Ew of `(NH_(4))_(2)C_(2)O_(4)=(124.1)/(2)=62.06g` <br/> Weight of `(NH_(4))_(2)C_(2)O_(4)=3.663xx10^(-3)xx62.05=0.2273g` <br/> `124.1 g of (NH_(4))_(2)C_(2)O_(4)` <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `14.0078 g of N`. <br/> `0.2273 g of (NH_(4))_(2)C_(2)O_(4)=(14.0078xx0.2273)/(124.1)=0.02565` g <br/> `% of N=(0.02565)/(0.4775)xx100=5.373%` <br/> `% of (NH_(4))_(2)C_(2)O_(4)=(0.2273)/(0.4775)xx100=5.373%`</body></html> | |
| 49575. |
A samll amount of solution containing Na^(24) radio nuclide with activity A = 2xx10^(3) dps was administeredinto blood of a patientin a hospital. Afer 5 hour a sample of the blooddrawn out form the patientshowed an activity of 16 dpm per cc t_(1//2) for Na^(24) = 15 hr. Find: (a) Volume of the blood in the patient. (b) Activity fo blood sample drawn after a further time fo 5 hr. |
| Answer» <html><body><p></p>Solution :Let `V mL` blood is present in patient. <br/> (a) `r_(0)` pf `Na^(24) = 2xx10^(3) <a href="https://interviewquestions.tuteehub.com/tag/dps-432950" style="font-weight:bold;" target="_blank" title="Click to know more about DPS">DPS</a> = 2xx10^(3) xx 60 dp m` <br/> `= 120xx10^(3) dp m` for `V mL` blood <br/> `r` of `Na^(24) = <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> dp m//mL` at `t = 5 <a href="https://interviewquestions.tuteehub.com/tag/hr-25457" style="font-weight:bold;" target="_blank" title="Click to know more about HR">HR</a>`<br/> `= 16xx V dp m//V mL`<br/> `:' (r_(0))/(r) = (N_(0))/(N)` <br/> `:. (N_(0))/(N) = (120xx10^(3))/(16V)` <br/> `:. t = (2.303)/(lambda) log_(10) ((N_(0)))/(N)` <br/> `5 = (2.303xx15)/(0.693) log_(10)((120xx10^(3)))/(16V)` <br/> `:. V = 5.95 xx10^(3) mL` <br/> (b) Activity of blood sample after`5 hr` more i.e., `t = 10 hrt` <br/> `t = (2.303)/(lambda) log_(10) (N_(0))/(N)` <br/> `10 = (2.303xx15)/(0.693) log_(10) (120xx10^(3))/(A)` <br/> `:. A = 75.6xx10^(3) dp m` per `5.95 xx10^(3)mL` <br/> `= (75.6xx10^(3))/(5.95xx10^(3)) = 12.71` dpm per `mL`. <br/> `= 0.2118` dpm per `mL`.</body></html> | |
| 49576. |
(A) : Salts of Mg does not impart any colour to the flame (R): Due to small size and high effective nuclear charge, 'Mg' requires a large amount of energy for excitation of electrons. |
| 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 correct 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/>R is False but A 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> | |
| 49577. |
A rrange s, p an d d subshells o f a shell in the in creasin g o rd er of effective n u clear charge (Z_("eff")) experienced by th e electron p resen t in them . |
| Answer» <html><body><p></p>Solution :s-orbital is spherical in shape, it shields the electrons from the nucleus m ore <a href="https://interviewquestions.tuteehub.com/tag/effectively-7267650" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECTIVELY">EFFECTIVELY</a> than p-orbital which in <a href="https://interviewquestions.tuteehub.com/tag/tu-1428586" style="font-weight:bold;" target="_blank" title="Click to know more about TU">TU</a> <a href="https://interviewquestions.tuteehub.com/tag/rn-614655" style="font-weight:bold;" target="_blank" title="Click to know more about RN">RN</a> shields m ore effectively than d-orbital. So, the effective nuclear charge `(Z_(eff))` experienced by electrons <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in them is `d lt p lt s`</body></html> | |
| 49578. |
A room of10 m xx 15m xx4m dimenstion having perfectly insulated walls, ceiling and floor has 60 students seated inside. The air inside the room is at 25^(@)C and 1 atm pressure. If each student loses 200 joules of heat in one second, calculate the rise in temperature noticedin 20 minutes ( Neglect loss of air to the outside as temperature is raised and C_(p) for air = ( 7)/( 2) R). |
| Answer» <html><body><p></p>Solution :Volume of room `= 10 m <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 15m xx4m = 600m^(3)= 600 xx 10^(3) L =6 xx 10^(5) L`<br/> Molesof air in the room at `25^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` at 1 atm pressure <br/> `n = ( PV)/( RT) = ( 1 xx 6 xx 10^(5))/( 0.0821 xx 298) = 2.45 xx 10^(4)` <br/>Heat producedin1 sec by each student `= 200 J` <br/> `:. `Heat produced in 1 sec by 60 students `= 200 xx 60 J = 12000J` <br/> Heat produced in <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> minutes `=12000 xx 20 xx60 J = <a href="https://interviewquestions.tuteehub.com/tag/32-144-273643" style="font-weight:bold;" target="_blank" title="Click to know more about 144">144</a> xx 10^(5)J` <br/> Change in enthalpy of air, `<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 = n C_(p) DeltaT` <br/>`:. 144 xx 10^(5) = ( 2.45 xx 10^(4)) = ( ( 7)/(2) xx 8.314)xx Delta T ` <br/>or ` Delta T = 20.2 K`</body></html> | |
| 49579. |
A rigid and insulated tank of 3m^(3) volume is divided into two compartments. One second compartment of volume of 2m^(3) contains an ideal gas at 0.8314 Mpa and 400K and while the second compartment of volume 1m^(3) contains the same gas at 8.314 MPa and 500K. If the partition between the two compartments isruptured. the final temperature of the gas is: |
| Answer» <html><body><p>420K<br/>450K<br/>480K<br/>none of these</p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> in the first compartment = `(PV)/(RT) = (8.314 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(5) xx 2)/(8.314 xx 400) = 500`. Number of moles in the second compartment `=(8.314 xx 10^(6) xx 1)/(8.314 xx 500) = 2000` <br/> <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> lost by the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> in second compartment = Heat gained by the gas in first compartment <br/> So `2000 xx (500-T) xx <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 500 xx (T - 400)xx C` <br/> On solving, T= 480K</body></html> | |
| 49580. |
A reversible reaction has DeltaG^(@)negative for forward reaction ? What will be sign of DeltaG^(@)for backworkreaction ? |
| 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/negative-570381" style="font-weight:bold;" target="_blank" title="Click to know more about NEGATIVE">NEGATIVE</a>.</body></html> | |
| 49581. |
A reversible reaction achieves equilibrium when the rates of forward and backward reactions equal. At equilibrium, the ratio of product of molar concentrations of prodcuts and the prodcut of molar concentration of reactants each raised to the powers equal to their stoichiometric coefficients, becomes constant. In case of gaseous reactios, the partial pressure of gases may be used in place of their molar concentrations. The equilibrium partial pressure of N_(2)O_(4)(g) and NO_(2)(g) are 4 and 2 atmm, respectively. Now, at constant temperature the pressure of system is increased to 60 atm. The new equilibrium partial pressure of N_(2)O_(4)(g) becomes. |
| Answer» <html><body><p>40atm<br/>46.6atm<br/>20atm<br/>33.4atm</p>Answer :<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> | |
| 49582. |
A reversible reaction achieves equilibrium when the rates of forward and backward reactions equal. At equilibrium, the ratio of product of molar concentrations of prodcuts and the prodcut of molar concentration of reactants each raised to the powers equal to their stoichiometric coefficients, becomes constant. In case of gaseous reactios, the partial pressure of gases may be used in place of their molar concentrations. If 1 mole of PCI_(5)(g) and 1 mole of CI_(2)(g) is taken in a 10L vessel, then the equilibrium concentration of PCI_(3)(g) will be : PCI_(3)(g)+CI_(2)(g)hArrPCI_(5)(g),""K_(c)=(20)/(3)M^(-1) |
| Answer» <html><body><p>`0.05`<br/>`0.04`<br/>`0.06`<br/>`0.025`</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> | |
| 49583. |
A reversible cyclic process involves6 steps. In step -1 and 3 system absorbs500 , 800 J of heat from a heat reservoir at temperature 250 K and 200 K respectively. Step 2,4,6 are adiabaticsuch that the temperature of one reservior changes to that of next. Total work done by the the system in whole cycle is 700 J. Find the temperature during step 5 if it exchanges heat from a reservoir at temperatureT_(5) |
| 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> :<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a></body></html> | |
| 49584. |
A reutral moleculeXF_(3)has zerodipole menent. The element X is is most likely |
| Answer» <html><body><p>chlorine <br/>boron <br/>nitrogen <br/><a href="https://interviewquestions.tuteehub.com/tag/bromine-904632" style="font-weight:bold;" target="_blank" title="Click to know more about BROMINE">BROMINE</a> </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> :` BF_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`is planar triangular</body></html> | |
| 49585. |
(A): Research must be carried in such a manner that there will not be any waste by product in the reactions. (R) : The reaction which gives no by product is an environment friendly reaction. |
| 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/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and (R) is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <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 true and (R) is not the correct explanation of (A)<br/> (A) is true but (R) is false <br/>(A) is false but (R) is true </p>Solution :Both (A) and (R) are true and (R) is the correct explanation.of (A)</body></html> | |
| 49586. |
A relation between vapour pressure and temperture is known as |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a> gas equation<br/>Boltzmam equation<br/>Clausious equation<br/>Clausius - <a href="https://interviewquestions.tuteehub.com/tag/clapeyron-420283" style="font-weight:bold;" target="_blank" title="Click to know more about CLAPEYRON">CLAPEYRON</a> equation<br/></p>Solution :`"<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>" (P_2)/(P_1) = (DeltaH_V)/(2.303R) (<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/(T_1) - 1/(T_2))`, clausius uapeyron equation.</body></html> | |
| 49587. |
A reflection from (111) planes of a cubic crystal was observed at a glancing angle of 11.2^@ when X-rays of wavelength 154 pm were used. What is the length of the side of the unit cell ? (sin 11.2^@=0.1944) |
| Answer» <html><body><p></p>Solution :Here, `lambda`=154 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a>, n=1 , `<a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>=11.2^@` <br/> Applying Bragg's equation `2 d <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> theta= n lambda` <br/> i.e., `d_111=(nlambda)/(2 sin theta)=(1xx154)/(2xx0.1944) ` pm =396 pm <br/> Further , the separation between the <a href="https://interviewquestions.tuteehub.com/tag/planes-1155607" style="font-weight:bold;" target="_blank" title="Click to know more about PLANES">PLANES</a> of a <a href="https://interviewquestions.tuteehub.com/tag/cubic-940181" style="font-weight:bold;" target="_blank" title="Click to know more about CUBIC">CUBIC</a> crystal is given by <br/> `d_"hkl"=a/(sqrt(h^2+k^2+l^2))` <br/> `therefore d_11=a/(sqrt(1^2+1^2+1^2))`=396 pm (calculated above ) <br/> or `a=396xxsqrt3`=686 pm</body></html> | |
| 49588. |
A reducing agent is a substance which can |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/accept-846678" style="font-weight:bold;" target="_blank" title="Click to know more about ACCEPT">ACCEPT</a> <a href="https://interviewquestions.tuteehub.com/tag/electrons-969138" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRONS">ELECTRONS</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/donate-432752" style="font-weight:bold;" target="_blank" title="Click to know more about DONATE">DONATE</a> electrons <br/>accept protons<br/>donate protons.</p>Answer :<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> | |
| 49589. |
A reducing agent is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sno-646494" style="font-weight:bold;" target="_blank" title="Click to know more about SNO">SNO</a><br/>`SnO_(2)`<br/>`SnCl_(2)`<br/>`SnCl_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`</p>Solution :`SnCl_(2)` is a poweful reducing <a href="https://interviewquestions.tuteehub.com/tag/agent-369049" style="font-weight:bold;" target="_blank" title="Click to know more about AGENT">AGENT</a></body></html> | |
| 49590. |
A redox reaction is always a // an "_____________". |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/proton-1170983" style="font-weight:bold;" target="_blank" title="Click to know more about PROTON">PROTON</a> <a href="https://interviewquestions.tuteehub.com/tag/transfer-1425652" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSFER">TRANSFER</a> reaction<br/>ion <a href="https://interviewquestions.tuteehub.com/tag/combination-922669" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINATION">COMBINATION</a> reaction<br/>reaction in solution<br/>electron transfer reaction</p>Answer :D</body></html> | |
| 49591. |
A redox reaction involves oxidation of reductant liberating electrons, which are then consumed by an oxidant. The sum of two half reactions give rise to net redox change. In half reaction charge and atoms are always conserved. In the reaction: As_(2) S_(3) + HNO_(3) rarrH_(2)AsO_(4) +H_(2)SO_4, +NOthe element oxidised is: |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/fe-460008" style="font-weight:bold;" target="_blank" title="Click to know more about FE">FE</a>^(+8//<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) to Fe^(+3) + 1/3 <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>`<br/>`Fe^(+8//3) to Fe^(+2) - 2/3 e `<br/>`(Fe^(+8//3))_3 to 3Fe^(+2)+ 2e`<br/>`2(Fe^(+8//3))_3 to 3 (Fe^(+3))_2 + 2e`</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> :Electrons and atoms are <a href="https://interviewquestions.tuteehub.com/tag/conserved-2539138" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVED">CONSERVED</a> in half reaction .</body></html> | |
| 49592. |
A red solide is insoluble in water. However it becomes soluble if some KI is added to water. On heating the red solid in a test tube, there is liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is : |
| Answer» <html><body><p>`HgO`<br/>`Pb_(3)O_(<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/>`(NH_(4))_(2) Cr_(2)O_(7)`<br/>`Hgl_(2)`</p>Solution :The red solide is `Hgl_(2)` <br/> `underset(("Red solide"))(Hgl_(2)) + 2 <a href="https://interviewquestions.tuteehub.com/tag/kl-528180" style="font-weight:bold;" target="_blank" title="Click to know more about KL">KL</a> rarr underset(("soluble"))(K_(2) [Hgl_(4)])` <br/> `Hgl_(2) overset("<a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a>")rarr underset(("Droplets"))(Hg(l)) + underset(("<a href="https://interviewquestions.tuteehub.com/tag/violet-1446649" style="font-weight:bold;" target="_blank" title="Click to know more about VIOLET">VIOLET</a> fumes"))(I_(2)(g))`</body></html> | |
| 49593. |
A real gas obeying van der Walls's equation will resemble ideal gas, if the: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/constants-20556" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANTS">CONSTANTS</a> a and <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> are small<br/>a is <a href="https://interviewquestions.tuteehub.com/tag/large-1066424" style="font-weight:bold;" target="_blank" title="Click to know more about LARGE">LARGE</a> and b is small<br/>a is <a href="https://interviewquestions.tuteehub.com/tag/smal-2256703" style="font-weight:bold;" target="_blank" title="Click to know more about SMAL">SMAL</a> and b is large<br/>constant a and b are large</p>Answer :a</body></html> | |
| 49594. |
A real gas most closely approaches the behaviour of an ideal gas at: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> and 200K<br/>1 atm and <a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> K<br/>0.5 atm and 500K<br/>15 atm and 500K</p>Answer :<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> | |
| 49595. |
A real gas most closely approaches the behaviour of an ideal gas at , |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> `&` low temperature<br/><a href="https://interviewquestions.tuteehub.com/tag/high-479925" style="font-weight:bold;" target="_blank" title="Click to know more about HIGH">HIGH</a> pressure `&` high temperature<br/>low pressure `&` high temperature<br/>high pressure `&` low temperature</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a></body></html> | |
| 49596. |
A real gas is subjected to an adibatic process causing in a change of state from (3 bar, 50 L, 500 K) to (5 bar, 40 L, 600 K) against a constant pressure of 4 bar. The magnitude of enthalpy change for the process is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4000-315574" style="font-weight:bold;" target="_blank" title="Click to know more about 4000">4000</a> J<br/>5000 J<br/>9000 J<br/>1000 J</p>Answer :<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> | |
| 49597. |
A real gas deviates most from ideal behaviour at |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>High <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> and Low pressure <br/>High pressure and Low temperature <br/>High pressure and High temperature <br/>Low pressure and Low temperature </p>Solution :At high P, low T, <a href="https://interviewquestions.tuteehub.com/tag/voluem-3265043" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUEM">VOLUEM</a> of less `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a>` more <a href="https://interviewquestions.tuteehub.com/tag/imf-498820" style="font-weight:bold;" target="_blank" title="Click to know more about IMF">IMF</a>.</body></html> | |
| 49598. |
A real gas can be liquefied: |
| Answer» <html><body><p>under adiabatic expansion<br/>above critical <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/>when <a href="https://interviewquestions.tuteehub.com/tag/cooled-7329693" style="font-weight:bold;" target="_blank" title="Click to know more about COOLED">COOLED</a> below critical temperature under applied <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a><br/>at temperature <a href="https://interviewquestions.tuteehub.com/tag/lower-1080637" style="font-weight:bold;" target="_blank" title="Click to know more about LOWER">LOWER</a> than critical temperature and pressure higher than critical pressure</p>Answer :A::C::D</body></html> | |
| 49599. |
A real gas acts as an ideal gas under which condition ? |
| Answer» <html><body><p><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>, <a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> <br/>Low temperature, high pressure <br/>High temperature, high pressure <br/>Low temperature, low pressure </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> | |
| 49600. |
A reactionwas nono-spontaneous at high temperature but became spontaneousat low temperature . The reaction is "…............" |
| 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/exothermic-2066005" style="font-weight:bold;" target="_blank" title="Click to know more about EXOTHERMIC">EXOTHERMIC</a></body></html> | |