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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.
| 51201. |
3.0 g sample of KOCl and CaOCl_(2) is dissolved in water to prepare 100 mL solution, which requried 100 " mL of " 0.15 M acidified K_(2)C_(2)O_(4). For the point. The clear solution is now treated with excess of AgNO_(3) solution which precipitates 2.87 g of AgCl. Calculate the mass percentage of KOCl and CaOCl_(2) in the mixture. |
| Answer» <html><body><p></p>Solution :(a). `2e^(-)+undersetunderset(x=+1)(x-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>=-1)(overset(+1)(ClO^(ɵ))toundersetx=-1)(Cl^(ɵ))(n=2)` <br/> (b). `2e^(-)+undersetnderset(2x=0)(2x-2=-2)(Cl_(2)O^(-2))tounderset(2x=-2)(Cl^(ɵ))(n=2)` <br/> (c). `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)(n=2)` <br/> Let a and b millimoles of KOCl and `CaCOCl_(2)` are present in the mixture. <br/> `m" Eq of "KOCl+m" Eq of "CaOCl_(2)=m" Eq of "K_(2)C_(2)O_(4)` <br/> `2a+2b=100xx0.15xx2` (n-factor) <br/> `therefore2a+2b=30`.(i) <br/> <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> millomoles of `Cl^(ɵ)` from `KOCl+` millimoles of `Cl^(ɵ)` from `CaOCl_(2)-=` millimoles of AgCl` <br/> `a(lCl^(ɵ)` ions) `+2b(2Cl^(ɵ)` ions)`=(2.87)/(143.5)xx10^(3)` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a> of AgCl=143.5]` <br/> `thereforea+2b=20`(ii) <br/> From <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (i) and (ii) `a=10,b=5` <br/> `% of KOCl=(10xx10^(-3)xx90.5xx100)/(3)` `[Mw of KOCl=90.5]` <br/> `=30.1%` <br/> `% of CaOCl_(2)=(5xx10^(-3)xx127)/(3)xx100``[Mw of CaOCl_(2)=127]` <br/> `=21.1%`</body></html> | |
| 51202. |
3.0 g of pyrolusite ore were treated with 20 g of pure ferrous ammonium sulphate (Mol.mass =392 " g mol"^(-1)) and dilute H_(2)SO_(4). After the reaction, the solution was diluted to 500 mL. 50 mL of diluted solution required 10 mL of 0.1 N K_(2)Cr_(2)O_(7) solution. Calculate the % of pure MnO_(2) in pyrolusite. |
| Answer» <html><body><p><br/></p>Solution :`MnO_(2)` present in pyrolusite oxidises ferrous ammonium <a href="https://interviewquestions.tuteehub.com/tag/sulphate-1234306" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHATE">SULPHATE</a> into ferric ammonium sulphate, i.e., `Fe^(2+) to Fe^(3+)`. <br/> Unreacted ferrous ammonium sulphate is estimated by potassium dichromate solution. <br/> `MnO_(2)+2Fe^(2+)+4H^(+) to Mn^(2+) +2H_(2)O+2Fe^(3+)` <br/> `Cr_(2)O_(7)^(2-)+6Fe^(2+) +14H^(+) to 2Cr^(3+)+7H_(2)O+6Fe^(3+)` <br/> 50 mL <a href="https://interviewquestions.tuteehub.com/tag/diluted-7675406" style="font-weight:bold;" target="_blank" title="Click to know more about DILUTED">DILUTED</a> ferrous ammonium sulphate solution <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> mL " of" 0.1 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> K_(2)Cr_(2)O_(7)` <br/> 500 mL diluted ferrous ammonium sulphate solution <br/> `=10xx10 mL " of" 0.1 N K_(2)Cr_(2)O_(7)` <br/> =100 mL of 0.1 N <a href="https://interviewquestions.tuteehub.com/tag/feas-3629986" style="font-weight:bold;" target="_blank" title="Click to know more about FEAS">FEAS</a> <br/> `=(0.1xx392)/(1000)xx100=3.92 g` <br/> Used FeAS=(20-3.92)=16.08 g <br/> `MnO_(2)` present in pyrolusite `=(87)/(392xx2)xx16.08=1.784 g` <br/> Percentage of pure `MnO_(2)=(1.784)/(3.0)xx100=59.4 %`</body></html> | |
| 51203. |
3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given K_(a) of acetic acid is 1.75xx10^(-5). |
| Answer» <html><body><p></p>Solution :Calculation of the concentration of acetic acid : <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a>/`<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)=Nxxg` eq. mass <br/> `"mass"//500cm^(3)=(Nxxg" eq. mass")/2rArrN_(CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/cooh-409857" style="font-weight:bold;" target="_blank" title="Click to know more about COOH">COOH</a>)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` <br/> Molecular wt. of `CH_(3)COOH=12xx2+1xx4+16xx2=24+04+32=60` <br/> `[CH_(3)COOH]=(3.0xx2)/60=6.0/60=0.1N` <br/> Calculation of concentration of sodium <a href="https://interviewquestions.tuteehub.com/tag/acetate-847375" style="font-weight:bold;" target="_blank" title="Click to know more about ACETATE">ACETATE</a>. `"mass"//500cm^(3)=(Nxxg" eq.mass")/2rArrN_(CH_(3)COONa)=("Mass"//500cm^(3)xx2)/("g.eq.mass")` <br/> Molecular wt. of `[CH_(3)COONa]=12xx2+1xx3+16xx2+1xx23=82` <br/> `[CH_(3)COONa]=(41xx2)/82=8.2/82=0.1N` <br/> `pH=pK_(a)+"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>"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `pH=-log_(10)(1.75xx10^(-5))+"log"([0.1])/([0.1])=-log1.75-log10^(-5)+log1` <br/> `pH=-0.2430+5+0=4.7570`</body></html> | |
| 51204. |
30 g of marble stone on heating produced 11g of CO_2. The percentage of CaCO_3 in marble is |
| Answer» <html><body><p>0.75<br/>0.8<br/>0.833<br/>0.866</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> | |
| 51205. |
3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of N/2 H_(2)SO_(4)mL of N/2 NaOH were required to neutralise excess of the acid. Calculate the percent purity of the given sample of ammonium chloride. |
| 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/82-339053" style="font-weight:bold;" target="_blank" title="Click to know more about 82">82</a> %`</body></html> | |
| 51206. |
3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of N//2 H_(2)SO_(4). 28 mL of N//2 NaOH were required to neutralise residual acid. Calculate the percentage of purity of the given sample of ammonium chloride. |
| 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> :`(120-28) mL N//<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> H_(2)SO_(4)-=92 mL N//2 NH_(4)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>`</body></html> | |
| 51207. |
30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) NaOH solutions are mixed and the resulting solution is |
| Answer» <html><body><p>2<br/>1<br/>3<br/>8</p>Solution :Total millimolar of `H^(+)=(30xx(1)/(3))+(20xx(1)/(2))` <br/> `=10+10=20` <br/> Total <a href="https://interviewquestions.tuteehub.com/tag/millimoles-1096825" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIMOLES">MILLIMOLES</a> of `OH^(-)=40xx(1)/(4)=10` <br/> `:. H^(+)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> left after neutralization = 10 <a href="https://interviewquestions.tuteehub.com/tag/millimole-2833492" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIMOLE">MILLIMOLE</a> <br/> Volume of solution `=1 dm^(3) = 1000cc` <br/> Hence, molarityof `H^(+)` ions `=(10)/(1000)M= 10^(-2)M` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =- log [H^(+)]=-log 10^(-2)=2`</body></html> | |
| 51208. |
3-Phenylpropene reacts with HBr in the presence of peroxide, the major product formed is |
| Answer» <html><body><p>2-bromo-1-phenylpropane<br/>1,2-dibromo-3-phenylpropane<br/>3-(o-bromophenyl)propane <br/>1-bromo-3-phenylpropane</p>Solution :`C_6H_5CH_2CH=CH_2+HBr <a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>"(Anti-Markovnikov <a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a>)"<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>"Peroxide effect"to underset"1-Bromo-3-phenylpropane"(C_6H_5CH_2CH_2CH_2Br)`</body></html> | |
| 51209. |
3-Phenylpropene on reaction with HBr gives (as a major product ) |
| Answer» <html><body><p>`C_6H_5CH_2CH(<a href="https://interviewquestions.tuteehub.com/tag/br-390993" style="font-weight:bold;" target="_blank" title="Click to know more about BR">BR</a>)CH_3`<br/>`C_6H_5CH(Br)CH_2CH_3`<br/>`C_6H_5CH_2CH_2CH_2Br`<br/>`C_6H_5CH(Br)CH=CH_2`</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> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E20_056_S01.png" width="80%"/></body></html> | |
| 51210. |
3 moles of water is boiled at 373 K and is changed to vapour state having the same temperature. What will be the change in entropy of the system ? [The molecules heat of vaporizations of water is 40.668 KJ/mol] |
| 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> :`DeltaS_("<a href="https://interviewquestions.tuteehub.com/tag/vap-1442817" style="font-weight:bold;" target="_blank" title="Click to know more about VAP">VAP</a>") = 0.327` KJ/Kelvin</body></html> | |
| 51211. |
3 moles of NH_3 are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of N_2 is 0.2 mole/lit . Then the total number of moles at equilibrium is |
| Answer» <html><body><p>2.5<br/>5<br/>1.5<br/>7.5</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> | |
| 51212. |
3 moles of ethanol reacts with one mole of phosphorous tribromide to form 3 moles of bromoethane and one mole of X. Which of the following is X? |
| Answer» <html><body><p>`H_(3)PO_(<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/>`H_(3)PO_(2)`<br/>`HPO_(3)`<br/>`H_(3)PO_(3)`</p>Solution :`3CH_(3)CH_(2)OH+PBr_(3) rarr 3CH_(3)CH_(2)Br+H_(3)PO_(3)(X)`</body></html> | |
| 51213. |
3 molesof anideal gas(gamma=5//3) is subjected tofollowing chargeof state.Identify thecorrectstatement:satete A(400 K,2 bar) underset("isothermalcooling") overset("Reversible")tostateB(T_(B),T_(B)) underset ("isothermal expansion") overset("Reversible")to State C (300 K, 1 "bar")underset("free expansion") overset("Asiabatic")to State D (T_(D), 0.5 "bar") |
| Answer» <html><body><p>`T_(B)` and `T_(D)` will <a href="https://interviewquestions.tuteehub.com/tag/besame-2466466" style="font-weight:bold;" target="_blank" title="Click to know more about BESAME">BESAME</a> and equal to 300 K<br/>`DeltaH_(<a href="https://interviewquestions.tuteehub.com/tag/ad-361679" style="font-weight:bold;" target="_blank" title="Click to know more about AD">AD</a>)=750 <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>`<br/>`DeltaS_(AC)=0`<br/>`DeltaE_(BD)=0`</p>Answer :a,d</body></html> | |
| 51214. |
3 moles of a mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2 M KMnO_(4) solution in acidic medium. Hence, mole fraction of FeSO_(4) in the mixture is |
| Answer» <html><body><p>`(1)/(3)`<br/>`(2)/(3)`<br/>`(2)/(5)`<br/>`(3)/(5)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> moles of `FeSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)=x` <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. `FeSO_(4)` = eq. `KMnO_(4)` <br/> `x xx1=0.1xx2xx5impliesx=1` <br/> `therefore` mole <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> =1/3</body></html> | |
| 51215. |
3 moles of a gas are present in a vessel at a temperature of 27^@C. What will be the value of R, the gas constant, in terms of the kinetic energy of the molecules of the gas ? |
| Answer» <html><body><p><br/></p>Solution :Kinetic energy of n moles of a gas is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by <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>.E. = n 3/2 RT` <br/> `:. " The kinetic energy of 3 moles of a gas at " 27^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> (300 K)` will be given by <br/> `K.E. = 3xx 3/2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> R xx 300` <br/>`:. "" R =(2 xx K.E.)/(3xx3xx300)= 7.41 xx 10&(-4) xx K.E`. <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, the value of R is `7.41 xx 10^(-4) " K.E. " K^(-1)`</body></html> | |
| 51216. |
3 moles of A and 4 moles of Bare mixed together and allowed to come into equilibrium according to the following reaction A(g) + 4 B(g) hArr 2 C (g) + 3 D (g) When equilirium is reached , there is 1 mole of C. The equilibriumextent of the reaction is |
| Answer» <html><body><p>`1//4`<br/>`1//3`<br/>`1//2`<br/>1</p>Solution :`{:(,A,+,4 B,hArr,2C,+,3D),("<a href="https://interviewquestions.tuteehub.com/tag/intial-2746226" style="font-weight:bold;" target="_blank" title="Click to know more about INTIAL">INTIAL</a> moles:" ,3,,4,,<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>,,0):}` <br/> 4 moles of B react with 1 mole of A. Hence , B is the limiting reactant . From 4 moles of B, if reaction were <a href="https://interviewquestions.tuteehub.com/tag/complete-423576" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETE">COMPLETE</a> , 2 moles of C would be formed . Actual C formed = 1 mole . Hence extent of reaction completed `=1/2` . <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> , when 1 mole of C is formed , out of 4 moles of B taken, 2 moles of B must have reacted . This also shows that the equilibrium extent is 1/2 .</body></html> | |
| 51217. |
3 mole of reactant A and one mole of reactant B are mixed in a vessel of volume 1 litre . The reaction taking place is A + B hArr 2 C . If 1.5 mol of C is formed at equilibrium , the value of K_(c) is |
| Answer» <html><body><p>`0.12`<br/>`0.50`<br/>`4.00`<br/>`0.25`</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> | |
| 51218. |
3-Methylpentane on monochlorination gives four possible products. The reaction follows free radical mechanism.The relative reactivities for replacement of -H are 3^(ul@):2^(ul@):1^(ul@) =6:4:1. Relative amounts of A,B,C and D formed are |
| Answer» <html><body><p>6/31,16/31,6/31,3/31<br/>16/31,6/31,6/31,3/31<br/>6/31,16/31,3/31,6/31<br/>6/31,3/31,6/31,16/31</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> | |
| 51219. |
3-methyl-pent-2-ene on reaction with HBr in the presence of peroxide forms an addition product. The number of possible stereoisomers for the product is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/four-464592" style="font-weight:bold;" target="_blank" title="Click to know more about FOUR">FOUR</a> <br/>six<br/>zero</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> | |
| 51220. |
3-Methyl-1-butene on oxymercuration demercuration yields............as the major product |
| Answer» <html><body><p>3-Methyl-2-butanol<br/>2-Methyl-2-butanol<br/>3-Methyl-1-butanol<br/>2-Methyl-1-butanol</p>Solution :`underset("2-Methyl-1-butene")(CH_(3)-underset(CH_(3))underset(|)(CH)-CH-CH_(2)) underset((ii) "<a href="https://interviewquestions.tuteehub.com/tag/demercuration-2570824" style="font-weight:bold;" target="_blank" title="Click to know more about DEMERCURATION">DEMERCURATION</a>")overset((i) "<a href="https://interviewquestions.tuteehub.com/tag/oxymercuration-1144684" style="font-weight:bold;" target="_blank" title="Click to know more about OXYMERCURATION">OXYMERCURATION</a>")<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> underset("3-Methyl-1-butanol")(CH_(3)-underset(CH_(3))underset(|)(CH)-underset(OH)underset(|)(CH)-CH_(3))` <br/> The addition of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> is in accordance with <a href="https://interviewquestions.tuteehub.com/tag/markownikoff-2170489" style="font-weight:bold;" target="_blank" title="Click to know more about MARKOWNIKOFF">MARKOWNIKOFF</a> rule.</body></html> | |
| 51221. |
3 litres each of Nitrogen and Hydrogen measured at STP are allowed to react together. Find the volumes of the gases after the reaction and also the weight of Ammonia formed in the reaction. |
| Answer» <html><body><p></p>Solution :Balanced equation is <br/> `N_(2(g))+3H_(2(g)) to 2NH_(3(g))` <br/> `"1 volt3 vol2 vol"` <br/> Ratio of <a href="https://interviewquestions.tuteehub.com/tag/volumes-1448111" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUMES">VOLUMES</a> of `N_(2), H_(2) and NH_(3)` is 1:3:2 <br/> 3 litres of `H_(2)` reacts with 1 litre of `N_(2)` and forms 2 litres of `NH_(3)`. Hence volumes of `N_(2) and H_(2)` after the <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> are 2 litres (3-1) and .0. litres <a href="https://interviewquestions.tuteehub.com/tag/respectively-1186938" style="font-weight:bold;" target="_blank" title="Click to know more about RESPECTIVELY">RESPECTIVELY</a>. <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of 22.4 litre of `NH_(3)` at STP =17g <br/> Weight of 2 lit of `NH_(3)` at STP `=(17 xx 2)/(22.4) =1.517gram`</body></html> | |
| 51222. |
3 g of an oxide of a metal Is converted to chloride completely and it yielded 5 g of chloride. The equivalent weight of the matel is |
| Answer» <html><body><p>33.25<br/>3.325<br/>12<br/>20<br/></p>Solution :a) `("Weight of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a>")/("Weight of metal chloride")` <br/> `= `("Eq.<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>. of metal" + "Eq.wt. of oxide")/("Eq.wt. of metal"+"Eq. wt. of chloride")` <br/> `3/5`=`(E+8)/(E+35.5)` <br/> Or <a href="https://interviewquestions.tuteehub.com/tag/5e-326283" style="font-weight:bold;" target="_blank" title="Click to know more about 5E">5E</a> = 40 = 3E + 106.5 or 2E = 66.5 <br/> `therefore` E=33.25</body></html> | |
| 51223. |
3 g of an oxide of a metal is converted to chloride completely and it yielded 5 g of chloride. Equivalent weight of the metal is : |
| Answer» <html><body><p>`33.25`<br/>`3.325`<br/>12<br/>20</p>Solution :N//A</body></html> | |
| 51224. |
3 g of activated chasrcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> mg <br/><a href="https://interviewquestions.tuteehub.com/tag/36-309156" style="font-weight:bold;" target="_blank" title="Click to know more about 36">36</a> mg <br/>42 mg <br/>54 mg </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> :Number of moles of acetic acid adsorbed `= (0.06xx(50)/(1000)-0.042xx(50)/(1000))`<br/>`= (0.9)/(1000)` moles. <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Weight of acetic acid adsorbed `= 0.9xx60 mg`<br/>= 54 mg<br/>Hence, the amount of acetic acid adsorbed per g of charcoal `= (54)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` mg = 18 mg<br/>Hence, option (A) is correct.</body></html> | |
| 51225. |
2xx10^(6) molecules of N_(2) gas enter into the vessel having volume 400 mL at 400 K temperature. Find the pressure of N_(2) gas. [R = 0.082 L atom mol^(-1)K^(-1)] [1 atom = 1.013 bar] |
| 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> :`27.233xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)` <a href="https://interviewquestions.tuteehub.com/tag/atom-887280" style="font-weight:bold;" target="_blank" title="Click to know more about ATOM">ATOM</a> ; `2.76xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)` <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a></body></html> | |
| 51226. |
2xx 10^(8)atoms of carbon arearranged side bysidecaculatethe radiusof carbonatom if thelengthof this arrangementis 2.4cm . |
| Answer» <html><body><p></p>Solution :Where`2xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)` carbonatom = 2.4 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a> <br/> 1carbonatom= (?)<a href="https://interviewquestions.tuteehub.com/tag/cam-18593" style="font-weight:bold;" target="_blank" title="Click to know more about CAM">CAM</a> <br/>diameter ofcarbon = `(1xx 2.4 cm)/(2xx 10^(8))= 1.2xx 10^(8) cm` <br/> radiusof carbon= `(1.2xx 10^(8) cm)/(2 ) ` <br/> `6.0 <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^(9) cm` <br/> `6.0 xx 10^(9) xx 10^(7) <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>` <br/> `-0.06 nm`</body></html> | |
| 51227. |
2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)), at 298 K of K_c is 7xx10^25 than calculate for SO_(3(g)) hArr SO_(2(g)) + 1/2O_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> :`1.1952xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)`</body></html> | |
| 51228. |
2SO_2(g) +O_2(g) leftrightarrow 2SO_3(g) IF the volume of reaction vessel is increased, what happens to Kc. Explain. |
| 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/kc-531574" style="font-weight:bold;" target="_blank" title="Click to know more about KC">KC</a> <a href="https://interviewquestions.tuteehub.com/tag/increases-1040626" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASES">INCREASES</a></body></html> | |
| 51229. |
2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)""DeltaHlt0 Which change(s) will increase the quantity of SO_(3)(g) at equilibrium? (P) Increasing the temperature. (Q) Reducing the volume of the container. (R) Adding He to increase the pressure keeping volume |
| 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>>P only<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> only<br/>P and Q only<br/>Q 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> only</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> | |
| 51230. |
2RMgCl + SiCl_(4) overset(H_(2)O)rarr B overset("polymerisation")rarr C. C is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/cyclic-2559324" style="font-weight:bold;" target="_blank" title="Click to know more about CYCLIC">CYCLIC</a> silicone<br/>linear silicone<br/>cross linked silicone<br/>`SiO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`</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::<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> | |
| 51231. |
2R - X + 2Na overset("dry ether") rarr R - R + 2Nax Which of the following alkanes is not obtained by Wurtz reaction ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/methane-1095077" style="font-weight:bold;" target="_blank" title="Click to know more about METHANE">METHANE</a><br/> <a href="https://interviewquestions.tuteehub.com/tag/ethane-975817" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANE">ETHANE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/propane-1169790" style="font-weight:bold;" target="_blank" title="Click to know more about PROPANE">PROPANE</a><br/> <a href="https://interviewquestions.tuteehub.com/tag/butane-906337" style="font-weight:bold;" target="_blank" title="Click to know more about BUTANE">BUTANE</a> </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> | |
| 51232. |
2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of NO_(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> :`2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-177" <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a>"` <br/> Decrease in pressure will <a href="https://interviewquestions.tuteehub.com/tag/favour-985586" style="font-weight:bold;" target="_blank" title="Click to know more about FAVOUR">FAVOUR</a> <a href="https://interviewquestions.tuteehub.com/tag/backward-391901" style="font-weight:bold;" target="_blank" title="Click to know more about BACKWARD">BACKWARD</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>, i.e. less `NO_(2)` will be formed.</body></html> | |
| 51233. |
2O_3 harr 3O_2 Ozone dissociates until the molecular weight of the equilibrium mixture becomes 40. Initial pressure of O_3 is 600 mm Hg. Then |
| Answer» <html><body><p>`K_p = 360` mm<br/>degree of <a href="https://interviewquestions.tuteehub.com/tag/dissociation-15685" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOCIATION">DISSOCIATION</a> = 0.4<br/>`P_(eq) = <a href="https://interviewquestions.tuteehub.com/tag/720-334419" style="font-weight:bold;" target="_blank" title="Click to know more about 720">720</a>` mm <br/>`K_p = p_(O)_3 = P_(O)_2`</p>Solution :`(D)/(d)=1+(n-1)alpha, (48)/(40)=1+((3)/(2)-1)alpha, alpha=40%` <br/> `2O_(3) harr 3O_(2)` <br/> at <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/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(=360)(600(1-alpha))""underset(360)((3alpha)/(2) xx 600)` <br/> `K_(P)=((360)^(3))/((360)^(2))=360`</body></html> | |
| 51234. |
2p, 3p,4p,5p arrangetheseorbitalsin increasingorder of energy ? |
| 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/2p-300467" style="font-weight:bold;" target="_blank" title="Click to know more about 2P">2P</a> to3p to4p to5p`energyincreases.</body></html> | |
| 51235. |
2NO(g)+O_(2)(g)hArr2NO_(2)(g) Which would increase the partial pressure of NO_(3)(g) at equlibrium? |
| Answer» <html><body><p>Decreasing the <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of the system<br/><a href="https://interviewquestions.tuteehub.com/tag/adding-2399902" style="font-weight:bold;" target="_blank" title="Click to know more about ADDING">ADDING</a> a mole <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> to <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> the pressure of the system<br/>Removing some NO(g) from the system <br/>Adding an appropriate catalyst</p>Answer :A</body></html> | |
| 51236. |
2NO_((g))+O_(2(g))hArr2NO_(2(g)),DeltaH=-117" kJ". Predict the effect of an increase in concentration of NO. |
| Answer» <html><body><p></p>Solution :`2NO_((g))+O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(g))hArr2NO_(2(g)),DeltaH=-177" <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a>"` <br/> If we <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> the concentration of NO, the rate of forward reaction will increase, i.e. more `NO_(2)` will be <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a>.</body></html> | |
| 51237. |
2NO_(2)(g)to N_(2)O_(4)(g) DeltaU_(f)^(@)[N_(2)O_(4)(g)]=2kcal//"moleand "DeltaU_(reaction)^(@)=-16kcal//mol, then calculate DeltaH_("formation" )^(@)=-16kcal//mol,then calculate DeltaH_("formation")^(@) of NO_(2)at 727^(@)C: |
| Answer» <html><body><p>9kcal/mol<br/>4.5kcal/mol<br/>8kcal/mol<br/>10kcal/mol</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> | |
| 51238. |
2NO(g)+O_(2)(g)hArr2NO_(2)(g) At a certain temperature the equilibrium concentration for this system are : [NO]=0.25M, [O_(2)]=0.24M,[NO_(2)]=0.18M. What is the value of K_(c) at this temperature? |
| Answer» <html><body><p>0.063<br/>`0.50`<br/>1.4<br/>`2.0`</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 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> | |
| 51239. |
2NO_(2(g)) (brown) hArr N_2O_(4(g)) , DeltaH=57.2 kJ mol^(-1) if the reaction vessel kept in ice and hot water than what is the change in colour ? |
| Answer» <html><body><p></p>Solution :This <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> is exothermic. So at lower temperature the forward reaction <a href="https://interviewquestions.tuteehub.com/tag/take-662846" style="font-weight:bold;" target="_blank" title="Click to know more about TAKE">TAKE</a> <a href="https://interviewquestions.tuteehub.com/tag/place-1155224" style="font-weight:bold;" target="_blank" title="Click to know more about PLACE">PLACE</a>. So in mixture `NO_2` decreases so brown colour. decreases. If the vessel kept in hot water then the reaction take place in <a href="https://interviewquestions.tuteehub.com/tag/reverse-1188140" style="font-weight:bold;" target="_blank" title="Click to know more about REVERSE">REVERSE</a> <a href="https://interviewquestions.tuteehub.com/tag/direction-1696" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTION">DIRECTION</a> and concentration of `NO_2` increases so brown colour increases.</body></html> | |
| 51240. |
2NH_(3)(g)hArrN_(2)(g)+3H_(2)(g) in a V lit container total x mol at eq. N_(2)H_(4)(g)hArrN_(2)+2H_(2)(g) in V lit (other) container total y mol at eq. If both are taken in same container (V lit) then at new equation total mols will be |
| Answer» <html><body><p>`x+y`<br/>`gtx+y`<br/>`ltx+y`<br/>No <a href="https://interviewquestions.tuteehub.com/tag/prediction-1163092" style="font-weight:bold;" target="_blank" title="Click to know more about PREDICTION">PREDICTION</a> is possible.</p>Solution :N//A</body></html> | |
| 51241. |
2NH_3(g) hArr N_2(g)+3H_2(g) If degree of dissociation of ammonia at equilibrium is 0.7 then observed molecular weight of reaction mixture at equilibrium : |
| Answer» <html><body><p>12<br/>10<br/>15<br/>17</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> | |
| 51242. |
2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI How many equivalents ofHypo is oxidised by one mole of Iodine? |
| 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> :`2Na_(2)overset(+2)(S_(2))O_(3)+overset(0)I_(2)rarrNa_(2)overset(+2.5)(S_(4))O_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)+2NaI^(-1)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> wt of `Na_(2)S_(2)O_(3)=("<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> wt of"Na_(2)S_(2)O_(3))/(1)` <br/> EQ wt of Iodine = `("Mol wt of iodine")/(2)` <br/> 1 mole iodine = 2 equivalents iodine <br/> = 2 equivalents hypo</body></html> | |
| 51243. |
2M(s)+H_(2)SO_(4)(aq)rarrM_(2)SO_(4)(aq)+H_(2)(g) Give therepresentation of the cell which involves the above redox reaction |
| 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> :`M (s) |M_(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>) (aq)||<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)(aq)|H_(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>)` pt</body></html> | |
| 51244. |
2 moles of pure KClO_(3) is decomposed to an extent of 66.6%. How many does of O_(2) is released? |
| 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> :`2KClO_(3)<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(Delta)rarr2KCl+3O_(2)<a href="https://interviewquestions.tuteehub.com/tag/uarr-3241817" style="font-weight:bold;" target="_blank" title="Click to know more about UARR">UARR</a>` <br/> `(2xx66.6)/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)-(2xx66.6)/(100)xx(3)/(2)=~~1.998=2`</body></html> | |
| 51245. |
2Mn_(2)O_7rarr4MnO_(2)+ 3O_2(If M is mol. wt. of Mn_(2)O_7). Find the equivalent, weight of Mn_2O_7in the above change. |
| Answer» <html><body><p><br/></p>Solution :`Mn_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)^(7+) +<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> e rarr 2Mn^(<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/> 2 `xx` <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `Mn_(2) O_(7) =4xx` mole of `MnO_2` <br/> `=4xx3 xx` eq. of `MnO_2 =<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> ` eq. of `MnO_2` <br/> `:.` Mole of `Mn_(2)O_(7) = 6 ` eq. of `MnO_2` <br/> `:. ` Eq. of `Mn_(2) O_(7) ` = Eq. of `MnO_7` <br/> `:.` Eq. wt. of `Mn_(2) O_(7) = M//6`</body></html> | |
| 51246. |
2KMnO_(4)+16HCl to2MnCl_(2)+2KCl+5Cl_(2)+8H_(2)O How many moles of HCl undergo oxidative in the above reaction ? |
| Answer» <html><body><p>14<br/>10<br/>5<br/>16</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_CHE_XI_P2_C08_E03_105_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> only 10 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of HCl are Oxidised to 5 moles of `Cl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` ____ (B) <br/> From 6HCl, `6Cl^(-)` are remaining as `6Cl^(-)` and the `6Cl^(-)` are not oxidised.</body></html> | |
| 51247. |
In the reaction, 2HgOrarr2Hg+O_(2),Hg^(2+) act as |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oxidising-2209005" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDISING">OXIDISING</a> <a href="https://interviewquestions.tuteehub.com/tag/agent-369049" style="font-weight:bold;" target="_blank" title="Click to know more about AGENT">AGENT</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/reducing-2981618" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCING">REDUCING</a> agent<br/>Oxidised<br/>none of the above</p>Answer :A</body></html> | |
| 51248. |
2H_(2)Orarr4e^(-)+O_(2)+4H^(+). The equivalent weight of molecules oxygen is |
| Answer» <html><body><p>32<br/>16<br/><a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a><br/><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 :n-f = no. 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> of `<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>^(-)` gained/mole <br/> `implies n-f" of "O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=4impliesE=(32)/(4)=8`</body></html> | |
| 51249. |
2H_(2)O(g) hArr 2H_(2)(g) + O_(2)(g) K_(C) = 4.1 xx 10^(-48) At 599 K N_(2)(g) +O_(2)(g) hArr hArr 2NO(g) K_(c)= 1 xx 10^(-30) at 1000 K Predict the extent of the above two reactions. |
| 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> :In the reactions , decomposition of water at 500 K and oxidation of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> at <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> K, the value, of `K_(c)` is very less `K_(c) lt 10^(-3)` . So reverse reaction is <a href="https://interviewquestions.tuteehub.com/tag/favoured-7681397" style="font-weight:bold;" target="_blank" title="Click to know more about FAVOURED">FAVOURED</a> <br/> `therefore` Products << <a href="https://interviewquestions.tuteehub.com/tag/reactants-1178112" style="font-weight:bold;" target="_blank" title="Click to know more about REACTANTS">REACTANTS</a></body></html> | |
| 51250. |
2H_(2)Orarr 4e^(-)+O_(2)+4H^(+).The equivalent weight of oxygen is |
| Answer» <html><body><p>32<br/>16<br/>8<br/>4</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> | |