Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51751. |
10 Its of an air sample with relative humidity 0.6 is compressed to 5 litres at same temperature. What is the partial pressure of the water vapour in the compressed air, if the pressure of saturated water vapour at that temperature is 2.4 K Pa |
|
Answer» 1.2 KPa `implies P_(H_2 O) = 0.6 xx 2.4 = 1.44 kPa` on making volume half, `P_(H_2 O)` should double, but R.H. `> 100% implies` saturation. `implies P_(H_2O) = 2.4 kPa` (LIMITING value) . |
|
| 51752. |
10 grams of steam at 100^(@)C is mixed with 50 gm of ice at 0^(@)C then final temperture is |
| Answer» SOLUTION :The kinetic energy of MOLECULES in STEAM will be GREATER than the kinetic energy of molecules in ICE. Hence internal energy of steam is greater than that of ice. | |
| 51753. |
10 grams of argon gas is compressed isothermally and reversibly at a temperature of 27^(@)C from 10L to 5L. Calculate the work done for this process in multiple of 10^(20 calories. |
|
Answer» `= (-10)/(40) xx 2 xx 300 xx "ln" (5)/(10) = 100` cal |
|
| 51754. |
10 grams of steam has higher internal energy than 10 grams of ice. Comment. |
| Answer» Solution :The kinetic energy of molecules in STEAM (GAS) will be GREATER than the kinetic energy of molecules in icc (SOLID). Hence internal energy of stcam is greater than that of ice. | |
| 51755. |
10 grams of a sample of potassium chlorate gave on complete decomposition 2.24 litre of oxygen at N.T.P. What is the percentage purity of the sample ? |
|
Answer» `underset(underset(=245g)(2xx(39+35.5+48)))(2KClO_(3)(s))overset("Heat")(RARR)2KCl(s)+underset(underset(=67.2L)(3xx22.4))(3O_(2)(G))` 67.2 L of `O_(2)` is evolved at N.T.P from `KClO_(3) 245 g` 2.24 L of `O_(2)` is evolved at N.T.P from `KClO_(3)=(245)/(67.5)xx2.24=8.167g` Percentage purity of sample `= ("Mass of pure "KClO_(3))/("Mass of the sample")xx100=(8.167)/(10.0)xx100=81.7%`. |
|
| 51756. |
10 grams of a mixture of CaCO_(3) and MgCO_(3) require equal weight of sulphuric acid for complete reaction. Calculate the percentage weight of MgCO(3) in the mixture. |
|
Answer» |
|
| 51757. |
10 grams of a hydrated sodium carbonate, Na_(2)CO_(3).xH_(2)O, on strong heating loses a weight of 6.29 grams. Report the value of x. |
|
Answer» SOLUTION :HYDRATED sodium carbonate LOSES water of HYDRATION on strong heating. `Na_(2)CO_(3).xH_(2)Orarr Na_(2)CO_(3)+xH_(2)O` 1 MOLE of `Na_(2)CO_(3)-=X` moles of `H_(2)O` 106 grams of `Na_(2)CO_(3)-=X xx18` grams of water 3.71 grams of `Na_(2)CO_(3)-=6.29` grams of water 3.71 grams of `Na_(2)CO_(3)` is related to 106 grams 6.29 grams of water is related to ? Weight of water in one formula unit `=(6.29)/(3.71)xx106` = 180 grams 18 grams of water = 1 mole 180 grams of water = ? Value of `X=(180)/(18)=10` |
|
| 51758. |
10 grams of 90% pure limestone is thermally decomposed. If the gas liberated is utilised for the conversion of aqueous sodium carbonate, how many moles of bicarbonate is formed? |
|
Answer» |
|
| 51759. |
1.0 gallon pure octane ( "density" 2.65 kg//"gallon") on combustion prodeces 11.53 kg CO,CO_(2) and H_(2)O.CO is formed partially due to combustion of octane which is responsible to decrease the efficiency of engine. If complete combustion of octane to CO_(2) and H_(2)O provide 100% efficiency to engine, calculate efficiency of engine in the above case. |
|
Answer» |
|
| 51760. |
1.0 g sample of the Rochelle salt [NaOOC-CHOH-CHOH-COOK] (NaKC_(4)H_(4)O_(6).4H_(2)O)(Mw=282), on ignition, is converted into NaKCO_(3)(Mw=122), which is titrated with 50 " mL of " 0.1 MH_(2)SO_(4). The excess of H_(2)SO_(4) requries 30 mL 0.2 M KOH. |
|
Answer» Solution :`NaKC_(4)H_(4)O_(6).4H_(2)Ooverset(Delta)toNaK.CO_(3)+3CO_(2)uarr+H_(2)Ouarr` Total m" Eq of "`H_(2)SO_(4)=50xx0.1xx2` (n factor) `=10.0` EXCESS m" Eq of "`H_(2)SO_(4)=m" Eq of "NAOH used` `=30xx0.2xx1` (n factor) `=6.0` m" Eq of "`H_(2)SO_(4)` used for `NaKCO_(3)` `=m" Eq of "H_(2)SO_(4)` (total)`-m" Eq of "H_(2)SO_(4)` used for NaOH `=10-6=4.0` `thereforem" Eq of "NaKCO_(3)=4.0` `implies` n-factor of `NaKCO_(3)=2`, during its neutralisation with `H_(2)SO_(4)(" Eq of "NaKCO_(3)=(122)/(2))` Weight of `NaKCO_(3)=4.0xx10^(-3)XX(122)/(2)=0.244g` 1 " MOL of "Rochelle salt`=1 " mol of "NaKCO_(3)` Therefore, weight of rochelle salt`=(282xx0.244)/(122)=0.364g` `%` purity of Rochelle salt`=(0.364xx100)/(1.0)=36.4%` |
|
| 51761. |
10 g sample of H_(2)O_(2) just decolorised 100 ml of 0.1 M KMnO_(4) in acidic medium % by mass of H_(2)O_(2) in the sample is |
|
Answer» `3.40` `(w)/(34)xx2=0.1xx0.1xx5impliesw=0.85gm` implies % purity = `(0.85)/(10)xx100=8.5%` |
|
| 51762. |
10 g of sample of mixture of CaCl_(2) and NaCl are treated to precipitate all the calcium as CaCO_(3). This CaCO_(3) is heated to convert all the Ca to CaO and the final mass of CaCl_(2) in the original mixture is |
|
Answer» 0.321 Let, weight of `CaCl_(2)` = XG `CaCl_(2) to CaCO_(3) to CaO` Moleof CaO = `(1.62)/(56) rArr x/111 = 1.62/56 x = 3.21g` `%` of `CaCl_(2)` = (3.21/10 x 100) = 32.1%` |
|
| 51763. |
10 g of sucrose are dissoled in 100 g of water. Find the mass percentage of sucrose in the solution. |
|
Answer» Solution :MASS of sucrose DISSOLVED = 10 G Mass of WATER (solvent) = 100 g `therefore` Total mass of the solution `=100 + 10 = 110 g` HENCE, mass percentage of sucrose `=("Mass of solute")/("Mass of solution")xx 100 = 10/110 xx 100 = 9.09` |
|
| 51764. |
1.0 g of Mg is burnt in a closed vessel which contains 0.5 g of O_2. (i) Which reactant is left in excess? (ii) Find the weight of excess reactant. (iii) How many millilitres of 0.5 N H_2 SO_4 will dissolve the residue in the vessel? |
|
Answer» Solution :`Mg+O_2 to MgO` Equivalent of `Mg = (1)/(12) = 0.0833""("eq. wt. of" Mg = ( "at. wt." )/( "VALENCY") = ( 24)/(2) )` Equivalent of oxygen `(0.5)/( 8) = 0.0625 "(eq. wt. of O = 8)"` (i) Mg is in excess because its eq. is GREATER than that of oxygen. (ii) Equivalent of Mg in excess `=0.0833 - 0.0625` `= 0.0208` `therefore` weight of Mg in excess `=` equivalent `XX` equivalent weight `= 0.0208 xx 12 = 0.25 g` (iii) The residue contains MgO and Mg which has not taken part in the reaction. Suppose `upsilon` mL of `H_2 SO_4` is required to DISSOLVE the residue. `therefore` m.e. of `H_2 SO_4` = m.e. of Mg + m.e. of MgO (not reacted) `therefore 0.5 upsilon` = (eq. of Mg + eq. of MgO) ` = 1000` (eq. of Mg `+` eq. of O) `=1000 (0.0208 + 0.0625)` `= 83.3` `therefore upsilon= 166.6` mL. |
|
| 51765. |
1.0 g of metal carbonate neutralises 200 mL of 0.1 N HCl. The equivalent mass of the metal will be : |
|
Answer» 50 |
|
| 51766. |
1.0 g of magnesium is burnt with 0.56 g of O_(2) in a closed vessel. Which reactant is left in excess and how much ? |
|
Answer» Mg , 0.44 g 32 g of `O_(2)` react with Mg = 48 g 0.56 g of `O_(2)` react with `Mg=((48G))/((32g))xx(0.56g)` Mass of mg TAKEN `= 1.0 g` Mass of Mg LEFT `= (1- 0.84) = 0.16 g`. |
|
| 51767. |
1.0 g of magnesium is burnt with 0.56 g O_(2) in a closed vessel. Which reaction is left in excess and how much ? ( At. wt. Mg = 24, O = 16) |
|
Answer» `MG, 0.16 g ` `(1.0)/(24) (0.56)/(32)` Oxygen is LIMITING reagent so `(0.5)/(12) - x(0.07)/(4) - x/2` `(0.07)/(4) - x/2 = 0` `x=(0.07)/(2)` excess `Mg=(0.5)/(12) - (0.07)/(2)` mole `rArr 1-0.7xx12= 0.16g` |
|
| 51768. |
1.0 g of magnesium is burnt with 0.56 g O_2 in a closed vessel. Which reactant is left in excess and how much? (At.wt: Mg=24, O = 16) |
|
Answer» MG, 0.16 G |
|
| 51769. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be |
|
Answer» 3 mole `2H_(2) + O_(2) to 2H_(2)O` 4 g of hydrogen react with 32 g of oxygen. So 10 g of hydrogen will react with 80G of oxygen. But we have GIVEN amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water. |
|
| 51770. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in the reaction will be : |
|
Answer» 3 mol 10 gof `H_(2)=((10G))/(("2 g mol"^(-1)))=5` mol 64 g of `O_(2)=((64g))/(("32 g mol"^(-1)))=2` mol In this reaction, `O_(2)` is LIMITING reactant 0.5 mol of OXYGEN from water = 1 mol 2 mol of oxygen from water `= (("1 mol"))/(("0.5 mol"))xx("2 mol")=4` mol. |
|
| 51771. |
10 g of hydrofluoric acid gas occupies 5.6 lit of volume at STP . If the empirical formula of the gas is HF , then its molecular formula in the gaseous state will be |
|
Answer» HF |
|
| 51772. |
1 g of Mg is burnt in a closed vessel containing 0.5 g of O_2. Which reactant is limiting reagent and how much of the excess reactant will be left? |
|
Answer» Solution :Carbon burns in oxygen according to the following EQUATION. `underset(12.01 G)C + underset(32.0 g)(O_(2)) to underset(12.01 + 32.0 = 44.01 g)(CO_(2))` `therefore` 12.01 g of carbon require for combustion, = 32.0 g `therefore 1.0` g of carbon will require `=(32.0)/(12.01) xx 1.0 = 2.7 g` Hence, the oxygen present in the closed vessel is much LESS than the required amount. The amount of `O_(2)`will thus be completely consumed and carbon will be left in excess. Thus, `O_2` is the LIMITING reagent in this case. `therefore 32.0` g of oxygen burn carbon = 12.01 g 1.5 g of oxygen will burn carbon `=12.01/32.0 xx 1.5 = 0.56 g` Hence, carbon left unreacted = `1.0 - 0.56 = 0.44` g Now, `therefore32.0` g of oxygen combine with carbon to GIVE `CO_2 = 44.01` g `therefore` 1.5 g of oxygen will give `CO_(2) = (44.01)/(32.0) xx 1.5 = 2.1 g` Hence, carbon is left in excess in the given reaction. The amount of excess reactant is 0.44 g and the mass of the product `CO_2` formed in the reaction is 2.1 g |
|
| 51773. |
10 g of argonis compressd isothermally and reversibly at a temperature of27^(@)C from10L to 5L . Calculate q,w,DeltaE and DeltaHfor this process. R =2.0 cal K^(-1)mol^(-1) , log= 2 = 0.30. Atomic wt. of Ar=40 |
|
Answer» Solution :`q= 2.303 nRT log. (V_(2))/(V_(1)) = 2.303 xx (10)/( 40) xx 300 xx log. (5)/(10)= -103.635 cal` For ISOTHERMAL expansion,`DeltaE= 0``w= DeltaE-q= 0-( - 103.635)= +103.635 cal` Also, when temperature is constant, `P_(1)V_(1) = P_(2)V_(2)` or `PV =` constant `DeltaH = DeltaE+Delta(PV) = 0+0=0` |
|
| 51774. |
10 grams of CaCO_(3) is completely decomposed to x and CaO. 'x' is passed into an aqueous solution containing 0.1mole of sodium carbonate. What is the number of moles of sodium bicarbonate formed? (mol. wts: CaCO_(3) = 100, NaCO_(3) = 106, NaHCO_(3) = 84) |
|
Answer» `0.2` `underset(0.1)(CO_(2))+Na_(2)CO_(3)+H_(2)Orarrunderset(0.2)(2NaHCO_(3))` |
|
| 51775. |
100 g of a pure sample of CaCO_3 is treated a with 500 cm^(3) of M/2HCI solution. Calculate the of CO_2 that will be evolved at S.T.P. |
|
Answer» Solution :The corresponding chemical EQUATION is: `underset(40.08 +12.01 + 48.0=100.1)(CaCO_(3)) + underset(2 xx 36.46= 72.92)(2HCl) to CaCl_(2) + H_(2)O + underset("22.4 L at S.T.P")(CO_(2))` The mass of HCl present in `50.0 cm^(3)` of `M/2` of HCl `=(1/2 xx 36.46 xx 50.0)/1000 = 0.911 g` From the equation it is clear that 1 g of `CaCO_3` will require 0.73 g of HCI. Therefore, the mass of HCI used is more than the required mass. This implies that `CaCO_3` is the limiting reagent. `therefore 100.1 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` `therefore 1.0 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` `therefore 1.0 g` of `CaCO_(3)` will give a volume of `CO_(2)` `=22.4/100.1 xx 1.0 = 0.22 L` Hence,the volumeof carbondioxide evolved =0.22 L. |
|
| 51776. |
1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^(3) of CO_(2) at N.T.P Calculate the percentage composition of the mixture. |
|
Answer» `:.` The mass of `MgCO_(3)` in the mixture `= (1-x)g` Step I. Volume of `CO_(2)` evolved from `CaCO_(3)` `underset(40+12+48+100g)(CaCO_(3)(s))overset("Heat")(rarr)CAO(s)+underset(22400"cc")(CO_(2)(g))` 100 g of `CaCO_(3)` evolve `CO_(2)` upon heating = 22400 cc xg of `CaCO_(3)` evolve `CO_(2)` upon upon heating `= (22400)/(100) xx x c c = (224x)c c`. Step II. Volume of `CO_(2)` evolved from `MgCO_(3)` `underset(24+12=48=84g)(MgCO_(3)(s))overset("Heat")(rarr)MgO(s)+underset("22400cc")(CO_(2)(g))` 84 g of `MgCO_(3)` evolve `CO_(2)` upon heating = 22400 cc `(1-x)`g of `MgCO_(3)` evolve `CO_(2)` upon heating `= 22400 xx ((1-x))/(84) c c. = 266.6 xx (1-x)c c` Step III. Percentage composition of the mixture According to available data : `224 x + 266.6 (1-x)=240` `224 x + 266.6 - 266.6 x = 240` `- 42.6 x= -26.6 or x = (26.6)/(42.6)=0.625` `:.` Percentage of `CaCO_(3)` in the mixture `= 0.625 xx 100 = 62.5 %` Percentage of `MgCO_(3)` in the mixture `= 100 - 62.5 = 37.5 %`. |
|
| 51777. |
1.0 g NaHSO_3 and Na_2SO_3 was dissolved in water to prepare a 200 mL solution. Two separate experiments were carried out. (a). 25 " mL of " sample was mixed with 25 " mL of " I_2 solution and excess of I_2 left after the reaction with NaHSO_3 and Na_2SO_3 was back titrated with 0.1002 N Na_2S_2O_3, 1.34 " mL of " which was required (25 " mL of " I_2 solution is equivalent to 24.20" mL of " Na_2S_2O_3 solution). (b). 50 " mL of " sample was oxidised to Na_2SO_4 by the action of H_2O_2,H_2SO_4 sormed (from NaHSO_3) was titrated with 22.3 " mL of " 0.1 N NaOH. Find percentage of NaHSO_3 and Na_2SO_3 in the original sample. |
|
Answer» Solution :1 G mixture is present in 200 mL (a). 25 " mL of " mixture] `impliesmEq` of `NaHSO_3+mEq` of `Na_2SO_3=mEq` of ` I_2` Used `mEq` of ` I_2` used`=24.20-1.34xx(0.1002)/(25)mL` `=22.905` m" Eq of "`I_2` in 200 mL`=(200)/(25)xx22.905=18.32` (b). 50 mL is oxidised to `Na_2SO_4` `H_2SO_4` formed by `NaHSO_3-=22.3mL` of `0.1` N `NaOH` `mEq` of `NaHSO_3=22.3xx0.1=(2.23)/(25)mL` `mEq` of ` NaHSO_2 i n 200mL=2.23xx(200)/(50)=8.92` `[{:("Ew of "NaHSO_(3)" in experiment"," "),((i)=(Mw)/(2)," "),(underset(x=4)underset(1+x-6=-1)(HSO_(3)^(ɵ)),tounderset(x=6)underset(x-8=-2)(SO_(4)^(2-)(x=2))):}]` `[{:("Ew of "NaHSO_(3) " in experiment"," "),((ii)=(Mw)/(1)," "),(HSO_(3)^(ɵ),toH^(o+)+SO_(3)^(2-)):}]` M" Eq of "`Na_(2)SO_(3)` in the original solution `=18.32-2xx8.92=0.48` Weight of `Na_2SO_3=8.92mEqxx(1)/(1000)xxEw of "NaHSO_(3)` `["Ew of "NaHSO_(3)=(Mw)/(1)=104]` `=(8.92)/(1000)xx104=0.92` `therefore%` of `NaHSO_(3)=(0.927)/(1)xx100=92.7%` [EQUIVALENT weight of `Na_(2)SO_(3)=(Mw)/(2)=(126)/(2)=63]` Weight of `Na_(2)SO_(3)=0.48mEqxx(1)/(1000)xx63=0.3g` `%` of `Na_(2)SO_(3)=(0.3)/(1)xx100=3%` `NaHSO_(3)=92.7%""Na_(2)SO_(3)=3%` |
|
| 51778. |
1.0 g carbonate of a metal was dissolved in 50 mL N//20 HCl solution. The resulting liquid required 25 mL of N//5 NaOH solution to neutralise it completely. Calculate the equivalent mass of the metal carbonate. |
|
Answer» |
|
| 51779. |
10 cm^(3) of 0.1 N monobasic acid requires 15 cm^(3) of sodium hydroxide solution whose normality is : |
|
Answer» Solution :`N_(1)V_(1)=N_(2)V_(2)` `0.1xx10=N_(2)xx15` `N_(2)=0.066` |
|
| 51780. |
10^(-6) M NaOH solution is diluted 100 times. Calculate the pH of the diluted base. |
|
Answer» |
|
| 51781. |
10^(-6)MNaOH is diluted 100 times. The pH of the diluted base is |
|
Answer» between 5 and 6 `:. OH^(-)` IONS from `H_(2)O` cannot be neglected. Total `[OH^(-)]=10^(-8)+10^(-7)=10^(-8)xx11` `:. pOH=-log(11xx10^(-8))=8-1.04=6.96` `:. pH = 14 - 6.96 = 7.04` |
|
| 51783. |
10^(-6) can be written as PREFIX : |
|
Answer» micro |
|
| 51784. |
10^(-4) g of gelatin is required to be added to 100 cm^(3) of a standard gold sol to just prevent its coagulation by the addition of 1 cm^(3) of 10% NaCl solution to it. Hence, the gold number of gelatin is |
|
Answer» 10 Here the amount of GELATIN required to protect 10 mL (or 10 `cm^(3)`) of gold sol `=(10^(-4)xx10)/(100)=10^(-5) g` `=10^(-5)xx10^(-3) mg=10^(-2)` mg =0.01 mg `therefore` Gold number =0.01 |
|
| 51785. |
10^(-3) gm equivalents of K_(2) Cr_(2) O_(7)in 50% H_(2) SO_(4) is needed to oxidise all the organic matter present in 1 lit of water. Then COD of water is |
|
Answer» 1 ppm |
|
| 51786. |
1. When 22.4 litres of H_2(g) is mixed with 11.2 litres of Cl_2 (g) each at 273 K at 1 atm the moles of HCl (g), formed is equal to....... |
|
Answer» 2 MOLES of HCL (G) Solutionn:` H_2 (g)+ Cl_2(g)to 2HCI (g)` 
|
|
| 51787. |
(1) What are auto redox reactions? Give example. (ii) Define orbital. What are the n and I values of 3p_x and 4d_(x^2-y^2) electron? |
Answer» Solution :Auto redox REACTIONS :Auto redor reactions. It is a rede reaction in which substance act as both the OXIDIZING agent and the REDUCING agent. (Reaction occur in the same substance)  (ii)1.Orbital is a three dimensional space which the probability of finding the electron is MAXIMUM 2. For `3p_x` eletron `""`n value =3 `"" "L value" =1 ` 3. For `4d_(x^2-y^2)` electron`""`n vlaue =4 `""` l value =2 |
|
| 51788. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Based on the given structure of the some of the molecules have been matched. Which is the incorrect matching ? |
|
Answer» `PCI_(5)^(-)` trigonal bipyramidal |
|
| 51789. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Square planar shape is predicted for: |
|
Answer» `ICI_(4)^(-), CIO_(3)^(-)` |
|
| 51790. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Pair of species with same shape and same state of hybridisation of the central atom is : |
|
Answer» `PCI_(5) , ICI_(4)^(-)` |
|
| 51791. |
1) The -=C-H unit of an alkyne is more acidic than a C – H unit of an alkene or alkane, allowingacetylene and terminal alkynes to be converted to their conjugate bases -= C. 2) Unlike alkenes, alkynes are reduced by metals, especially Li, Na and K. 3) Unlike alkenes, alkynes can undergo nucleophilic as well as electrophilic addition. Which of the following best describes what happens in the first step in the mechanism of the reactionshown? CH_3 CH_2CH_2CHBr_2 + 3NaNH_2 overset(NH_3)to CH_3CH_2C -=CNa +2NaBr +3NH_3 |
|
Answer»
|
|
| 51792. |
1) The -=C-H unit of an alkyne is more acidic than a C – H unit of an alkene or alkane, allowingacetylene and terminal alkynes to be converted to their conjugate bases -= C. 2) Unlike alkenes, alkynes are reduced by metals, especially Li, Na and K. 3) Unlike alkenes, alkynes can undergo nucleophilic as well as electrophilic addition. Which of the following best describes what happens in the first step in the mechanism of the hydrogen- deuterium exchange reaction shown? CH_3 -= CD underset(NH_3)overset(NaNH_2)to CH_3C -= CH |
|
Answer»
|
|
| 51793. |
1-Propanol and 2-propanol can be best distinguished by |
|
Answer» Oxidation with `KMnO_(4)` follwed by reaction with Fehling solution (Give positive TEST with Fehling solution) `2-"Propanol" underset(-H_(2))overset(Cu//573K)rarr "Propanone"` Oxidation with `KMnO_(4)` or `K_(2)Cr_(2)O_(7)/H^(+)` will give propanoic acid in case of 1-propanol which can not be tested with Fehlingsolution. Reaction with conc. `H_(2)SO_(4)` will LEAD to dehydration and as such no oxidation is possible |
|
| 51794. |
1-Propanol and 2-Propanol can be best distinguished by : |
|
Answer» OXIDATION with alkaline `KMnO_(4)` followed by reaction with Fehling solution |
|
| 51795. |
1-Phenylethanol can be prepared by reaction of benzaldehyde with |
|
Answer» METHYL iodide and Magnesium |
|
| 51796. |
1) N_2+2O_2 |
|
Answer» `NO_2` |
|
| 51797. |
1 mole SO_(2)=.......... |
|
Answer» `6.4g SO_(2)` |
|
| 51799. |
1 mole of water vapour is condensed to liquid at 25^(@)C. Now this water contains i) 3 moles of atoms ii) 1 mole of hydrogen molecules iii) 10 moles of electrons iv) 16 g of oxygen The correct combination is |
|
Answer» (i) & (ii) are CORRECT |
|
| 51800. |
1 mole of photons, each of frequency 250 s^(-1) would have approximately a total energy of _____ erg. |
|
Answer» |
|
