94420 + Interview Questions in Chemistry in Class 12 Page 1791 InterviewSolution

89501.	A gaseous mixture contains 56 g of N_2 ,44 g CO_2and 16 g of CH_4The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_4 is
Answer» 180MM 360MM 540 mm 720mm ANSWER :A

Discussion

89502.	A gaseous mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture?
Answer» `19.97%` `20.05%` `50%` `75%` Solution :`("Vol. of He")/("Vol. of" CH_(4)) = (50)/(50) = (1)/(1)` `("No. of molecules of He")/("No. of molecules of" CH_(4))=(1)/(1)` `("MASS of He")/("Mass of" CH_(4))=(4)/(16)=(1)/(4)` `:.` Percentage of `CH_(4)` in mixture `(4)/(5)xx100=80%`

Discussion

89503.	A gaseous mixture contains 40% H_(2) and 60% He, by volume. What is the average molecular mass of mixture ?
Answer» Solution :`M_("av")=(SUM(%"by VOL."XX"MOLECULES"))/(100)=(40xx2+60xx4)/(100)=3.20`

Discussion

89504.	A gaseous mixture contains 220 g of carbon dioxide and 280 g of nitrogen gas. If the partial pressure of nitrogen gas in the mixture is 1.5 atm then the partial pressure of carbon dioxide gas in the mixture will be
Answer» 1.25 atm 0.75 atm 0.50 atm 3 atm Answer :D

Discussion

89505.	A gaseous mixture contain four gases A, B, C and D. The mole fraction of "B" is 0.5. The mole fraction of "A" is
Answer» `0.525` `0.375` `0.625` `0.732` ANSWER :B

Discussion

89506.	A gaseous mixture contain 50% He and 50% CH_4by volume. What is the percent by weight of CH_4in the mixture ?
Answer» 0.1997 0.2005 0.5 0.75 Solution :`("VOL. of He")/("vol. of " CH_4) = 50/50 = 1/1 ` ` ("no. of MOLECULES of He")/("no. of molecules of " CH_4) = 1/1` `(" mass of He")/("mass of " CH_4) = 4/16 = 1/4` % of `CH_4` in mixture `= (4 xx 100)/(5) = 80.03%`

Discussion

89507.	A gaseous hypothetical chemical equations ,is carried out in a closed vessel . The concentration of B is found to increase by 5 xx 10^(-3) mol l^(-1) in 10 second . The rate of appearance of B is
Answer» `5 xx 10^(-4) mol L^(-1) sec^(-1)` `5 xx 10^(-5) mol l^(-1) sec^(-1)` `6 xx 10^(-5) mol l^(-1) sec^(-1)` `4 xx 10^(-4) mol l^(-1) sec^(-1)` Solution :Increase in CONCENTRATION of `B = 5 xx 10^(-3)` mol`l^(-1)` Time = 50 sec Rate of apperance ofB = `("Increase of conc. B")/("Time TAKEN")` `= (5xx 10^(-3) mol l^(-1))/(10"sec") = 5 xx 10^(-4) mol l^(-1) sec^(-1)`.

Discussion

89508.	A gaseoushydrocarbon'X'on reactionwithbromineinlight formsa mixtureof twomonobromo alkanesand HBr.The hydrocarbon'X'is :
Answer» `C_(2) H_(6) ` `C_(3) H_(6)` `C_(3)H_(8)` `C_(4)H_(10)` Answer :C

Discussion

89509.	A gaseous hydrocarbon was exploded with excess of oxygen. On cooling a contraction of 1.5 times the volume of the hydrocarbon was observed. When treated with NaOH a further contraction of double the volume of the hydrocarbon was noted. The molecular formula of the hydrocarbon is:
Answer» `C_2 H_4` `C_3 H_6` `C_2 H_2` `C_5 H_12` ANSWER :C

Discussion

89510.	A gaseous hydrocarbon requires 6 times its own volume of O_(2) for complete oxidation and produces 4 times its volume of CO_(2). What is its formula ?
Answer» Solution :The balanced EQUATION for combustion `C_(x) H_(y)+(x+(y)/(4))O_(2) rarr xCO_(2) +(y)/(2) H_(2)O` 1 VOLUME `(x+(y)/(4))` volume `therefore x+(y)/(4)=6` (by equation) or 4x+y=24 Again x=4 since evolved `CO_(2)` is 4 times that of hydrocarbon `therefore 16 +y=24" or "y=8 therefore` formula of hydrocarbon `C_(4)H_(8)`

Discussion

89511.	A gaseous hydrocarbon has 85%C and its vapour density is 28. The possible formula of the compound is
Answer» `C_(3)H_(6)` `C_(2)H_(2)` `C_(3)H_(8)` `C_(4)H_(8)` SOLUTION :`%C=85`, `%=H=15` `{:("Element","Percentage","Mole","Mole ratio"),(C,85,85//12=7.08,1),(H,15,15//1=15,2):}` Empirical formula `=CH_(2)` Empirical formula mass `=12+2=14` Molecular mass `=28xx2=56` `n=(56)/(14)=4` Molecular formula `=(CH_(2))_(4)=C_(4)H_(8)`

Discussion

89512.	A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is :
Answer» `C_(2)H_(4)` `C_(3)H_(4)` `C_(6)H_(5)` `C_(7)H_(8)` SOLUTION :`C_(x)H_(y) + (x+(y)/(4))O_(2) RARR xCO_(2) + (y)/(2) H_(2)O` Moles of `H_(2)O = (0.72)/(18) = 0.04` Moles of `CO_(2)= (3.08)/(44) = 0.07` `-""x : (y)/(2) = 0.07 : 0.04` or `x : y = 0.07 : 0.08 or 7 : 8` `therefore""C : H = 7 : 8` Therefore, the empirical formula is `C_(7)H_(8)`.

Discussion

89513.	A gaseous hydrocarbon given upon combustion, 0.72g water and 3.08g of CO_(2). The empirical formula of the hydrocarbon is:
Answer» `C_(6)H_(5)` `C_(7)H_(8)` `C_(2)H_(4)` `C_(3)H_(4)` SOLUTION :Number of moles of `H_(2)O=(0.72)/(18)=0.04` Number of moles of `CO_(2)=(3.08)/(44)=0.07` `n_(H)=2n_(H_(2))=2xx0.04=0.08` `n_(C)=n_(CO_(2))=0.07` `C:H=0.07:0.08=7:8` `THEREFORE` Empirical =`C_(7)H_(8)`.

Discussion

89514.	A gaseous hydrocarbon gives upon combustion, 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is
Answer» `C_(2)H_(4)` `C_(3)H_(4)` `C_(6)H_(5)` `C_(7)H_(8)` Solution :`18g H_(2)O` CONTAINS `2g H` `:. 0.72 g H_(2)O` contains `0.08 g H` `44 g CO_(2)` contains `12 g C` `:. 3.08 g CO_(2)` contains `0.84 g C` `:. C : H = (0.84)/(12): (0.08)/(1) = 0.07:00.08 = 7 : 8` `:. ` Empirical formula `= C_(7)H_(8)`.

Discussion

89515.	A gaseous hydrocarbon contaisn 85.7% carbon and 14.3% hydrogen 1 litre of the hydrocarbon weights 1.26g at NTP. Determine the molecular formula of the hydrocarbon.
Answer» ANSWER :`C_(2)H_(4)`

Discussion

89516.	A gaseous compound which contains only C, H and S is burnt with oxygen under such conditions that individual volumes of the reactant and product can be measured at the same temperature and pressure. It is found that 3 volumes of the compound react with oxygen to yield 3 volumes of CO_(2), 3 volumes of SO_(2) and 6 volumes of water vapour. What volume of oxygen is required for teh combustion? What is the formula of the compound ? Is this an empirical formula or molecular formula ?
Answer» Solution :`{:("COMPOUND",+,O_(2),rarr,CO_(2),+,SO_(2),+,H_(2)O),(3 "vol",,v"vol(say)",,3 "vol",,3"vol",,6"vol"),(or " 3 moles",,v"moles",,3"moles",,3"moles",,6"moles"):}` Applying POAC for O atoms, moles of O in `O_(2)` =moles of O in `CO_(2) +` moles of O in `SO_(2)`+ moles of O in `H_(2)O 2xx` moles of `O_(2)` `=2 xx` moles of `CO_(2)+2 xx` moles of `SO_(2) +1 xx` moles of `H_(2)O` `2v=2 xx 3+2 xx3 +1 xx 6= 18, v= 9` `therefore` volume of `O_(2)` required for combustion is 9. Again, moles of C in `CO_(2)=1 xx` moles of `CO_(2)=3` moles of S in `SO_(2)=1XX` moles of `SO_(2)=3` moles of H in `H_(2)O=2 xx` moles of `H_(2)O=12` `therefore` formula is `C_(3)S_(3)H_(12) or CSH_(4)` Now we see 3 moles of the compound contain 3 moles of C atoms in `C_(3)S_(3)H_(12)`. Hence i mole of the compound should contain 1 mole of C atoms and therefore, the molecular formula is `CSH_(4)`.

Discussion

89517.	A gaseous compound of carbon and nitrogen containing 53.8% by weight of nitrogen was found to have a vapour density of 25.8. What is the molecular formula of the compound.
Answer» ANSWER :`C_(2)N_(2)`

Discussion

89518.	A gaseous compound A reacts by three independent first order process (as shown in figure) with rate constant 2xx10^(-3),3xx10^(-3) and 1.93xx10^(-3) sec^(-1) for products B, C and D respectively, If initially pure A was taken in a closed container with p=8 atm then the partial pressure of B (in atm) after 100 sec from starting the experiment.
Answer» `0.288` `0.577` `1.154` NONE of these Answer :C

Discussion

89519.	A gaseous compound is composed of 85.7% by weight of C and 14.3% by weight of H. Its density is 2.28 g/L at 300 K and 1 atm pressure. Calculate the molecular formula of the compound.
Answer» SOLUTION :`C_(4)H_(8)`

Discussion

89520.	A gaseous carbon compound is soluble in dilute HCl. The solution on treating with NaNO_(2) gives off nitrogen leaving behind a solution which smells of wood spirit. The carbon compound is
Answer» A)`HCHO` B) `CO` C)`C_(6)H_(5)NH_(2)` D) `CH_(3)NH_(2)` Solution :The carbon compound is `CH_(3)NH_(2)`. As, `CH_(3)NH_(2) underset(NaNO_(2))overset(HCL)to CH_(3)OH+N_(2)+H_(2)O`.

Discussion

89521.	A gaseous alkane (C_(n)H_(2n+2)) is exploded with oxygen. The volume of O_(2) and CO_(2) formed are in the ratio of 7:4, Deduce the value of n.
Answer» Solution :`C_(N)H_(2n+2)+[n+(n+1)/(2)]O_(2)rarrnCO_(2)+(n+1)H_(2)O` `(n+((n+1))/(2))/(n)=(7)/(4)` or n=2

Discussion

89522.	A gaseous A_(2)(g) rarr B(g) +(1)/(2) C(g), shows increase in pressure from 100 mm to 120 mm in 5 min. The rate of disappearance of A_(2) is
Answer» 4 mm `"MIN"^(-1)` 8 mm `"min"^(-1)` 16 mm `"min"^(-1)` 2 mm `"min"^(-1)` Solution :`{:(,A_(2)(g)rarr,B(g)+,(1)/(2)C(g)),("Initially",100mm,0,0),("After 5min",(100-p)min,p,(1)/(2)p):}` Total pressure = `100 - p + p +(1)/(2) p = 100 +(1)/(2) p` `therefore` Increase in pressure =` (1)/(2) p = 120 - 100 = 20 mm` `therefore` Decrease in pressure of `A_(2) = 2 xx 20 = 40 mm` `therefore` Rate of disappearance of `A_(2)=(DX)/(dt)=(40)/(5)=8 mm " min"^(-1)`

Discussion

89523.	A gas X turns line water milky. The milkiness disappears if excess of 'X' is passed. Milkiness reappears on heating the colourless solution. The gas is
Answer» `CO_(2)` `NO_(2)` `NH_(3)` `SO_(2)` SOLUTION :`Ca(OH)_(2)+CO_(2) to CaCO_(3)underset("Excess") overset(CO_(2))to Ca(HCO_(3))_(2)` `Ca(OH)_(2)+SO_(2) tounderset("Milky white")(CaSO_(3)) underset("Excess "H_(2)O) overset(SO_(2))to Ca(HSO_(3))_(2)`

Discussion

89524.	A gas X is passedthroughwater to forms a saturated solution .The aqaeoussolutionon tretment aqeoussolutionalsodissolve magnassium ribbon with the evolation of a colourless gas Y identify X and Y
Answer» `X = CO_(2),Y = Cl_(2)` `X = Cl_(2),Y = CO_(2)` `X = Cl_(2),Y = H_(2)` `X = H_(2),Y = Cl_(2)` Solution :The first gas is CHLORIDE and the secondgas ishydrogen `CI_(2) + H_(2)O rarr HCI +HOCI` `HCI + AgNO_(3) rarr underset(WHITE)(AgCI darr ) + HNO_(3)` `Mg + 2HCI rarr MgCI_(2) + H_(2) uarr `

Discussion

89525.	A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y'. Identify 'X' and 'Y':
Answer» `X = CO_(2), Y = CI_(2)` `X = Cl,_(2), Y = CO_(2)` `X = Cl_(2), Y = H_(2)` `X = H_(2), Y = CI_(2)` Answer :C

Discussion

89526.	A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white ppt. The saturated aqueous solution also dissolve magnesium ribbon with evolution of a colourless gas 'Y'. Identify 'X' and 'Y'.
Answer» `X=CO_(2), Y=Cl_(2)` `X=Cl_(2), Y=CO_(2)` `X=Cl_(2), Y=H_(2)` `X=H_(2), Y=Cl_(2)` ANSWER :C

Discussion

89527.	A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatement with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y' Identify 'X' and 'Y'
Answer» `X=CO_(2),Y=Cl_(2)` `X=Cl_(2),Y=CO_(2)` `X=Cl_(2),Y=H_(2)` `X=H_(2),Y=Cl_(2)` SOLUTION :Since the SATURATED aqueous solution of (X) give white ppt with `AgNO_(3)`,so (X) may be `Cl_(2)`. Hence `underset((X))(Cl_(2))+H_(2)OtoHOCl+HCl,` `HCl+AgNO_(3) to underset(White)(AgCl)darr+HNO_(3)` `2HCl+Mgto MgCl_(2)+underset((Y))(H_(2))UARR`

Discussion

89528.	A gas 'X' diffuses five times as rapidly as another gas 'Y'. Calculate the ratio of molecular weights of 'X' and 'Y'.
Answer» ANSWER :`1//25`

Discussion

89529.	A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of colourless gas Y , X and Y are :
Answer» `X=CO_2 , Y=Cl_2` `X=Cl_2, Y=CO_2` `X=Cl_2, Y=H_2` `X=H_2, Y=Cl_2 ` Solution :`UNDERSET((X))(Cl_2) + H_2O to HCL + HOCL` `AgNO_3 + HCl tounderset ("WHITE ppt")(AGCL) + HNO_3 ` `Mg+2HCl to Mg Cl_2 + underset((Y))(H_2)uarr`

Discussion

89530.	A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y^(-) and 1 M Z^(-) at 25^(@)C. If the order of reduction potentials is Z gt Y gt X, then
Answer» Y will OXIDISE X and not Z Y will oxidise Z and not X Y will oxidise both Z and X. Y will oxidise both X and Z. Solution :(a) is the CORRECT ANSWER.

Discussion

89531.	A gas X at 1 atm is bubbles through a solution containing a mixture of 1M Y^(-) and MZ^(-) at 25^(@)C If the reduction potential of ZgtYgtX, then,
Answer» Y will oxidize X and not Z Y will oxidize Z and not X Y will oxidize both X and Z Y will reduce both X and Z Solution :The GIVEN order of REDUCTION potential is `Z gt Y gt X`. A spontaneous reaction will have the following characteristics Z REDUCED and Y oxidized, Z reduced and X oxidised, Y reduced and X oxidized. Hence, Y will oxidize X and not Z.

Discussion

89532.	A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y^(-) and 1MZ^(-) ions at 25^(@)C. If the reduction potential of ZgtYgtX, then,
Answer» Y will OXIDIZE but not Z. Y will oxidize both X and Z Y will oxizide Z but not X Y will REDUCE BOTHX and Z. Solution :REDUCTION potentials in the ORDER ZgtYgtX means that Z can be reduced most easily and X least easiliy, i.e., their oxidizing powers are: ZgtYgtX. Thus, Y will oxidize X but not Z. lt

Discussion

89533.	A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y and 1MZ ions at 25^(@)C if the reduction potential of ZgtYgtX, then
Answer» Y will oxidise X but not Z Y will oxidise both X and Z Y will oxidise Z but not X Y will reduce both X and Z. Solution :Reduction potentials in the order `ZgtYgtX` means that Z can be reducedmost easily and X LEAST easily i.e., their oxidising powers are `ZgtYgtX ` Hence Y will oxidise X but not Z.

Discussion

89534.	A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y^(-) and 1 M Z^(-) at 25^(@)C. If the reduction potential of ZgtYgtX, then
Answer» Y will OXIDISE X and not Z Y will oxidise Z and not X Y will oxidise both X and Z Y will REDUCE both X and Z Answer :A

Discussion

89535.	A gas X at 1 atm is bubbled through a solution containg a mixture of 1 M Y^- and 1M Z^- at 25^@C. If the reduction potential of Z gt Y gt X, then :
Answer» Y will oxidise X and not Z Y will oxidise Z and not X Y will oxidise both X and Z Y will REDUCE both X and Z Answer :A

Discussion

89536.	A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^(-) and 1MZ^(-) at 25^@C . If the reduction potential of Z > Y > X, then
Answer» Y will oxidized X and not Z Y will OXIDIZE Z and not X Y will oxidize X and Z Y will REDUCE both X and Z ANSWER :A

Discussion

89537.	A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^(-)"and " 1MZ^(-) at 25^(@)C. If the reduction potential of Z gt Y gt X, then
Answer» Y will OXIDIZE X and not Z Y will oxidize Z and not X Y will oxidize both X and Z Y will REDUCE both X and Z Solution :Y will oxidize X and not Z

Discussion

89538.	A gas X at 1 atm is bubblcd through a soluution containing a mixture of 1 M Y ^(-) and 1 M Z ^(-) at 25^(@)C. If the rcduction potential tial of Z gt Y gt X, then,
Answer» Y will oxidize X and not Z Y will oxidize Z and not X Y will oxidize both X and Z Y will reduce both X and Z Solution :The given ORDER of reduction POTENTIALS (or tendencies) IS `ZGT Y gt X. A` spontacous reaction will have the followig characteristices Z reduced andY oxidised Z reduced and X oxidised Y reduced and X oxidised Hence, Y will oxidise X and not Z.

Discussion

89539.	A gas will approach ideal behaviour at :
Answer» LOW T and HIGH P Low T and low P High T and low P High T and high P Answer :C

Discussion

89540.	A gas will approach ideal behaviour at
Answer» LOW TEMPERATURE and low pressure low temperature and HIGH pressure high temperature and low pressure high temperature and high pressure Answer :C

Discussion

89541.	A gas which is used as anaesthetic in dental surgery is
Answer» `N_(2)` CO `N_(2)O` `NH_(3)` ANSWER :3

Discussion

89542.	A gas which burns with blue flame is :
Answer» `O_(2)` `CO_(2)` `N_(2)` `CO`. ANSWER :D

Discussion

89543.	A gas when passed through the K_(2)Cr_(2)O_(7) and dil. H_(2)SO_(4) solution, turns it green, the gas is
Answer» `H_(2)S` `NH_(3)` `Cl_(2)` `SO_(2)` ANSWER :B::D

Discussion

89544.	A gas which burns with a blue flame is :
Answer» CO `O_2` `N_2` `CO_2` ANSWER :A

Discussion

89545.	A gas when passed through K_(2)Cr_(2)O_(7) and dil. H_(2)SO_(4) solution turns it green, the gas is
Answer» `CO_(2)` `NH_(3)` `SO_(2)` `Cl_(2)` ANSWER :B::C

Discussion

89546.	A gas turns lime water milky and acidified K_(2)Cr_(2)O_(7) solution green then gas is :
Answer» `HCL` `H_(2)S` `SO_(2)` `CO_(2)` ANSWER :C

Discussion

89547.	A gas that cannot be collected over water is :
Answer» `N_(2)` `SO_(2)` `O_(2)` `PH_(3)`. Answer :B

Discussion

89548.	A gas reacts with CaO but not with NaHCO_3. The gas is
Answer» `CO_2` `Cl_2` `N_2` `O_2` ANSWER :B

Discussion

89549.	A gas reacts with CaOand not with NaHCO_3 is:
Answer» `CO_2` `Cl_2` `O_2` `N_2` ANSWER :A

Discussion

89550.	A gas phase reaction has energy of activation 200 "kJ mol"^(-1).If the frequency factor of the reaction is 1.6xx10^(13) s^(-1). Calculate the rate constant at 600 K . (e^(-40.09)=3.8xx10^(-18)) .
Answer» SOLUTION :`k=Ae^((-(E_a)/(RT)))` `k=1.6xx10^(13)s^(-1)_e^(-((200xx103"J mol"^(-1))/(8.314"JK "^(-1)mol^(-1)xx600K)))` `k = 1.6xx10^(13)s^(-1)e^(-(40.1))` `k=1.6xx10^(13)s^(-1)xx3.8xx10^(-18)` `k=6.21xx10^(-5)s^(-1)`

Discussion

Explore topic-wise InterviewSolutions in .

A gaseous mixture contains 56 g of N_2 ,44 g CO_2and 16 g of CH_4The total pressure of the mixture is 720 mm Hg. The partial pressure of CH_4 is

A gaseous mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture?

A gaseous mixture contains 40% H_(2) and 60% He, by volume. What is the average molecular mass of mixture ?

A gaseous mixture contains 220 g of carbon dioxide and 280 g of nitrogen gas. If the partial pressure of nitrogen gas in the mixture is 1.5 atm then the partial pressure of carbon dioxide gas in the mixture will be

A gaseous mixture contain four gases A, B, C and D. The mole fraction of "B" is 0.5. The mole fraction of "A" is

A gaseous mixture contain 50% He and 50% CH_4by volume. What is the percent by weight of CH_4in the mixture ?

A gaseous hypothetical chemical equations ,is carried out in a closed vessel . The concentration of B is found to increase by 5 xx 10^(-3) mol l^(-1) in 10 second . The rate of appearance of B is

A gaseoushydrocarbon'X'on reactionwithbromineinlight formsa mixtureof twomonobromo alkanesand HBr.The hydrocarbon'X'is :

A gaseous hydrocarbon was exploded with excess of oxygen. On cooling a contraction of 1.5 times the volume of the hydrocarbon was observed. When treated with NaOH a further contraction of double the volume of the hydrocarbon was noted. The molecular formula of the hydrocarbon is:

A gaseous hydrocarbon requires 6 times its own volume of O_(2) for complete oxidation and produces 4 times its volume of CO_(2). What is its formula ?

A gaseous hydrocarbon has 85%C and its vapour density is 28. The possible formula of the compound is

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is :

A gaseous hydrocarbon given upon combustion, 0.72g water and 3.08g of CO_(2). The empirical formula of the hydrocarbon is:

A gaseous hydrocarbon gives upon combustion, 0.72 g of water and 3.08 g of CO_(2). The empirical formula of the hydrocarbon is

A gaseous hydrocarbon contaisn 85.7% carbon and 14.3% hydrogen 1 litre of the hydrocarbon weights 1.26g at NTP. Determine the molecular formula of the hydrocarbon.

A gaseous compound of carbon and nitrogen containing 53.8% by weight of nitrogen was found to have a vapour density of 25.8. What is the molecular formula of the compound.

A gaseous compound is composed of 85.7% by weight of C and 14.3% by weight of H. Its density is 2.28 g/L at 300 K and 1 atm pressure. Calculate the molecular formula of the compound.

A gaseous carbon compound is soluble in dilute HCl. The solution on treating with NaNO_(2) gives off nitrogen leaving behind a solution which smells of wood spirit. The carbon compound is

A gaseous alkane (C_(n)H_(2n+2)) is exploded with oxygen. The volume of O_(2) and CO_(2) formed are in the ratio of 7:4, Deduce the value of n.

A gaseous A_(2)(g) rarr B(g) +(1)/(2) C(g), shows increase in pressure from 100 mm to 120 mm in 5 min. The rate of disappearance of A_(2) is

A gas X turns line water milky. The milkiness disappears if excess of 'X' is passed. Milkiness reappears on heating the colourless solution. The gas is

A gas X is passedthroughwater to forms a saturated solution .The aqaeoussolutionon tretment aqeoussolutionalsodissolve magnassium ribbon with the evolation of a colourless gas Y identify X and Y

A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y'. Identify 'X' and 'Y':

A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white ppt. The saturated aqueous solution also dissolve magnesium ribbon with evolution of a colourless gas 'Y'. Identify 'X' and 'Y'.

A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatement with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y' Identify 'X' and 'Y'

A gas 'X' diffuses five times as rapidly as another gas 'Y'. Calculate the ratio of molecular weights of 'X' and 'Y'.

A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of colourless gas Y , X and Y are :

A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y^(-) and 1 M Z^(-) at 25^(@)C. If the order of reduction potentials is Z gt Y gt X, then

A gas X at 1 atm is bubbles through a solution containing a mixture of 1M Y^(-) and MZ^(-) at 25^(@)C If the reduction potential of ZgtYgtX, then,

A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y^(-) and 1MZ^(-) ions at 25^(@)C. If the reduction potential of ZgtYgtX, then,

A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y and 1MZ ions at 25^(@)C if the reduction potential of ZgtYgtX, then

A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y^(-) and 1 M Z^(-) at 25^(@)C. If the reduction potential of ZgtYgtX, then

A gas X at 1 atm is bubbled through a solution containg a mixture of 1 M Y^- and 1M Z^- at 25^@C. If the reduction potential of Z gt Y gt X, then :

A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^(-) and 1MZ^(-) at 25^@C . If the reduction potential of Z > Y > X, then

A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^(-)"and " 1MZ^(-) at 25^(@)C. If the reduction potential of Z gt Y gt X, then

A gas X at 1 atm is bubblcd through a soluution containing a mixture of 1 M Y ^(-) and 1 M Z ^(-) at 25^(@)C. If the rcduction potential tial of Z gt Y gt X, then,

A gas will approach ideal behaviour at :

A gas will approach ideal behaviour at

A gas which is used as anaesthetic in dental surgery is

A gas which burns with blue flame is :

A gas when passed through the K_(2)Cr_(2)O_(7) and dil. H_(2)SO_(4) solution, turns it green, the gas is

A gas which burns with a blue flame is :

A gas when passed through K_(2)Cr_(2)O_(7) and dil. H_(2)SO_(4) solution turns it green, the gas is

A gas turns lime water milky and acidified K_(2)Cr_(2)O_(7) solution green then gas is :

A gas that cannot be collected over water is :

A gas reacts with CaO but not with NaHCO_3. The gas is

A gas reacts with CaOand not with NaHCO_3 is:

A gas phase reaction has energy of activation 200 "kJ mol"^(-1).If the frequency factor of the reaction is 1.6xx10^(13) s^(-1). Calculate the rate constant at 600 K . (e^(-40.09)=3.8xx10^(-18)) .