52671 + Interview Questions in Chemistry in Class 11

1.	The % composition by volume of Cl_(2), H_(2) and N_(2) are in 1 : 2 : 7 by proportion. If the total pressure is 40 bar, Find the partial pressures of each gas ?[Partial pressure =("Volume of "% xx "Total pressure")/(100)]
Answer» Answer :`P_(Cl_(2)), P_(H_(2))` and `P_(N_(2))=4, 8, 28` bar respectively

Discussion

2.	The compositin of a sample of wustite isFe_(0-93)O_(1-00) , what precentage of the iron is present in the form ofFe(III) ?
Answer» Solution :The composition is ` Fe_(0-93)`intereadof FeO because some ` Fe^(2+)`ions have been replaced by ` Fe^(3+)`ions. Let us first calcuate the number of ` Fe^(2+) and Fe^(3+)`ions present. The FORMULA ` Fe_(0.93) O _(1.00)`implies that93 Fe ATOMS are combined with 100O - atoms. Out of 93 Fe atoms , suppose Fe atoms present as ` Fe^(3+)`= x , then`Fe^(2+) = 93 -x `, as the compoundis neutral, totalcharge on ` Fe^(2+) and Fe^(3+)`ions = total CHARGE on ` O^(2-)`ions. Thus, ` 3 xx x + 2( 9 3-x)= 2xx 100 or 3 x +1 86-2x = 200 orx=14 , i.e. Fe^(3+) =14` Hence, ` Fe^(2+) = 93 - 14=79` Thus, out of 93 Fe atoms, Fe PRESNET as ` Fe^(3+)=14 ` :% ageof Fe presnet as Fe ( III)= ` 14/93 xx 100 = 15%`

Discussion

3.	The component that distinguishes classical smog from photochemical smog is__________.
Answer» ANSWER :D

Discussion

4.	The component of mixture can be seperated by ..........
Answer» filtration CRYSTALIZATION DISTILLATION all of the above Answer :A::B::C::D

Discussion

5.	The component of cement that has properly of setting instantenously in the presence of H_(2)O and imparting internal strength to cement is
Answer» TRI CALCIUM ALUMINATE Tri calcium silicate Tetra calcium silicate Zypsum Solution :TRICALCIUM aluminate

Discussion

6.	The complete transfer of one or more valence electron from one atom to another leads to the formation of …………………………
Answer» SOLUTION :lonic BOND

Discussion

7.	The commonly used antiknock compound is ........
Answer» SOLUTION : TETRAETHYL LEAD

Discussion

8.	The common salt was obtained from two different sources. In one sample, the percentage of chlorine was found to be 60.75 %. In the second sample, 3.888 g of chlorine were present in 6.4 g of the salt. Show that these data are in accordance to the law of constant proportion.
Answer» Solution :The first SAMPLE of COMMON salt contains 60.75% CHLORINE. In the second sample, 6.4 g of salt contains chlorine= 3.888g `therefore` 100g of salt will contain chlorine `=(3.888 xx 100)/(6.4)` Hence, the PERCENTAGE of chlorine in second sample = 60.75 % Since, both the SAMPLES of common salt contain the same percentage of chlorine, the given data are in accordance to the law of constant proportion.

Discussion

9.	The common oxidation state exhibited by inner transition elements usually in their compounds is
Answer» `+2` `+3` `+5` ZERO Answer :B

Discussion

10.	The common name of sodium hexamet phosphate is
Answer» Calgon Zeolit Resin Permutit ANSWER :A

Discussion

11.	The common method used to prepare ethane is X,while ethene and ethyne is Y. Now, X and Y respectively are
Answer» decarboxylation , Wurtz REACTION KOLBE's electrolysis , SABATIER reaction Kolbe's electrolysis , dehydrohalogenation Kolbe's electrolysis , decarboxylation Answer :C

Discussion

12.	The common featrues among the species CN^(-), CO and CO^(+) are
Answer» Bondorder three and ISOELECTRONIC Bond ORDER three andweak-fieldligands Bond ordertwo and `PI-`acceptor Isoelectronic and weak-field ligands Solution :Each of them has 14 ELECTRONS and bond order = 3

Discussion

13.	The common features among the species CN^(-1) , CO and NO^(+) are
Answer» bond order THREE and isoelctronic bond order three and weak field ligands bond order TWO and pi` electron ACCEPTORS ISOELECTRONIC and weak field ligands Answer :A

Discussion

14.	The common characteristic of cis 2-butene and trans-2-butene is
Answer» Solubility Melting POINT and BOILING point Heat of hydrogenation Product of hydrogenation Answer :D

Discussion

15.	The common components of photo chemical smog are
Answer» `O_3 , NO_2 ` , ACROLEIN , HCHO, PAN `O_3,O_(2),H_(2), Ni, Cu` `NO+O_3` OnlyPAN, acrolein Answer :A

Discussion

16.	The commercial production of 'Water gas' utilzes the endothermic reaction C(s)+H_(2)O(g)to H_(2)(g)+CO(g) the heat required for this reaction is generated by combustion of coal to CO_(2) using stoichiometric amount of air (79% N_(2) by volume and 21%O_(2) by volume ). the superheated steam undergoes 75% conversion . usingthe following data ,answer the question that follows : DeltaH_(f)[CO(g)]=-110.53KJmol DeltaH_(f)[H_(2)O(g)]=-241.81KJ//molDeltaH_(f)[CO_(2)(g)]=-314.0 Kj//mol Match the gas and percentage of each gas in one litre product gases. {:(Gas ,"percentage"),((a)N_(2),(p)~~23.1),((b)CO_(2),(Q)~~36.4),((C)H_(2),(R)~~7.7),((d)H_(2)O,(S)~~9.7):}
Answer» (A-P),(B-Q),(C-R),(D-S) (A-Q),(B-p),(c-S),(D-R) (A-R),(B-s),(C-P),(D-R) (A-Q),(B-S),(C-P),(D-R) ANSWER :d

Discussion

17.	The commercial production of 'Water gas' utilzes the endothermic reaction C(s)+H_(2)O(g)to H_(2)(g)+CO(g) the heat required for this reaction is generated by combustion of coal to CO_(2) using stoichiometric amount of air (79% N_(2) by volume and 21%O_(2) by volume ). the superheated steam undergoes 75% conversion . usingthe following data ,answer the question that follows : DeltaH_(f)[CO(g)]=-110.53KJmol DeltaH_(f)[H_(2)O(g)]=-241.81KJ//molDeltaH_(f)[CO_(2)(g)]=-314.0 Kj//mol THe amount of heat liberated when one litre of product gases are burnt at 373 K and one atm is:
Answer» `~=3.36KJ` `~=3.9KJ` `~=4.43KJ` `~=5.34Kj` ANSWER :a

Discussion

18.	Thecommericalproducationof sodiumcarbonateis doneby _____.
Answer» LEAD- chamberprocess Haber'sprocess Solvay process Castner'sprocess Answer :C

Discussion

19.	The commercial production of sodium carbonate is done by...
Answer» Lead-chamber process Haber's process Solvay's process Castner's process Answer :C

Discussion

20.	The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO_(2(g)) and H_(2) O_((l)) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, Delta_(r) H^( Theta ) of benzene. Standard enthalpies of formation of CO_(2(g)) and H_(2) O_((l)) are -393.5 kJ "mol"^(-1) and -285.83 "kJ mol"^(-1) respectively.
Answer» Solution :The formation reaction of benzene is given by `6C_("(graphite)") + 3H_(2(g)) to C_(6) H_(6(l)), Delta_(f) H^( Theta )= (?)….(i)` Then enthalpy of combustion of 1 MOL of benzene is : `C_(6) H_(6(l)) + (15)/(2) O_(2) to 6CO_(2(g)) + 3H_(2) O_((l)),` `Delta_(C) H^( Theta ) = - 3267 "kJ mol"^(-1) ...(ii)` The enthalpy of formation of 1 mol of `CO_(2(g))`, `C_("(graphite)") + O_(2(g)) to CO_(2(g)) ,` `Delta_(f) H^( Theta ) = -393.5 "kJ mol"^(-1) ...(III)` The enthalpy of formation of 1 mol of `H_(2) O_((l))` is `H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O_((l)),` `Delta_(f) H^( Theta ) = -285.83 "kJ mol"^(-1) ....(iv)` Multiplying eq. (iii) by 6 and eq. (iv) by 3 we GET: `6C_("(graphite)") + 6O_(2(g)) to 6CO_(2(g)),` `Delta_(f) H^( Theta ) =-2361 "kJ mol"^(-1)` `3H_(2(g)) + (3)/(2) O_(2(g)) to 2H_(2) O_((l)),` `Delta_(f) H^( Theta )= -857.49 "kJ mol"^(-1)` SUMMING up the above two equations : `6C_("(graphite)") + 3H_(2(g)) + (15)/(2) O_(2(g)) to 6CO_(2(g)) + 3H_(2)O_((l)),` `Delta_(f) H^( Theta ) = -3218.49 "kJ mol"^(-1) .....(V)` Reversing equation (ii), `6 CO_(2(g)) + 3H_(2) O_((l)) to C_(6) H_(6(l)) + (15)/(2) O_(2(g)),` `Delta_(f) H^( Theta ) = 3267.0 "kJ mol"^(-1) ....(vi)` Adding equations (v) and (vi), we get, `6C_("(graphite)") + 3H_(2(g)) to C_(6)H_(6(l)),` `Delta_(f) H^( Theta ) = 48.51 "kJ mol"^(-1)`

Discussion

21.	The combustion of benzoic acid occurs according to expression C_(6)H_(5)COOH(s) + 7(1)/(2)O_(2)(g) hArr 7CO_(2) + 3H_(2)O (l) in a bomb calorimeter at 25^(@) C . If it liberates Q_(v) "mol"^(-1) , then DeltaH for reaction is
Answer» `-Q_(v)+(RT)^(1//2)` `-Q_(v)+(RT)^(-1//2)` `-Q_(v)+(1)/(2)RT` `-Q_(v)-(1)/(2)RT` SOLUTION :`DeltaH=DeltaE+DeltanRT`, `Deltan` for reaction =7-7.5=-0.5 `DeltaH=-Q_(v)-(1)/(2)RT` (In bomb calorimeter the reaction is done at a constant VOLUME)

Discussion

22.	The combustion of benzene (l) gives CO_(2)(g)and H_(2)O(l) . Given that heat of combustion of benzene at constant volumeis - 3263.9 Kj mol^(-1) at 25^(@)C, heatof combustion( in kJ mol^(-1))of benzene at constant pressure will be( R = 8.314JK^(-1) mol^(-1))
Answer» 4152.6 `-452.46` `3260` `-3267.6` SOLUTION :`C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr6CO_(2)(g)+3H_(2)O(l)` `Deltan_(g)= 6 - ( 15)/( 2)= - (3)/(2)` `DeltaH=DeltaU+ Deltan_(g) RT` `= - 3263.9 + ( - (3)/(2)) XX ( 8.314xx 10^(-3)) xx ( 298)` `= - 3263.9 + ( -3.71)= - 3267 .6 kJ mol^(-1)`

Discussion

23.	The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l).Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol^(-1) at 25^(@)C, heat of combustion (in kJ mol^(-1)) of benzene at constant pressure will be (R = 8.314 JK^(-1) mol^(-1)).
Answer» 4152.6 `-452.46` 3260 `-3267.6` SOLUTION :`C_(6)H_(6)(l) + (15)/(2) O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l)` `Deltan_(g) = 6-15/2 = -3/2` `DeltaH =DeltaU - Deltan_(g)RT` `= -3263.9 + (-3/2) xx 8.314 xx 10^(-3) xx 298` ` = -3263.9 - 3.716` ` = - 3267.678`.

Discussion

24.	The combustion of benzene (l) gives CO_2(g)and H_2O (l) . Given that heat of combustion of benzene at constant volume is -3263.9kJ "mol"^(-1)at 25^@C , heat of combustion (in kJ "mol"^(-1)) of benzene at constant pressure will be
Answer» 4152.6 452.46 3260 -3267.6 Solution :Value of COMBUSTION ENTHALPY `Delta H`MUST be -ve. So option (B) or (D) can be POSSIBLE The value of option (B) is not possible because of (-452.46) >>> -3267.6 Only option (D) is possible

Discussion

25.	the combustion of 2- propanol (M=60.0gxxmol^(-1)) occurs accordingto the equation , 2CH_(3)CHOHCH_(3)(l)+90_(2)(g)to6CO_(2)(g)+8H_(2) O(l) What is q for the combustion of 15.0 g of 2- propanol?
Answer» `-5.01xx10^(2)KJ` `-1.00xx10^(3)KJ` `-2.01xx10^(3)KJ` `-4.01xx10^(3)Kj` ANSWER :a

Discussion

26.	The combustion of 1.00 mole of mehtane , CH_(4),producescarbon dioxide and water releases 802Kjxxmol^(-1) .when 3.00 mol oxygenreacts with a stoichiometric quantity o fmethane , what is DeltaH for the reaction ?
Answer» `-1.20xx10^(3)KJ` `-8.02xx10^(2)KJ` `8.02xx10^(2)Kj` `1.20xx10^(3)Kj` ANSWER :a

Discussion

27.	The combustion of 1 mole of benzene takes placeat 298K and 1 atm. After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0kJ of heat is liberated. Calculate the standard enthalpy of formation, Delta _(f) H^(@) of benzene. Standard enthalpies of formation of CO_(2)(g) and H_(2)O(l) are -393.5kJ mol^(-1) and -258.83kJ mol^(_1) respectively
Answer» Solution :Aim `: 6C(s) + 3H_(2)(g) rarr C_(6)H_(6)(l),DeltaH= ?` Given `: (i) C_(6) H_(6)(l) + (15)/(2) O_(2)(g) rarr 6 CO_(2)(g) + 3H_(2)O(l), Delta H = - 3267.0 kJ mol^(-1)` (ii) `C(s) + O_(2)(g) rarr CO_(2)(g), DeltaH = -393.5 kJ mol^(-1)` (III) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l) , Delta H = -285.83 kJ mol^(_1)` In order to get th requiredthermochemical equation, multiply Eq. (ii) 6 and Eq. (iii) by 3 and subtract Eq. (i) from THEIRSUM, i.e., operating 6 `xx` Eqn. (ii) `+ 3 xx `Eqn (iii)- Eqn (i), we get `6 C(s) + 3H_(2) (g) rarr C_(6) H_(6)(l) , DeltaH = 6(-393.5) + 3(-285.83) - ( -3267.0)` `= - 2361- 857.49 + 3267.0` `= - 48.51kJ mol^(-1)` Thus, the enthalpy of formation of BENZENE is `Delta _(f) H= - 48.51kJ mol^(-1)`

Discussion

28.	The combination of smoke and fog known as …………….
Answer» Fog SMOG MIST All of these SOLUTION :Smog

Discussion

29.	The combination of elements form compounds is governed by the laws of chemical combination What is meant by limiting reagent in a chemical reaction?
Answer» SOLUTION : A REACTANT which is completely used up in a reaction is called the LIMITING reactant.

Discussion

30.	The combination of elements form compounds is governed by the laws of chemical combination 28 g of N_2 is mixed with 12g of H_2 to form ammonia as per the reaction,N_2+3H_2rarr2NH_3. Which is the limiting reagent in this reaction?
Answer» Solution :`N_2+3H_2to2NH_(3)28g6g to 34g28g12g toN_2` is the limiting REAGENT (`N_2` is completely USED up in the reaction)

Discussion

31.	The combination of Boyl's law, Charle's law and Avogadro's law is known as ideal gas equation. But the real gases deviate from ideal behaviour. Name the above equation.
Answer» SOLUTION :VAN DER Waal.s EQUATION

Discussion

32.	Give the name of modified ideal gas equation and write down it.
Answer» <P> SOLUTION : `(P+(n^2a)/V_2)(v-nb)` = NRT

Discussion

33.	The combination of Boyl's law, Charle's law and Avogadro's law is known as ideal gas equation. But the real gases deviate from ideal behaviour.
Answer» <P> Solution :Modified ideal GAS EQUATION `(P+(n^2a)/V_2)(v-nb)` = NRT

Discussion

34.	The colours emitted by excited atoms are characteristics of element. The element famous for the red emission of fireworks and warning flares is
Answer» Pb Mg Sr Ba ANSWER :C

Discussion

35.	The colourless solution of 'D' contains:
Answer» `[FeF_(6)]^(3-)` `[Fe(H_(2)O)_(5)(SCN)]^(2+)` `[Fe(H_(2)O_(6))]^(3+)` `FeS_(4)O_(6)` Answer :A

Discussion

36.	The colour of the solution/precipitate obtained in the elemental analysis of an organic compound and the molecule/ion responsible for the colour are given below. Choose the incorrectly matched pair:
Answer» Prussian blue - `Fe_(4)[Fe(CN)_(6)]_(3).xH_(2)O` Black - PbS Violet colour - `[Fe(CN)_(5)NOS]^(4-)` Yellow - `(NH_(4))_(2)MoO_(4)` Solution :During the detection of P, the yellow PPT. obtained are due to `(NH_(4))_(3)PO_(4).12 MoO_(3)` (amm. Phosphomolybdate) and not due to `(NH_(4))_(2)MoO_(4)` (AMMONIUM molybdate). Thus, OPTION (e) is incorrect.

Discussion

37.	The colour of the solution of alkali metal in ammonia can be
Answer» BLUE violet bronze LIKE red. SOLUTION :Solution of alkalli metal in AMMONIA can be blue COLOURED (if not very concentrated) or bronzelike (if concentrated).

Discussion

38.	The colour of the dye zirconium-Alizarin-s fades when added to water containing fluoride ions because of the formation of
Answer» `ZrF_3` `ZrF_6^(-2)` `ZrF_4` `ZrF_4^(-2)` ANSWER :B

Discussion

39.	The colour of sulphates of II A group metals are
Answer» YELLOW BLACK WHITE Blue Answer :3

Discussion

40.	The colour of sky is due to
Answer» absorption of LIGHT by atmospheric gases transmission of light wavelength of scattered light All of these Solution :According to Rayleigh scattering law, scattering light `PROP 1//lambda`. As BULE colour has maximum wavelenght in the visible range of light, it is least scattered. Hence sky LOOKS blue.

Discussion

41.	The colour of a colloidal solution depends upon
Answer» particle size temperature the direction in which it is being viewed both A and C Solution :Colour of a colloidal solution sometimes also depends UPON the direction in which it is BRING viewed. For example, dilute MILK gives a bluish tinge in the reflected LIGHT and a reddish tinge in the transmitted light.

Discussion

42.	The colour developed for Co^(2+) basic radical in borax bead test is
Answer» GREEN VIOLET YELLOW BLUE ANSWER :D

Discussion

43.	The colloidal solution of mercury in water can be easily obtained by:-
Answer» MECHANICAL precipitation Bredig's ARC method Repeated WASHING ULTRASONIC dispersion Answer :D

Discussion

44.	The colloidal sols are purified by
Answer» Peptisation coagulation Dialysis flocculation Answer :C

Discussion

45.	The coefficients x,y, and z in the following balanced equation : xZn+yNO_3^(-) rarr zZn^(2+)+NH_4^(+) (in basic medium) are
Answer» 4,1,4 2,2,2 4,2,4 4,4,4 Answer :A

Discussion

46.	The coefficient of PH, in the following balanced equation will be: P_(4(s)) +OH^(-) rarr PH_(3_((g))) + HPO_(2(aq))^(-)
Answer» <P> Solution :`P_(4)+3H_(2)O+3OH^(-) rarr3H_(2)PO_(2)^(-) +PH_3`

Discussion

47.	The COD values of three water samples A,B and C are 60 ppm, 990 ppm and 120 ppm respectively. The most polluted water sample is
Answer» A B C All are EQUALLY POLLUTED ANSWER :C

Discussion

48.	The COD value of a water sample is 40 ppm. Calculate the amount of acidified K_(2)Cr_(2)O_(7) required to oxidise the organic matter present in 500 ml of that water sample.
Answer» Solution :COD value is 40 ppm. It means 106g of water sample require 40 g of oxygen to oxidise the organic matter in it. ` 500g " water " to (40 XX 500)/(10^6)=2 xx 10^(-2) g " of " O_2` 500 mL water sample requires ` 2 xx 10^(-2) g ` of `O_2` , to oxidise the organic matter present in it . `8g O_2 -= 49 g K_2Cr_2O_7` ` 2 xx 10^(-2) g " of " O_2 -= (49 xx 2 xx 10^-2)/(8) g " of " K_2Cr_2O_7` Amount of `K_2Cr_2O_7` REQUIRED to oxidise the organic matter present in the water sample is 0.1225 g .

Discussion

49.	The COD of pure sample of water is ______ppm.
Answer» ANSWER :4

Discussion

50.	The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As_(2)S_(3) are given below I. (NaCl)=2 II. (BaCl_(2))=0.69 III. (MgSO_(4))=0.22
Answer» IIIgtIgtII IgtIIgtIII IIgtIgtIII IIIgtIIgtI Solution :COAGULATION POWER `prop(1)/("Coagulation value")` Higher the coagulation power, lower is coagulation values in millimoles PER litre `MgSO_(4) gt BaCl_(2) gt NaCl`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The % composition by volume of Cl_(2), H_(2) and N_(2) are in 1 : 2 : 7 by proportion. If the total pressure is 40 bar, Find the partial pressures of each gas ?[Partial pressure =("Volume of "% xx "Total pressure")/(100)]

The compositin of a sample of wustite isFe_(0-93)O_(1-00) , what precentage of the iron is present in the form ofFe(III) ?

The component that distinguishes classical smog from photochemical smog is__________.

The component of mixture can be seperated by ..........

The component of cement that has properly of setting instantenously in the presence of H_(2)O and imparting internal strength to cement is

The complete transfer of one or more valence electron from one atom to another leads to the formation of …………………………

The commonly used antiknock compound is ........

The common salt was obtained from two different sources. In one sample, the percentage of chlorine was found to be 60.75 %. In the second sample, 3.888 g of chlorine were present in 6.4 g of the salt. Show that these data are in accordance to the law of constant proportion.

The common oxidation state exhibited by inner transition elements usually in their compounds is

The common name of sodium hexamet phosphate is

The common method used to prepare ethane is X,while ethene and ethyne is Y. Now, X and Y respectively are

The common featrues among the species CN^(-), CO and CO^(+) are

The common features among the species CN^(-1) , CO and NO^(+) are

The common characteristic of cis 2-butene and trans-2-butene is

The common components of photo chemical smog are

Thecommericalproducationof sodiumcarbonateis doneby _____.

The commercial production of sodium carbonate is done by...

The combustion of benzoic acid occurs according to expression C_(6)H_(5)COOH(s) + 7(1)/(2)O_(2)(g) hArr 7CO_(2) + 3H_(2)O (l) in a bomb calorimeter at 25^(@) C . If it liberates Q_(v) "mol"^(-1) , then DeltaH for reaction is

The combustion of benzene (l) gives CO_(2)(g)and H_(2)O(l) . Given that heat of combustion of benzene at constant volumeis - 3263.9 Kj mol^(-1) at 25^(@)C, heatof combustion( in kJ mol^(-1))of benzene at constant pressure will be( R = 8.314JK^(-1) mol^(-1))

The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l).Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol^(-1) at 25^(@)C, heat of combustion (in kJ mol^(-1)) of benzene at constant pressure will be (R = 8.314 JK^(-1) mol^(-1)).

The combustion of benzene (l) gives CO_2(g)and H_2O (l) . Given that heat of combustion of benzene at constant volume is -3263.9kJ "mol"^(-1)at 25^@C , heat of combustion (in kJ "mol"^(-1)) of benzene at constant pressure will be

the combustion of 2- propanol (M=60.0gxxmol^(-1)) occurs accordingto the equation , 2CH_(3)CHOHCH_(3)(l)+90_(2)(g)to6CO_(2)(g)+8H_(2) O(l) What is q for the combustion of 15.0 g of 2- propanol?

The combustion of 1.00 mole of mehtane , CH_(4),producescarbon dioxide and water releases 802Kjxxmol^(-1) .when 3.00 mol oxygenreacts with a stoichiometric quantity o fmethane , what is DeltaH for the reaction ?

The combination of smoke and fog known as …………….

The combination of elements form compounds is governed by the laws of chemical combination What is meant by limiting reagent in a chemical reaction?

The combination of elements form compounds is governed by the laws of chemical combination 28 g of N_2 is mixed with 12g of H_2 to form ammonia as per the reaction,N_2+3H_2rarr2NH_3. Which is the limiting reagent in this reaction?

The combination of Boyl's law, Charle's law and Avogadro's law is known as ideal gas equation. But the real gases deviate from ideal behaviour. Name the above equation.

Give the name of modified ideal gas equation and write down it.

The combination of Boyl's law, Charle's law and Avogadro's law is known as ideal gas equation. But the real gases deviate from ideal behaviour.

The colours emitted by excited atoms are characteristics of element. The element famous for the red emission of fireworks and warning flares is

The colourless solution of 'D' contains:

The colour of the solution/precipitate obtained in the elemental analysis of an organic compound and the molecule/ion responsible for the colour are given below. Choose the incorrectly matched pair:

The colour of the solution of alkali metal in ammonia can be

The colour of the dye zirconium-Alizarin-s fades when added to water containing fluoride ions because of the formation of

The colour of sulphates of II A group metals are

The colour of sky is due to

The colour of a colloidal solution depends upon

The colour developed for Co^(2+) basic radical in borax bead test is

The colloidal solution of mercury in water can be easily obtained by:-

The colloidal sols are purified by

The coefficients x,y, and z in the following balanced equation : xZn+yNO_3^(-) rarr zZn^(2+)+NH_4^(+) (in basic medium) are

The coefficient of PH, in the following balanced equation will be: P_(4(s)) +OH^(-) rarr PH_(3_((g))) + HPO_(2(aq))^(-)

The COD values of three water samples A,B and C are 60 ppm, 990 ppm and 120 ppm respectively. The most polluted water sample is

The COD value of a water sample is 40 ppm. Calculate the amount of acidified K_(2)Cr_(2)O_(7) required to oxidise the organic matter present in 500 ml of that water sample.

The COD of pure sample of water is ______ppm.

The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As_(2)S_(3) are given below I. (NaCl)=2 II. (BaCl_(2))=0.69 III. (MgSO_(4))=0.22