Quiz about Chemistry in Class 12

Question 1

The condition of vitamin deficiency is known as

Accepted Answer

NA

Question 2

The condition for spontaneity of a process is

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NA

Question 3

The condition of spontaneity of process is

Accepted Answer

NA

Question 4

The condition for minimum change in pH for a buffer solution is

Accepted Answer

NA

Question 5

The condition for a reaction to occur spontaneously is

Accepted Answer

NA

Question 6

The condition for a compound to be precipitated is

Accepted Answer

NA

Question 7

The condensation reaction of esters with nitriles containing alpha- hydrogen is known as…………………

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NA

Question 8

The condensation polymer among the following is

Accepted Answer

NA

Question 9

The condensation polymer among the following is

Accepted Answer

NA

Question 10

The condensation of two amino acids glycine and (pm) alanine yields total products x. Find value of x ?

Accepted Answer

NA

Question 11

The concept of resonance explains various properties of compounds.The molecules with conjugated system of pi bonds are stabilized by resonace and have low heat of hydrogenation.Hyperconjugative stabilization also decreases heat f hydrogenation.In aromic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like -NO -NC -CH=CH_2 can function both as electron releasing (+m+R) or electron withdrawing (-m-R) groups.More extended conjugation provides more stabilization.  The most stable carbocation is

Accepted Answer

NA

Question 12

The concept of resonance explains various properties of compounds.The molecules with conjugated system of pi bonds are stabilized by resonace and have low heat of hydrogenation.Hyperconjugative stabilization also decreases heat f hydrogenation.In aromic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like -NO -NC -CH=CH_2 can function both as electron releasing (+m+R) or electron withdrawing (-m-R) groups.More extended conjugation provides more stabilization.  The correct heat of hydrogenation order is  (p)13-Pentadiene (q)13-Butadiene (r)23-Dimethyl-13-butadiene (s)Propadiene

Accepted Answer

NA

Question 13

The concept of resonance explains various properties of compounds.The molecule with conjugated system of pi bonds are stabilized by resonance and have low heat of hydrogenation .Hyperconjugative stabilization also decreases heat of hydrogenation. In aromatic rings a functional group with a lone pair of electron exerts +m effect.Some functional groups like -NO -NC -CH=CH_2 can function both as electrons releasing (+m+R) or electron withdrawing (-m-R) groups.More extended conjugation provides more stabilization.  The most stable resonating structure of following compound is

Accepted Answer

NA

Question 14

The concept of resonance explains various properties of compounds.The molecule with conjugated system of pi bonds are stabilized by resonance and have low heat of hydrogenation .Hyperconjugative stabilization also decreases heat of hydrogenation. In aromatic rings a functional group with a lone pair of electron exerts +m effect.Some functional groups like -NO -NC -CH=CH_2 can function both as electrons releasing (+m+R) or electron withdrawing (-m-R) groups.More extended conjugation provides more stabilization.  The most stable carbocation is

Accepted Answer

NA

Question 15

The concept of resonance explains various properties of compounds.The molecule with conjugated system of pi bonds are stabilized by resonance and have low heat of hydrogenation .Hyperconjugative stabilization also decreases heat of hydrogenation. In aromatic rings a functional group with a lone pair of electron exerts +m effect.Some functional groups like -NO -NC -CH=CH_2 can function both as electrons releasing (+m+R) or electron withdrawing (-m-R) groups.More extended conjugation provides more stabilization.  The correct heat of hydrogenation order is  (p)13-Pentadiene (q)13-Butadiene (r)23-Dimethyl-13-butadiene (s)Propadiene

Accepted Answer

NA

Question 16

The concept of electrode potential is useful in the isolation of metals.

Accepted Answer

NA

Question 17

The concentraton of hydrochloric acid in agiven solutioon is 10^(-8)M what is the vlaue of pH for this solution?

Accepted Answer

NA

Question 18

The concept of critical temperature for a gas was given by

Accepted Answer

NA

Question 19

The concentrations of N_(2)O_(5) decomposingin firstorder kinetics after 800 s is1.45 mol L^(-1) andafter1600 s is 0.88 mol L^(-1). Calculate the rateconstant.

Accepted Answer

NA

Question 20

The concentration term independent of temperature is molaritymolality.

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NA

Question 21

The concentration R in the reaction RtoP was measured as a function of time and the following data is obtained.    The order of reaction is

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NA

Question 22

The concentrationfo R in the reaction RtoP was measured as a function of time and the following data is obtained  [R] (molar)1.00.750.400.10  t(min)0.00.050.120.18  The order of the reactionis

Accepted Answer

NA

Question 23

The concentration of which ion is to be decreased when NH_(3) solution is added

Accepted Answer

NA

Question 24

The concentration of the reactant A in the reaction AtoB at different times are given below  Concentration (M)Time(seconds)  0.0690  0.05217  0.03534  0.01851  The rate constant of the reaction according to the correct order of reaction is

Accepted Answer

NA

Question 25

The concentration of the aqueous solutions of each of the following salts is 0.01 (M)  KFCH_3N^(o+)H_3C^(o+)lKClCH_3COOH  If K_a(HF)=7.0xx10^(-4)K_b(CH_3NH_2)=5xx10^(-4) and K_a(CH_3COOH)=1.8xx10^(-5) then increasing order of the pH values of the solutions-

Accepted Answer

NA

Question 26

The concentration of sulphide ion in 0.1 M HCI solution saturated with hydrogen sulphide is 1.0 xx10^(-19) M. If 10 mL of this solution is added to 5 ml of 0.04 M solution of FeSO_4 MnCl_2 ZnCl_2 and CdCl_2 in which solutions precipitation will take place? (Given K_(sp)for FeS = 63 xx10^(-18) MnS = 2.5 xx10^(-13) ZnS = 1.6 xx10^(-24) and CdS = 8.0 xx10^(-27).)

Accepted Answer

NA

Question 27

The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M_((s))|M^(+)(aq0.05 molar)||M^(+)(aq1 molar)|M_((s))  For the above electrolytic cell the magnitude of the cell potential |E_(cell)|=70mV.  Q. For the above cell

Accepted Answer

NA

Question 28

The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M_((s))|M^(+)(aq0.05 molar)||M^(+)(aq1 molar)|M_((s))  For the above electrolytic cell the magnitude of the cell potential |E_(cell)|=70mV.  Q. if the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solutionthen the magnitude of the cell potential would be

Accepted Answer

NA

Question 29

The concentration of R in the reaction R to P was measured as a funcation of time and the following data is obtained

Accepted Answer

NA

Question 30

The concentration of pure solid and liquid phase is not inculded in the expression of equilibrium constant because

Accepted Answer

NA

Question 31

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such concentration cell involving a metal M is M_((s)) |M_((aq 0.05( molar))^(+) ||M_((aq 1 molar))^(+) | M_(s) For the above electrolytic cell the magnitude of the cell potential |E_(cell)|=70mV. If the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solution then the magnitudethe cell potential would be

Accepted Answer

NA

Question 32

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such concentration cell involving a metal M is M_((s)) |M_((aq 0.05( molar))^(+) ||M_((aq 1 molar))^(+) | M_(s) For the above electrolytic cell the magnitude of the cell potential |E_(cell)|=70mV.  For the above cell

Accepted Answer

NA

Question 33

The concentration ofpotassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple mode for such a concentration cell involving a metal M is  M_((s))|M_((aq.))^(@) 0.05molar||M_((aq.))^(@) 1molar|M_((s))  For the above electrolytic cell the magnitude of the cell potential |E_(cell) | = 70mV  If the 0.05 molar solution of M^(+) is replaced by 0.0025 molar M^(+) solution then the magnitude of the cell potential would be

Accepted Answer

NA

Question 34

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in serveral processes such as transmission of nerve impules and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M_((s))|M^(+)(aq 0.05 molar)||M^(+)(aq 1 molar)|M_((s))  For the above electroltic cell the magnitude of the cell potential |E_(cell)|=70 m V.  For the above cell

Accepted Answer

NA

Question 35

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in serveral processes such as transmission of nerve impules and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M_((s))|M^(+)(aq 0.05 molar)||M^(+)(aq 1 molar)|M_((s))  For the above electroltic cell the magnitude of the cell potential |E_(cell)|=70 m V.  If the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solution then the magnitude of the cell potential would be

Accepted Answer

NA

Question 36

The concentration of potassium ions inside a bilological cell is at least twenty times higher than the outisde. The resulting potential difference across the cell is important in several process such as transmission of nerve impuses and maintaning the ion balance. A simple model for such a concentration cell involving a metal M is M(s)|M^(+)(aq0.05 molar)||M^(+)(aq1 molar)|M(s)  For the above electrolytic cell the magnitude of the cell potential |E_(cell)|=70mV.  Q. For the above cell

Accepted Answer

NA

Question 37

The concentration of NaOH solution is 10^(-8)M . Find out the (OH^(-)) concentration

Accepted Answer

NA

Question 38

The concentration of K^(+) inside a biological cell is at least 20 times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M(s)|M^(+)(aq0.05M)||M^(+)(aq1M)|M(s)  For the above galvanic cell the magnitude of the cell potential |E_(cell)|=70mV.  If the 0.05 M solution of M^(+) is replaced by 0.0025 M M^(+) solution then the magnitude of the cell potential would be

Accepted Answer

NA

Question 39

The concentration of KI and KCl in certain solution containing both is 0.001M each. If 20 mL of this solution is added to 20 ml of a saturated of AgI in water ? What will happen

Accepted Answer

NA

Question 40

The concentration of Kl and KCl in a certain solution containing both is 0.001 M each . If 20 mL of this solution is added to 20 mL of a saturated solution of Agl in water . What will happen ?  K_(sp) ofAgCl = 10 ^(-10) K_(sq)ofAgI = 10^(-16)

Accepted Answer

NA

Question 41

The concentration of KI and KCl in centain solution containing both is 0.001 M each. If 20 ml of this solution is added to 20 ml of a saturated solution of AgI in water . What will happen ?

Accepted Answer

NA

Question 42

The concentration of K^(+) inside a biological cell is at least 20 times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is  M(s)|M^(+)(aq0.05M)||M^(+)(aq1M)|M(s)  For the above galvanic cell the magnitude of the cell potential |E_(cell)|=70mV.  For the above cell

Accepted Answer

NA

Question 43

the concentration of hydroxide ion in a water sample is found to be 2.5 times 10^-6 M. Identify the nature of the solution.

Accepted Answer

NA

Question 44

The concentration of hydroxide ion in a water sample is found to be 2.5xx10^(-6)M. Identify the nature of the solution.

Accepted Answer

NA

Question 45

The concentration of hydronium (H_(3)O^(+)) ion in water is

Accepted Answer

NA

Question 46

The concentration of hydrogen ion is a measure of acidity or alkalinity of a solution. A convenient way of expressing the hydrogen ion concentration of a solution is in terms of the pH scale devised by Sorensons . The pH of a solutionis defined as the negative logarithm of active concentration of H^(+) ions to the base 10 i.e.  pH =- log_(10)[HI^(+)] or [H^(+)]=10^(-pH)  At 25^(@)C  (i) if pH lt1 then solution is acidic  (ii) if pH=7 then solution is neutral  (iii) if pH gt 7 then the solution is basic  Total [H^(+)]and [OH^(-)] in a mixture of strong acids or bases are represented in terms normality which is equal to final number of milliequivalents of H^(+) or OH^(-) present in millilitre of solution Since H_(2)O ionizes as  H_(2)O hArrH^(+)+OH^(-) Ionic product of water K_(w)=[H^(+)][OH^(-)] ...(i)  Taking log on both sidesin equation (i) we find pH +pOH=pK_(w)  For neutral solution pH =pOH. 2pH =pK_(w)  . pH =1.2pK_(w)  Since dissociationof water is an endothermic process so temperature willhave great effect on K_(w) as well as on the pH of solution i.e. the nature of the soluiton . K_(w) at different temperature are related as under.  ln (K_(w2))(K_(w1))=(DeltaH)(R)[(1)(T_(1))-(1)(T_(2))]  At T_(1)K=T_(2)K K_(w_(2))=K_(w1)  (i) pH mixture of monoprotic weak acid is calculated as under pH of monoprotic weak acid i.e. CH_(3)COOH  (CH_(3)COOHhArrCH_(3)COO^(-)+H^(+))(at t=0C00)(at t=tC-alphaC alphaCalpha)  . K_(a)=(C alpha xx C alpha)(C-Calpha)=(alpha^(2)C)(1-alpha)  When a le0.1 100 alpha =1-alpha=1  . K_(a)=(alpha^(2)C)(1) [When (1-alpha)=1]  alpha=sqrt((K_(a))(C)). [H^(+)]=alphaC=sqrt(K_(a)xxC)  (ii) pH of a mixture of monoprotic weak acids is calculated as follows  [H^(+)]=sqrt(K_(a1)C_(1)+K_(a2)C_(2)).  where K_(a1) and K_(a2) are dissociation constants of monoprotic weak acids HA and HB acid C_(1) and C_(2) are their concentrations respectively.  (ii) pH of polyproticweak acid say H_(3)A-a triprotic weak acid having dissociation constants K_(a1).K_(a2) and K_(a3) where K_(a1) gt gt K_(a2) gt gt K_(a3). Then in that case maximum [H^(+)] will be contributed from step I and neglibily small from concentration of species producing from step III will be negligible with respect to steop II and similarly concentration of species producig from step II will be negligible with respect to step I. Based upon the above discussion answer the following questions. 100mLM10HA (K_(a1) =8 xx 10^(-5)) and 200 mL of HB (K_(a2)=10^(-5)) acid mixture is diluted to one litre then what will be the pH of the resulting solution ?

Accepted Answer

NA

Question 47

The concentration of hydrogen ion is a measure of acidity or alkalinity of a solution. A convenient way of expressing the hydrogen ion concentration of a solution is in terms of the pH scale devised by Sorensons . The pH of a solutionis defined as the negative logarithm of active concentration of H^(+) ions to the base 10 i.e.  pH =- log_(10)[HI^(+)] or [H^(+)]=10^(-pH)  At 25^(@)C  (i) if pH lt1 then solution is acidic  (ii) if pH=7 then solution is neutral  (iii) if pH gt 7 then the solution is basic  Total [H^(+)]and [OH^(-)] in a mixture of strong acids or bases are represented in terms normality which is equal to final number of milliequivalents of H^(+) or OH^(-) present in millilitre of solution Since H_(2)O ionizes as  H_(2)O hArrH^(+)+OH^(-) Ionic product of water K_(w)=[H^(+)][OH^(-)] ...(i)  Taking log on both sidesin equation (i) we find pH +pOH=pK_(w)  For neutral solution pH =pOH. 2pH =pK_(w)  . pH =1.2pK_(w)  Since dissociationof water is an endothermic process so temperature willhave great effect on K_(w) as well as on the pH of solution i.e. the nature of the soluiton . K_(w) at different temperature are related as under.  ln (K_(w2))(K_(w1))=(DeltaH)(R)[(1)(T_(1))-(1)(T_(2))]  At T_(1)K=T_(2)K K_(w_(2))=K_(w1)  (i) pH mixture of monoprotic weak acid is calculated as under pH of monoprotic weak acid i.e. CH_(3)COOH  (CH_(3)COOHhArrCH_(3)COO^(-)+H^(+))(at t=0C00)(at t=tC-alphaC alphaCalpha)  . K_(a)=(C alpha xx C alpha)(C-Calpha)=(alpha^(2)C)(1-alpha)  When a le0.1 100 alpha =1-alpha=1  . K_(a)=(alpha^(2)C)(1) [When (1-alpha)=1]  alpha=sqrt((K_(a))(C)). [H^(+)]=alphaC=sqrt(K_(a)xxC)  (ii) pH of a mixture of monoprotic weak acids is calculated as follows  [H^(+)]=sqrt(K_(a1)C_(1)+K_(a2)C_(2)).  where K_(a1) and K_(a2) are dissociation constants of monoprotic weak acids HA and HB acid C_(1) and C_(2) are their concentrations respectively.  (ii) pH of polyproticweak acid say H_(3)A-a triprotic weak acid having dissociation constants K_(a1).K_(a2) and K_(a3) where K_(a1) gt gt K_(a2) gt gt K_(a3). Then in that case maximum [H^(+)] will be contributed from step I and neglibily small from concentration of species producing from step III will be negligible with respect to steop II and similarly concentration of species producig from step II will be negligible with respect to step I. Based upon the above discussion answer the following questions. What will be the value of [H^(+)] of a 5 xx 10^(-6)M CH_(3)COOH solution (K_(a)=1.8 xx 10^(-5)) ?

Accepted Answer

NA

Question 48

The concentration of hydrogen ion is a measure of acidity or alkalinity of a solution. A convenient way of expressing the hydrogen ion concentration of a solution is in terms of the pH scale devised by Sorensons . The pH of a solutionis defined as the negative logarithm of active concentration of H^(+) ions to the base 10 i.e.  pH =- log_(10)[HI^(+)] or [H^(+)]=10^(-pH)  At 25^(@)C  (i) if pH lt1 then solution is acidic  (ii) if pH=7 then solution is neutral  (iii) if pH gt 7 then the solution is basic  Total [H^(+)]and [OH^(-)] in a mixture of strong acids or bases are represented in terms normality which is equal to final number of milliequivalents of H^(+) or OH^(-) present in millilitre of solution Since H_(2)O ionizes as  H_(2)O hArrH^(+)+OH^(-) Ionic product of water K_(w)=[H^(+)][OH^(-)] ...(i)  Taking log on both sidesin equation (i) we find pH +pOH=pK_(w)  For neutral solution pH =pOH. 2pH =pK_(w)  . pH =1.2pK_(w)  Since dissociationof water is an endothermic process so temperature willhave great effect on K_(w) as well as on the pH of solution i.e. the nature of the soluiton . K_(w) at different temperature are related as under.  ln (K_(w2))(K_(w1))=(DeltaH)(R)[(1)(T_(1))-(1)(T_(2))]  At T_(1)K=T_(2)K K_(w_(2))=K_(w1)  (i) pH mixture of monoprotic weak acid is calculated as under pH of monoprotic weak acid i.e. CH_(3)COOH  (CH_(3)COOHhArrCH_(3)COO^(-)+H^(+))(at t=0C00)(at t=tC-alphaC alphaCalpha)  . K_(a)=(C alpha xx C alpha)(C-Calpha)=(alpha^(2)C)(1-alpha)  When a le0.1 100 alpha =1-alpha=1  . K_(a)=(alpha^(2)C)(1) [When (1-alpha)=1]  alpha=sqrt((K_(a))(C)). [H^(+)]=alphaC=sqrt(K_(a)xxC)  (ii) pH of a mixture of monoprotic weak acids is calculated as follows  [H^(+)]=sqrt(K_(a1)C_(1)+K_(a2)C_(2)).  where K_(a1) and K_(a2) are dissociation constants of monoprotic weak acids HA and HB acid C_(1) and C_(2) are their concentrations respectively.  (ii) pH of polyproticweak acid say H_(3)A-a triprotic weak acid having dissociation constants K_(a1).K_(a2) and K_(a3) where K_(a1) gt gt K_(a2) gt gt K_(a3). Then in that case maximum [H^(+)] will be contributed from step I and neglibily small from concentration of species producing from step III will be negligible with respect to steop II and similarly concentration of species producig from step II will be negligible with respect to step I. Based upon the above discussion answer the following questions. What is the sulphide ion concentration of a dilute solution that has been saturated with 0.1 M of H_(2)S If pH of the solution is 3 ? (K_(a1)= 1.0 xx 10^(-7) K_(a2) =1.3 xx 10^(-13))

Accepted Answer

NA

Question 49

The concentration of hydrogen ions ina sample of soft drink is 3.8 times 10^-3m. What is the pH value? Whether the soft drink is acidic (or) basic?

Accepted Answer

NA

Question 50

The concentration of hydrogen ion is a measure of acidity or alkalinity of a solution. A convenient way of expressing the hydrogen ion concentration of a solution is in terms of the pH scale devised by Sorensons . The pH of a solutionis defined as the negative logarithm of active concentration of H^(+) ions to the base 10 i.e.  pH =- log_(10)[HI^(+)] or [H^(+)]=10^(-pH)  At 25^(@)C  (i) if pH lt1 then solution is acidic  (ii) if pH=7 then solution is neutral  (iii) if pH gt 7 then the solution is basic  Total [H^(+)]and [OH^(-)] in a mixture of strong acids or bases are represented in terms normality which is equal to final number of milliequivalents of H^(+) or OH^(-) present in millilitre of solution Since H_(2)O ionizes as  H_(2)O hArrH^(+)+OH^(-) Ionic product of water K_(w)=[H^(+)][OH^(-)] ...(i)  Taking log on both sidesin equation (i) we find pH +pOH=pK_(w)  For neutral solution pH =pOH. 2pH =pK_(w)  . pH =1.2pK_(w)  Since dissociationof water is an endothermic process so temperature willhave great effect on K_(w) as well as on the pH of solution i.e. the nature of the soluiton . K_(w) at different temperature are related as under.  ln (K_(w2))(K_(w1))=(DeltaH)(R)[(1)(T_(1))-(1)(T_(2))]  At T_(1)K=T_(2)K K_(w_(2))=K_(w1)  (i) pH mixture of monoprotic weak acid is calculated as under pH of monoprotic weak acid i.e. CH_(3)COOH  (CH_(3)COOHhArrCH_(3)COO^(-)+H^(+))(at t=0C00)(at t=tC-alphaC alphaCalpha)  . K_(a)=(C alpha xx C alpha)(C-Calpha)=(alpha^(2)C)(1-alpha)  When a le0.1 100 alpha =1-alpha=1  . K_(a)=(alpha^(2)C)(1) [When (1-alpha)=1]  alpha=sqrt((K_(a))(C)). [H^(+)]=alphaC=sqrt(K_(a)xxC)  (ii) pH of a mixture of monoprotic weak acids is calculated as follows  [H^(+)]=sqrt(K_(a1)C_(1)+K_(a2)C_(2)).  where K_(a1) and K_(a2) are dissociation constants of monoprotic weak acids HA and HB acid C_(1) and C_(2) are their concentrations respectively.  (ii) pH of polyproticweak acid say H_(3)A-a triprotic weak acid having dissociation constants K_(a1).K_(a2) and K_(a3) where K_(a1) gt gt K_(a2) gt gt K_(a3). Then in that case maximum [H^(+)] will be contributed from step I and neglibily small from concentration of species producing from step III will be negligible with respect to steop II and similarly concentration of species producig from step II will be negligible with respect to step I. Based upon the above discussion answer the following questions.  What will be the [S^(2-)] and pH for 0.1 M H_(2)S solution respectively ? (Given that for H_(2)S K_(a1)=1.0 xx 10^(-7) K_(a2)=1.3 xx 10^(-13))

Accepted Answer

NA

Explore topic-wise InterviewSolutions in Current Affairs.

The condition of vitamin deficiency is known as

The condition for spontaneity of a process is

The condition of spontaneity of process is

The condition for minimum change in pH for a buffer solution is

The condition for a reaction to occur spontaneously is

The condition for a compound to be precipitated is

The condensation reaction of esters with nitriles containing alpha- hydrogen is known as…………………

The condensation polymer among the following is

The condensation polymer among the following is :

The condensation of two amino acids, glycine and (pm) alanine yields total products x. Find value of x ?

The concept of electrode potential is useful in the isolation of metals.

The concentraton of hydrochloric acid in agiven solutioon is 10^(-8)M what is the vlaue of pH for this solution?

The concept of critical temperature for a gas was given by :

The concentrations of N_(2)O_(5) decomposingin firstorder kinetics after 800 s is1.45 mol L^(-1) andafter1600 s is 0.88 mol L^(-1). Calculate the rateconstant.

The concentration term independent of temperature is molarity/molality.

The concentration R in the reaction RtoP was measured as a function of time and the following data is obtained. The order of reaction is

The concentrationfo R in the reaction RtoP was measured as a function of time and the following data is obtained: [R] (molar)""1.0""0.75""0.40""0.10 t(min)""0.0""0.05""0.12""0.18 The order of the reactionis

The concentration of which ion is to be decreased, when NH_(3) solution is added

The concentration of the reactant A in the reaction AtoB at different times are given below: "Concentration "(M)"""Time(seconds)" 0.069""0 0.052""17 0.035""34 0.018""51 The rate constant of the reaction according to the correct order of reaction is

The concentration of the aqueous solutions of each of the following salts is 0.01 (M) KF,CH_3N^(o+)H_3C^(o+)l,KCl,CH_3COOH If K_a(HF)=7.0xx10^(-4),K_b(CH_3NH_2)=5xx10^(-4) and K_a(CH_3COOH)=1.8xx10^(-5) then increasing order of the pH values of the solutions-

The concentration of R in the reaction R to P was measured as a funcation of time and the following data is obtained:

The concentration of pure solid and liquid phase is not inculded in the expression of equilibrium constant because:

The concentration of NaOH solution is 10^(-8)M . Find out the (OH^(-)) concentration

The concentration of KI and KCl in certain solution containing both is 0.001M each. If 20 mL of this solution is added to 20 ml of a saturated of AgI in water ? What will happen

The concentration of Kl and KCl in a certain solution containing both is 0.001 M each . If 20 mL of this solution is added to 20 mL of a saturated solution of Agl in water . What will happen ? K_(sp) ofAgCl = 10 ^(-10) , K_(sq)ofAgI = 10^(-16)

The concentration of KI and KCl in centain solution containing both is 0.001 M each. If 20 ml of this solution is added to 20 ml of a saturated solution of AgI in water . What will happen ?

the concentration of hydroxide ion in a water sample is found to be 2.5 times 10^-6 M. Identify the nature of the solution.

The concentration of hydroxide ion in a water sample is found to be 2.5xx10^(-6)M. Identify the nature of the solution.

The concentration of hydronium (H_(3)O^(+)) ion in water is

The concentration of hydrogen ions ina sample of soft drink is 3.8 times 10^-3m. What is the pH value? Whether the soft drink is acidic (or) basic?

1.	The condition of vitamin deficiency is known as
Answer» Vitaminosis AVITAMINOSIS Both 1 and 2 Anemea ANSWER :B

2.	The condition for spontaneity of a process is
Answer» <P>lowering of entropy at constant temperature and PRESSURE lowering of Gibbs free ENERGY of system at constant temperature and pressure increase of entropy of system at constant temperature and pressure increase of Gibbs free energy of the universe at constant temperature and pressure. Solution :`dG_(P,T) = -ve` is the criterion for SPONTANEITY .

3.	The condition of spontaneity of process is
Answer» LOWERING of entropy at constant TEMPERATURE and pressure Lowering of Gibbs free energy of system at constant temperature and pressure Increase of entropy of system at constant temperature and pressure Increase of Gibbs free energy of the UNIVERSE at constant temperature and pressure Solution :`dG_(P,T)=-ve` is the CRITERION for SPONTANEITY.

4.	The condition for minimum change in pH for a buffer solution is
Answer» Isoelectronic species are ADDED CONJUGATE acid or base is added `pH = pK_(a)` NONE of these Answer :3

5.	The condition for a reaction to occur spontaneously is
Answer» `DELTA H`must be negative `Delta S`must be negative `(Delta H - T Delta S)`must be negative `(Delta H + T Delta S)`must be negative. Solution : For the spontaneity of a reaction, `Delta G` must be -ve. Now,`Delta G = Delta H - T Delta S ` ` therefore `For a SPONTANEOUS reaction, `(Delta H - T Delta S)`should be negative.

6.	The condition for a compound to be precipitated is
Answer» IONIC product= SOLUBILITY product Ionic product`lt` solubility procut Ionic product`GT` solubility procut Ionic product`le` solubility procut Answer :C

7.	The condensation reaction of esters with nitriles containing alpha- hydrogen is known as…………………
Answer» ANSWER :LEVINE and HAUSER

8.	The condensation polymer among the following is
Answer» Protein PVC Polythene Rubber Solution :Condensation polymers are formed by the COMBINATION oif MONOMERS with the elimination of simple molecules LIKE water, or alcohol, e.g., polyesters and POLYAMIDES. Thus, out of the given options, proteins are condensation polymers of amino ACIDS.

9.	The condensation polymer among the following is :
Answer» PROTEIN PVC Polythene Rubber Answer :A

10.	The condensation of two amino acids, glycine and (pm) alanine yields total products x. Find value of x ?
Answer» SOLUTION :`H_2N-CH_2-COOH("Glycine")+H_2N-undersetunderset(CH_3)(\|)CH-COOH` (1)Gly-Gly (2)Gly-ALA(+) (3)Gly-Ala(-) (4)Ala(+)-Gly (5)Ala(-)-Gly (6)Ala(+)-Ala(+) (7)Ala(-)-Ala(-) (8)Ala(+)-Ala(-) (9)Ala(-)-Ala(+)

17.	The concentraton of hydrochloric acid in agiven solutioon is 10^(-8)M what is the vlaue of pH for this solution?
Answer» salts have ORDERED packing arrrangements called lattices salts have LOW melting points but high boiling points salts are brittle salts conduct electricity when dissolved in WATER or even oin the moten state Solution :salts are ionic compounds in which the constituent ions have a fixed POSITION ie arranged in a definite into PIECES) and can conduct electricity in aqueous solution or in the molten state being ionic they have high melting and boiling points

23.	The concentration of which ion is to be decreased, when NH_(3) solution is added
Answer» `OH^(-)` `NH_(4)^(+)` `H_(3)O^(+)` `O_(2)^(-)` SOLUTION :Because `NH_(3)` acts as LEWIS acid any they give electron pair to `H_(3)O^(+)` ION. `H_(3)O^(+)` is a Lewis BASE. Which accept the electron pair from `NH_(3)`.

26.	The concentration of sulphide ion in 0.1 M HCI solution saturated with hydrogen sulphide is 1.0 xx10^(-19) M. If 10 mL of this solution is added to 5 ml of 0.04 M solution of FeSO_4, MnCl_2, ZnCl_2 and CdCl_2, in which solutions precipitation will take place? (Given K_(sp)for FeS = 6,3 xx10^(-18) , MnS = 2.5 xx10^(-13), ZnS = 1.6 xx10^(-24) and CdS = 8.0 xx10^(-27).)
Answer» `FeSO_4` `MnCI_2` `ZnCI_2` `CdCI_2` Solution :When `K_(ip)gtK_(sp)` precipition takesplace, as10 MLOF solutioncontaining`S^(2-)`ionis mixedwith5mLofmetalsaltsolution, aftermixing `[s^(2-)]=1.0 xx10^(-19) xx (10)/(15 ) =6.67 xx 10^(-20)` `[Fe^(2+)] = [Mn^(2+)] = [Zn^(2+)] = [Cd^(2+)]` `=(5)/( 15) xx 0.04 = 1.33 xx 10^(-2)M` ` therefore ` Ionicproduct FOREACH of thesewill be ` =[M^(2+)] [S^(2-)]` `=(1.33 xx 10^(-2) ) ( 6.67 xx 10^(-20)) = 8.87 xx 10^(-22)` asthisvalueis greater than the solubilityproductof ` ZnS ` andCdstherefore`,ZnCI_2`and `CdCI_2`solutionswillbe precipitatedwhereasFeSand MnSis lesserthansolubilityproductthereforeprecipitationdoes nottakeplacein thiscase .

37.	The concentration of NaOH solution is 10^(-8)M . Find out the (OH^(-)) concentration
Answer» `10^(-8)` Greater than `10^(-6)` `10^(-6)` Lies between `10^(-6)` and `10^(-7)` Solution :Concentrate of `[OH^(-)]` in `NAOH = 10^(-8)` and in `H_(2)O = 10^(-7)`. Total `[OH^(-)] = 10^(-8) + 10^(-7) = 1.1 XX 10^(-7)` pOH = 7 - log 1.1 PH = 14-7 + log 1.1 `pH = 7+ log 1.1 gt 7`.

11.	The concept of resonance explains various properties of compounds.The molecules with conjugated system of pi bonds, are stabilized by resonace and have low heat of hydrogenation.Hyperconjugative stabilization also decreases heat f hydrogenation.In aromic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like -NO, -NC, -CH=CH_2 can function both as electron releasing (+m,+R) or electron withdrawing (-m,-R) groups.More extended conjugation provides more stabilization. The most stable carbocation is :
Answer» SOLUTION :

12.	The concept of resonance explains various properties of compounds.The molecules with conjugated system of pi bonds, are stabilized by resonace and have low heat of hydrogenation.Hyperconjugative stabilization also decreases heat f hydrogenation.In aromic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like -NO, -NC, -CH=CH_2 can function both as electron releasing (+m,+R) or electron withdrawing (-m,-R) groups.More extended conjugation provides more stabilization. The correct heat of hydrogenation order is (p)1,3-Pentadiene , (q)1,3-Butadiene , (r)2,3-Dimethyl-1,3-butadiene , (s)Propadiene
Answer» pgtqgtrgts sgtqgtpgtr qgtsgtpgtr sgtpgtqgtr Solution :

16.	The concept of electrode potential is useful in the isolation of metals.
Answer» ANSWER :1

18.	The concept of critical temperature for a gas was given by :
Answer» Andrew Boyle Charles None Answer :A

19.	The concentrations of N_(2)O_(5) decomposingin firstorder kinetics after 800 s is1.45 mol L^(-1) andafter1600 s is 0.88 mol L^(-1). Calculate the rateconstant.
Answer» ANSWER :`6.2 XX 10^(-4) s^(-1)`

20.	The concentration term independent of temperature is molarity/molality.
Answer» SOLUTION :MOLALITY

21.	The concentration R in the reaction RtoP was measured as a function of time and the following data is obtained. The order of reaction is
Answer» Solution :This follows rate equation for ZERO order REACTIONS. `(d[R])/(dt) = (1-0.75)/(0.05-0) = 0.25/0.05 =5` `(d[R])/(dt) = (0.75-0.40)/(0.12-0.05) = 0.35/0.07=5`

22.	The concentrationfo R in the reaction RtoP was measured as a function of time and the following data is obtained: [R] (molar)""1.0""0.75""0.40""0.10 t(min)""0.0""0.05""0.12""0.18 The order of the reactionis
Answer» SOLUTION :`r_(1)=0.25/0.05=25/5=5` M/MIN, `r_(2)=0.35/0.07=35/7=5M`/min, `r_(3)=0.30/0.06=30/6=5`M/min r=K = constant so ZERO ORDER

24.	The concentration of the reactant A in the reaction AtoB at different times are given below: "Concentration "(M)"""Time(seconds)" 0.069""0 0.052""17 0.035""34 0.018""51 The rate constant of the reaction according to the correct order of reaction is
Answer» `0.001 M^(-2)s^(-1)` `0.001 M^(-2)s^(-1)` `0.001 s^(-1)` `0.001Ms^(-1)` ANSWER :D

25.	The concentration of the aqueous solutions of each of the following salts is 0.01 (M) KF,CH_3N^(o+)H_3C^(o+)l,KCl,CH_3COOH If K_a(HF)=7.0xx10^(-4),K_b(CH_3NH_2)=5xx10^(-4) and K_a(CH_3COOH)=1.8xx10^(-5) then increasing order of the pH values of the solutions-
Answer» `KClltKFltCH_3N^(o+)H_3Cl^(o+)ltCH_3COOH` `KClltKClltCH_3N^(o+)H_3Cl^(o+)ltCH_3COOH` `CH_3COOHltCH_3N^(o+)H_3C^(o+)lltKClltKF` `CH_3COOHltKClltCH_3N^(o+)H_3C^(o+)lltKF` ANSWER :C

27.	The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M_((s))\|M^(+)(aq,0.05" molar")\|\|M^(+)(aq,1" molar")\|M_((s)) For the above electrolytic cell the magnitude of the cell potential \|E_(cell)\|=70mV. Q. For the above cell
Answer» `E_(cell) LT 0, DELTAG gt 0` `E_(cell) gt 0,DeltaG lt 0` `E_(cell) lt 0, DeltaG^(@) gt 0` `E_(cell) gt 0, DeltaG^(@) lt 0` SOLUTION :`E_(cell)=0-(2.303RT)/(F)"log"(0.05)/(1)` =a POSITIVE value =70mV Hence `DeltaG lt 0`. [GIVEN]

28.	The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M_((s))\|M^(+)(aq,0.05" molar")\|\|M^(+)(aq,1" molar")\|M_((s)) For the above electrolytic cell the magnitude of the cell potential \|E_(cell)\|=70mV. Q. if the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solution,then the magnitude of the cell potential would be
Answer» 35 mV 70 mV 140 mV 700 mV Solution :`E_(CELL)=0-(2.303RT)/(F)"LOG"(0.0025)/(1)` `=(2.303RT)/(F)log(0.05)^(2)=2xx70mV=140mV`.

29.	The concentration of R in the reaction R to P was measured as a funcation of time and the following data is obtained:
Answer» ANSWER :0

30.	The concentration of pure solid and liquid phase is not inculded in the expression of equilibrium constant because:
Answer» SOLID and liquid CONC. are independent of their quarries. solid and liquid react slowly. solid and liquid at equilibrium don't INTERACT with gaseous phase the molecules of solid and liquid cannot MIGRATE to the gaseous phase. Answer :A

31.	The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such concentration cell involving a metal M is : M_((s)) \|M_((aq , 0.05(" molar"))^(+) \|\|M_((aq, 1 " molar"))^(+) \| M_(s) For the above electrolytic cell the magnitude of the cell potential \|E_("cell")\|=70mV. If the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solution, then the magnitudethe cell potential would be
Answer» `35mV ` `70 mV ` `140 mV ` `700 mV ` SOLUTION :`E_1 = E_0^(0) - (2.303RT)/(NF) log(0.05), 70 =0 -(2.303RT)/(nF) log 0.05` `((2.303RT)/(nF)) = ((70)/(1.301)), " But " E_2 = E^0 - ((2.303 RT)/(nF)) XX log(2.5 xx 10^(-4))` `E_2 = 0 - ((70)/(1.303)) xx (-2.602) , E_2 = 140 mV `

32.	The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such concentration cell involving a metal M is : M_((s)) \|M_((aq , 0.05(" molar"))^(+) \|\|M_((aq, 1 " molar"))^(+) \| M_(s) For the above electrolytic cell the magnitude of the cell potential \|E_("cell")\|=70mV. For the above cell
Answer» `E_("cell") lt 0 , DELTAG GT 0` `E_("cell") gt 0 , DeltaG lt 0` `E_("cell") lt 0 , DeltaG^(0) gt 0` `E_("cell") gt 0 , DeltaG^(@) lt 0` SOLUTION :`E_("cell") = E^(0) - (0.059)/(1) LOG""(0.05)/(1) , E_("cell") gt 0 , DeltaG lt 0`

33.	The concentration ofpotassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple mode for such a concentration cell involving a metal M is : M_((s))\|M_((aq.))^(@) 0.05"molar"\|\|M_((aq.))^(@) 1"molar"\|M_((s)) For the above electrolytic cell the magnitude of the cell potential \|E_(cell) \| = 70mV If the 0.05 molar solution of M^(+) is replaced by 0.0025 molar M^(+) solution, then the magnitude of the cell potential would be :
Answer» `35 mV` `70 mV` `140 mV` `700 mV` Solution :In this case `E_(cell) = (0.059)/(1)"log"_(10)([1.0])/(2.5 xx 10^(-3))` `= 0.152 V` or `APPROX 140 mV`

36.	The concentration of potassium ions inside a bilological cell is at least twenty times higher than the outisde. The resulting potential difference across the cell is important in several process such as transmission of nerve impuses and maintaning the ion balance. A simple model for such a concentration cell involving a metal M is M(s)\|M^(+)(aq,0.05" molar")\|\|M^(+)(aq,1" molar")\|M(s) For the above electrolytic cell, the magnitude of the cell potential \|E_(cell)\|=70mV. Q. For the above cell
Answer» `E_(cell)lt0,DELTAGGT0` `E_(cell)gt0,DeltaGlt0` `E_(cell)lt0,DeltaG^(@)gt0` `E_(cell)gt0,DeltaG^(@)lt0` SOLUTION :For the given concentration cell, `E_(cell)=(2.303RT)/(nF)"log"(C_(2))/(C_(1))` `=(2.303RT)/(F)"log"(1)/(0.05)=+ve,` i.E., `E_(cell)gt0`. For the reaction to be spontaneous, `DeltaG lt 0`.

39.	The concentration of KI and KCl in certain solution containing both is 0.001M each. If 20 mL of this solution is added to 20 ml of a saturated of AgI in water ? What will happen
Answer» AgCl will be precipitated AGI will be precipitated Both AgCl and AgI will be precipitated There will be no precipitated Solution :Because of IONIC PRODUCT of `AgI gt gt` solubility product of its.

40.	The concentration of Kl and KCl in a certain solution containing both is 0.001 M each . If 20 mL of this solution is added to 20 mL of a saturated solution of Agl in water . What will happen ? K_(sp) ofAgCl = 10 ^(-10) , K_(sq)ofAgI = 10^(-16)
Answer» AGL will be PRECIPITATED AGCL will be precipitated Both AgCl and Agl will be precipitated There will be no PRECIPITATE ANSWER :A