Quiz about Chemistry in Class 12

Question 1

The process of getting freshwater from sea water is known as

Accepted Answer

NA

Question 2

The process of froth floatation and chromatography are based on

Accepted Answer

NA

Question 3

The process of formation of RNA from DNA is known as

Accepted Answer

NA

Question 4

The process of extraction of sodium on a commercial scale by the electrolysis of fused sodium chloride is called

Accepted Answer

NA

Question 5

The process of extracting the metal from its ore is called

Accepted Answer

NA

Question 6

The process of evaporation of a liquid is accompanied by

Accepted Answer

NA

Question 7

The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol turpentine oil caustic potash conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2) O_(3) CO_(2) and H_(2)O (v) respectively.  A 2.0 mole mixture of H_(2)(g)O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions.  The mole fraction of H_(2)(g) present in the initial mixture is

Accepted Answer

NA

Question 8

The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol turpentine oil caustic potash conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2) O_(3) CO_(2) and H_(2)O (v) respectively.  A 2.0 mole mixture of H_(2)(g)O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions.  The moles of O_(2)(g) introduced in the vessel after first electrical spark is

Accepted Answer

NA

Question 9

The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol turpentine oil caustic potash conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2) O_(3) CO_(2) and H_(2)O (v) respectively.  A 2.0 mole mixture of H_(2)(g)O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions.  The moles of He (g) present in the mixture samples is

Accepted Answer

NA

Question 10

The process of employed for the concentration of sulphide ore is

Accepted Answer

NA

Question 11

The process of covering iron sheet with a layer of zinc is called

Accepted Answer

NA

Question 12

The process of converting precipitate into colloidal solution on adding an electrolyte is called

Accepted Answer

NA

Question 13

The process of converting hydrated alumina to anhydrous alumina is called

Accepted Answer

NA

Question 14

The process of converting hydrated alumina into anhydrous alumina is called

Accepted Answer

NA

Question 15

The process of converting alkyl halides into alcohols involves _________reaction.

Accepted Answer

NA

Question 16

The process of converting alkyl halides into alcohols involves_____

Accepted Answer

NA

Question 17

The process of converting alkyl halide into alcohols involves .........

Accepted Answer

NA

Question 18

The process of converign one enatiomer of an optically active compound inot racemic mixture is called

Accepted Answer

NA

Question 19

The process of conversion of colloidal solution into precipitate is called .............

Accepted Answer

NA

Question 20

The process of conversion of colloidal solution into precipitate is known as .....

Accepted Answer

NA

Question 21

The process ofconcentrating silver are is based on its solubility in

Accepted Answer

NA

Question 22

What is the method of concentrating iron pyrite ore?

Accepted Answer

NA

Question 23

The process of combining metals present in their native ores with mercury to form an alloy is called

Accepted Answer

NA

Question 24

The process of calcination and roasting are carried out in

Accepted Answer

NA

Question 25

The process of AgCN+KCNtoK[Ag(CN)_(2)] involves the oxidation of Ag.

Accepted Answer

NA

Question 26

The process of bringing the metal or its ore into solution by the action of a suitable chemical reagnetfollowing by extraction of the metal either by electrolysis or by suitable precipitating agentis called

Accepted Answer

NA

Question 27

The process of a protein losing its higher order structure without losing the primary structure is called……………..

Accepted Answer

NA

Question 28

The process involving reduction of metal oxide with coke or carbon monoxide is called………..

Accepted Answer

NA

Question 29

The process is spontaneous at the given temperature if

Accepted Answer

NA

Question 30

The process involving transfer of five electrons is

Accepted Answer

NA

Question 31

The process involving heating of rubber with sulphur is called

Accepted Answer

NA

Question 32

The process involving heating of natural rubber with sulphur is known as

Accepted Answer

NA

Question 33

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. Which one of the following statements is correct regarding the experiment ?

Accepted Answer

NA

Question 34

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. In experiment -2what % of the metal ions get precipitated at which S^(2-) get saturated with another ion ?

Accepted Answer

NA

Question 35

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. In experiment 02 what is the maximum concentration of S^(2-)ion at which one of the two metal ions gets maximum precipitation ?

Accepted Answer

NA

Question 36

The process in which the concentrated ore is strongly heated in the absence of air is called as………

Accepted Answer

NA

Question 37

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. What %of the (ion) will get precipitate when the second ion start precipitating in experiment -1 ?

Accepted Answer

NA

Question 38

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. What percent of the anions Br^(-) and I^(-) get precipitated respectively when the third ion starts precipitating in experiment -1 ?

Accepted Answer

NA

Question 39

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. Which of the following will precipitate first in experiment -1 ?

Accepted Answer

NA

Question 40

The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the saltnot the solubility product of the salt by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example in a solution containing Cl^(-) Br^(-) and I^(-) ions if Ag^(+) ions are added then out of three the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.  If the stiochiometry of the salt is same then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same then from the solubility product data alone we cannot predict which ions will precipitate first. For example a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together.  Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount.  Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-)Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2)K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2))  Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively.  (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. Which of the following is the correctorder of appearance of coloour of the precipitate in experiment ?

Accepted Answer

NA

Question 41

The process in which pressure remains constant throughout a change is

Accepted Answer

NA

Question 42

The process in which no heat enters or leaves the system is termed as

Accepted Answer

NA

Question 43

The process in which metalsurface is madeinactiveis called

Accepted Answer

NA

Question 44

The process in which metal surface is made inactive is called

Accepted Answer

NA

Question 45

The process in which higher hydrocarbons are broken down into lower hydrocarbons by controlled pyrolysis is called

Accepted Answer

NA

Question 46

The process in which chemical change occurs on passing electricity is termed as __________.

Accepted Answer

NA

Question 47

The process in which chemical change occurs on passing electricity is termed

Accepted Answer

NA

Question 48

The process in which chemical change accompanies the passage of current is called

Accepted Answer

NA

Question 49

The process in which lighter earthy particles are freed fromthe heavier particle by washing with water is called

Accepted Answer

NA

Question 50

The process (es) by which lighter earthy particles are freed from the heavier particles using water is are

Accepted Answer

NA

1.	The process of getting freshwater from sea water is known as:
Answer» Osmosis Filtration Diffusion Reverse Osmosis Answer :D

2.	The process of froth floatation and chromatography are based on :
Answer» Emulaification Adsorption Absorption Either of them Answer :D

3.	The process of formation of RNA from DNA is known as :
Answer» Translation Transcription Replication Mutation Answer :A

4.	The process of extraction of sodium on a commercial scale by the electrolysis of fused sodium chloride is called :
Answer» ANSWER :A

5.	The process of extracting the metal from its ore is called :
Answer» REFINING CONCENTRATION LEACHING METALLURGY ANSWER :D

6.	The process of evaporation of a liquid is accompanied by:
Answer» INCREASE in enthalpy Decrease in FREE energy Increase in entropy All Answer :D

7.	The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2), O_(3), CO_(2) and H_(2)O (v) respectively. A 2.0 mole mixture of H_(2)(g),O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions. The mole fraction of H_(2)(g) present in the initial mixture is:
Answer» 0.05 0.7 0.5 1.25 Solution :Let the no. of moles of `H_(2),O_(2)` and He are x,y and z respectively `therefore x+y+z=20` Pressure exerted by 2 moles of gaseous MIXTURE is 50 atm. After the first electric spark decrease in pressure =50-12.5=37.5 atm `therefore` decrease in no. of moles of gaseous mixture is 1.5 But from the given information limiting reagent in first step is `O_(2)` `therefore" from "H_(2)(g)+(1)/(2)O_(2)(g) RARR H_(2)O(l)` `therefore H_(2) and O_(2)` REACTED are in the ratio of 2:1 `therefore` 0.5 mole of `O_(2)` should be reacted with 1.0 mole of `H_(2)` `therefore` No. of mole of `O_(2)` in the mixture (y) =0.5 When again `O_(2)` is passed, pressure increased from 12.5 to 25.0 atm `therefore` Change in pressure =12.5 atm `therefore` No. of moles of `O_(2)` added `=(12.5xx2)/(50)=0.5` mole Now `H_(2)` will be completely reacted No. of moles `H_(2)` actually left after first electric spark =x-1 `H_(2)+(1)/(2)O_(2) rarr H_(2)O` `(x-1)((x-1)/(2))` Now change in pressure in 25-10 =15 atm `therefore` change in no. of moles is `(15xx2)/(50)=0.6` `therefore (x-1)+((x-1)/(2))=0.6` `rArr x=1.4` `therefore` no of moles `H_(2)=1.4` `therefore` no. of moles of `He=2.0-(1.4+0.5)=0.1` `therefore` mole fraction of `H_(2)=(1.4)/(2.0)=0.7`

8.	The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2), O_(3), CO_(2) and H_(2)O (v) respectively. A 2.0 mole mixture of H_(2)(g),O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions. The moles of O_(2)(g) introduced in the vessel after first electrical spark is :
Answer» 0.25 0.5 0.1 0.2 Solution :Let the no. of moles of `H_(2),O_(2)` and He are x,y and z respectively `therefore x+y+z=20` Pressure exerted by 2 moles of gaseous mixture is 50 atm. After the first electric spark DECREASE in pressure =50-12.5=37.5 atm `therefore` decrease in no. of moles of gaseous mixture is 1.5 But from the given information limiting reagent in first STEP is `O_(2)` `therefore" from "H_(2)(G)+(1)/(2)O_(2)(g) rarr H_(2)O(l)` `therefore H_(2) and O_(2)` reacted are in the ratio of 2:1 `therefore` 0.5 MOLE of `O_(2)` should be reacted with 1.0 mole of `H_(2)` `therefore` No. of mole of `O_(2)` in the mixture (y) =0.5 When again `O_(2)` is passed, pressure increased from 12.5 to 25.0 atm `therefore` Change in pressure =12.5 atm `therefore` No. of moles of `O_(2)` added `=(12.5xx2)/(50)=0.5` mole Now `H_(2)` will be completely reacted No. of moles `H_(2)` actually left after first electric spark =x-1 `H_(2)+(1)/(2)O_(2) rarr H_(2)O` `(x-1)((x-1)/(2))` Now change in pressure in 25-10 =15 atm `therefore` change in no. of moles is `(15xx2)/(50)=0.6` `therefore (x-1)+((x-1)/(2))=0.6` `rArr x=1.4` `therefore` no of moles `H_(2)=1.4` `therefore` no. of moles of `He=2.0-(1.4+0.5)=0.1` `therefore` mole fraction of `H_(2)=(1.4)/(2.0)=0.7`

9.	The process of eudiometry is used to calculate the volume of the various reacting gases present together in a closed vessel, it is done by sparking the content and noting the volume change or pressure change depending upon the conditions mentioned. Various reagents like alkaline pyrogallol, turpentine oil, caustic potash, conc. H_(2)SO_(4) are used to absorb theselective gases like O_(2), O_(3), CO_(2) and H_(2)O (v) respectively. A 2.0 mole mixture of H_(2)(g),O_(2)(g) and He(g) are placed together in a closed container at pressure equal to 50 atm. An electric spark is passed and pressure is noted as 12.5 atm after cooling the content to original temp. (room temp). Oxygen gas is introduced for pressure to change to 25 atm, keeping volume and temp constant. Again electrical spark is passed and pressure drops to 10 atm under original temp. conditions. The moles of He (g) present in the mixture samples is
Answer» 0.2 0.1 0.5 1.4 Solution :Let the no. of moles of `H_(2),O_(2)` and He are x,y and z respectively `therefore x+y+z=20` Pressure EXERTED by 2 moles of gaseous MIXTURE is 50 atm. After the FIRST electric spark decrease in pressure =50-12.5=37.5 atm `therefore` decrease in no. of moles of gaseous mixture is 1.5 But from the given information limiting reagent in first step is `O_(2)` `therefore" from "H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l)` `therefore H_(2) and O_(2)` reacted are in the ratio of 2:1 `therefore` 0.5 MOLE of `O_(2)` should be reacted with 1.0 mole of `H_(2)` `therefore` No. of mole of `O_(2)` in the mixture (y) =0.5 When again `O_(2)` is passed, pressure increased from 12.5 to 25.0 atm `therefore` Change in pressure =12.5 atm `therefore` No. of moles of `O_(2)` added `=(12.5xx2)/(50)=0.5` mole Now `H_(2)` will be completely reacted No. of moles `H_(2)` actually left after first electric spark =x-1 `H_(2)+(1)/(2)O_(2) rarr H_(2)O` `(x-1)((x-1)/(2))` Now change in pressure in 25-10 =15 atm `therefore` change in no. of moles is `(15xx2)/(50)=0.6` `therefore (x-1)+((x-1)/(2))=0.6` `rArr x=1.4` `therefore` no of moles `H_(2)=1.4` `therefore` no. of moles of `He=2.0-(1.4+0.5)=0.1` `therefore` mole fraction of `H_(2)=(1.4)/(2.0)=0.7`

10.	The process of employed for the concentration of sulphide ore is
Answer» Froth-floatation Roasting Electrolysis Calcination Answer :A

17.	The process of converting alkyl halide into alcohols involves .........
Answer» ADDITION reaction SUBSTITUTION reaction Dehydrohalogenation reaction Rearrangement reaction Solution :`R- X OVERSET(NaOH)to R- OH+ NAX`

24.	The process of calcination and roasting are carried out in:
Answer» BLAST furnace Muffle furnace Reverberatory furnace Open HEARTH furnace Answer :C

29.	The process is spontaneous at the given temperature, if
Answer» `Delta H" is " +ve and Delta S " is " -ve` `Delta H " is " -ve and Delta S " is " +ve` `Delta H " is " +ve and Delta S " is " +ve` `Delta H "is " and DeltaS` is EQUAL to zero. Solution :`Delta G = Delta H - T Delta S` (Gibbs ENERGY EQUATION) When, `Delta H` is -ve, but `T Delta S` is `+ve rarr Delta G` will be (-ve) [Highly spontaneous].

30.	The process involving transfer of five electrons is:
Answer» `CrO_4^(2-) rarrCr^(3+) `MnO_4^- rarrMn^(2+)` `MnO_4^- rarrMnO_2` `Cr_2O_7^(2-) rarr2Cr^(3+)` ANSWER :A

11.	The process of covering iron sheet with a layer of zinc is called:
Answer» Galvanizing Zinc plating Tempering Rusting Answer :A

12.	The process of converting precipitate into colloidal solution on adding an electrolyte is called
Answer» Peptisation Dialysis Electro-osmosis Electrophoresis Solution :The conversion of a FRESH PPT. into colloidal STATE by ADDITION of small amount of a suitable electrolyte is called peptization e.g. i.e., ions are adsorbed on particles to give colloidal SIZE.

13.	The process of converting hydrated alumina to anhydrous alumina is called
Answer» Calcination Smelting Rosting Concentration Answer :A

14.	The process of converting hydrated alumina into anhydrous alumina is called :
Answer» Roasting Smelting Dressing Calcination Answer :D

15.	The process of converting alkyl halides into alcohols involves _________reaction.
Answer» addition substitution dehydrohalogenation rearrangement Answer :B

18.	The process of converign one enatiomer of an optically active compound inot racemic mixture is called:
Answer» Resolution Inversion Epimerisation Racemission Solution :NA

19.	The process of conversion of colloidal solution into precipitate is called .............
Answer» SOLUTION :COAGULATION

20.	The process of conversion of colloidal solution into precipitate is known as .....
Answer» PEPTISATION dispersion COAGULATION decomposition Solution :coagulation

21.	The process ofconcentrating silver are is based on its solubility in :
Answer» HCL `HNO_3` KCN NaoH Answer :C

22.	What is the method of concentrating iron pyrite ore?
Answer» ANSWER :C

Explore topic-wise InterviewSolutions in Current Affairs.

The process of getting freshwater from sea water is known as:

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The process of converting hydrated alumina into anhydrous alumina is called :

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The process of converting alkyl halides into alcohols involves_____

The process of converting alkyl halide into alcohols involves .........

The process of converign one enatiomer of an optically active compound inot racemic mixture is called:

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The process of conversion of colloidal solution into precipitate is known as .....

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The process of combining metals present in their native ores with mercury to form an alloy is called

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The process of bringing the metal or its ore into solution by the action of a suitable chemical reagnetfollowing by extraction of the metal either by electrolysis or by suitable precipitating agentis called

The process of a protein, losing its higher order structure without losing the primary structure is called……………..

The process involving reduction of metal oxide with coke or carbon monoxide is called………..

The process is spontaneous at the given temperature, if

The process involving transfer of five electrons is:

The process involving heating of rubber with sulphur is called

The process involving heating of natural rubber with sulphur is known as:

The process in which the concentrated ore is strongly heated in the absence of air is called as………

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The process, in which no heat enters or leaves the system, is termed as

The process in which metalsurface is madeinactiveis called

The process in which metal surface is made inactive is called

The process in which higher hydrocarbons are broken down into lower hydrocarbons by controlled pyrolysis, is called

The process in which chemical change occurs on passing electricity is termed as __________.

The process in which chemical change occurs on passing electricity is termed:

The process in which chemical change accompanies the passage of current is called :

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The process (es) by which lighter earthy particles are freed from the heavier particles using water is / are :

23.	The process of combining metals present in their native ores with mercury to form an alloy is called
Answer» ALLOYING GALVANISATION amalgamation crystallisation Answer :C

25.	The process of AgCN+KCNtoK[Ag(CN)_(2)] involves the oxidation of Ag.
Answer» SOLUTION :The OXIDATION NUMBER is not CHANGING for AG.

26.	The process of bringing the metal or its ore into solution by the action of a suitable chemical reagnetfollowing by extraction of the metal either by electrolysis or by suitable precipitating agentis called
Answer» Electrometallurgy Electro-refining Hydrometallurgy Zone-refining Answer :C

27.	The process of a protein, losing its higher order structure without losing the primary structure is called……………..
Answer» SOLUTION :DENATURATION

28.	The process involving reduction of metal oxide with coke or carbon monoxide is called………..
Answer» SOLUTION :SMELTING

31.	The process involving heating of rubber with sulphur is called
Answer» galvanisation VULCANIZATION bessemerisation sulphonation. Solution :Vulcanization

32.	The process involving heating of natural rubber with sulphur is known as:
Answer» VULCANISATION GALVANISATION SULPHONATION BESSEMERISATION Answer :B

33.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing Cl^(-), Br^(-) and I^(-) ions, if Ag^(+) ions are added, then out of three , the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first, we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage, practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-),Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2)) Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively. (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. Which one of the following statements is correct regarding the experiment ?
Answer» First of all YELLOW ppt. will appear First of all black ppt. will appear Appearance of yellow ppt. and black ppt. will OCCU simultaneously Cannot be PREDICTED Answer :A

34.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing Cl^(-), Br^(-) and I^(-) ions, if Ag^(+) ions are added, then out of three , the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first, we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage, practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-),Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2)) Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively. (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. In experiment -2what % of the metal ions get precipitated at which S^(2-) get saturated with another ion ?
Answer» `10%` `80%` `20%` `90%` ANSWER :A

35.	The process in which solution containing more than one typeof ions (either cation or anion) and in which one ion undergoes almst complete precipitation followed by the second ion and similarly second ion undergoes complete precipitation followed by the third ion is known as preferentialprecipitation. It is the solubility of the salt,not the solubility product of the salt, by means of which one can predict the preferential precipitation of the salt or ion among the lot of ions in the solution. For example , in a solution containing Cl^(-), Br^(-) and I^(-) ions, if Ag^(+) ions are added, then out of three , the less soluble salt is precipitated first . If the addition of Ag^(+) ions is continued, eventually , a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stiochiometry of the salt is same, then the salt with the minimum K_(sp) ( solubility product ) will have minimum solubility and will precipitate first, followed by the salt of next higher solubility product and so on. If the stiochiometry of teh salt is not same, then from the solubility product data alone, we cannot predict which ions will precipitate first. For example, a solution containing Cl^(-) and CrO_(4)^(2-). In order to predict which ion will precipitate first, we have to calculate the amount of Ag^(+) ions needed to start precipitation through the solubility productdata given. When AgNO_93) is added to the solution, the minimum of the two concentrations needed to start the precipitation will be reached and thus corresponding [Ag^(+)] to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO_(4)^(2-) . At this stage, practically almost the whole of Cl^(-) ions get precipitated . Addition of more AgNO_(3) causes simultaneous precipitation of both the ions together. Suppose a student of chemistry performs the following two sets of experiments in order to known the preferential precipitation of ions as well as its precipitated percentage amount. Experiment 1. He prepare a solution of anions composed of 0.1 M each of Cl^(-),Br^(-) and I^(-) . Further he adds gradually solid AgNO_(3) to this solution. He assumes thatvolume of the solution does not change after the addition of solid AgNO_(3)(K_(sp)AgCl=1.0 xx 10^(-9)M^(2),K_(sp) AgBr=10^(-10)M^(2) and K_(sp) AgI=10^(-12)M^(2)) Experiment -2. He prepares a solution of cations Cd^(2+)(0.2 M) and Bi^(3+) (0.3 m) . Now he adds S^(2-) ions into the solution of cations in order to separate them by selective precipitationCd^(2+) forms yellow precipitate of CdSand Bi^(3+) forms black precipitate of Bi_(2)S_(3) with S^(2-) ions respectively. (K_(sp)Bi_(2)S_(3) = 9 xx 10^(-25) M^(5) and K_(sp)CdS =2 xx 10^(-20)M^(2)).Answer the following questions ont he basis of the above write up. In experiment 02, what is the maximum concentration of S^(2-)ion at which one of the two metal ions gets maximum precipitation ?
Answer» `10^(-8)` M `2 XX 10^(-7)M` `2 xx 10^(-9)M` None Answer :A