94420 + Interview Questions in Chemistry in Class 12

1.	Silver nitrate is prepared by
Answer» The action of only CONC. `HNO_(3)` on silver Heating silver oxide with `NO_(2)` The action of HOT DIL. `HNO_(3)` on silver Dissolve Ag in aqua-regia Solution :`3Ag+underset("DILUTE")(4HNO_(3))OVERSET("heat")rarr3AgNO_(3)+NO+2H_(2)O`

Discussion

2.	Silver Mirror test is given by which one of the following compounds ?
Answer» Acetaldehyde Acetone Formaldehyde Benzophenone Solution :

Discussion

3.	Silver metal is extracted from Ag_(2) S by cyanide process. The one is concentrated through froth floation process. The concentrated ore is the leached and solution is r ed with reducing agents to get spongy silver which then purified by fusion with an oxidant followed by electrolysis. Which of the following statements about electrolytic refining of silver is not true
Answer» Anode consists of impure silver cathode consists of PURE silver Electrolytic solution consists of `AgNO_(3) and HNO_(3)` Elecipitated solution consists of `AgNO_(3) and HCl` Solution :Pure Ag is obtained by the electrolysis of `AgNO_(3)` SOL containing 10% `HNO_(3)` and using the impure silver as anode and pure silver plate as cathode.

Discussion

4.	Silver nitrate is mainly used
Answer» In photography In MODEL formation As REDUCING agent As DEHYDRATING agent Solution :In photography as it is sensitive towards light.

Discussion

5.	Silver metal is extracted from Ag_(2) S by cyanide process. The one is concentrated through froth floation process. The concentrated ore is the leached and solution is r ed with reducing agents to get spongy silver which then purified by fusion with an oxidant followed by electrolysis. Ag is precipitated from an aqueous solution of sodium argentocynide by adding
Answer» Zn dust Cu power NaHg `Na_(2)S_(2)O_(3)` Solution :Recovery of Ag : `UNDERSET("Sodium ARGENTO cyanide")(2Na[Ag(CN)_(2)]+Zn) to underset("Sodium Zincoyanide")(Na_(2)[Zn(CN)_(4)]+2Ag)`

Discussion

6.	Silver metal is extracted from Ag_(2) S by cyanide process. The one is concentrated through froth floation process. The concentrated ore is the leached and solution is r ed with reducing agents to get spongy silver which then purified by fusion with an oxidant followed by electrolysis. Leaching of Ag_(2)S is carried out by heating it with a dilute solution of
Answer» NACN alone NaCN in presence of `O_(2)` HCI NaOH Solution :Leaching of `Ag_(2)S: 2NA[AG(CN)_(2)]` `4g_(4)S+4NaCN hArr 2Na Ag(CN)_(2)+Na_(2)S` `4Na_(2)S+5O_(2)+2H_(2)O to 2Na_(2)SO_(4)+4NaOH+2S`

Discussion

7.	Silver metal is extracted by cyanide process from silver glance (Ag_(2)S). The ore is concentrated through forth floatation process. The concentrated ore is then leached and solution is treated with reducing agent to get spongy silver which is then purified by fusion with an oxidsing agent followed by electrolysis.
Answer» Silver is precipitated from an aqueous solution of sodium argntocyanide by adding Zinc dust Copper powder Sodium amalgam Solution :`2Na[Ag(CN)_(2)]+UNDERSET("dust")(Zn) to underset(["VIA Hydrometallurgical process"])(Na[Zn(CN)_(4)]+2Ag)`

Discussion

8.	Silver metal is extracted by cyanide process from silver glance (Ag_(2)S). The ore is concentrated through forth floatation process. The concentrated ore is then leached and solution is treated with reducing agent to get spongy silver which is then purified by fusion with an oxidsing agent followed by electrolysis. Leaching of Ag_(2)S is carried out by heating it with a dilute solution of
Answer» NaCN alone NaCN in presence of `O_(2)` HCl NaOH Solution :Leaching of `Ag_(2)S` measn DISSOLUTION. `2Ag_(2)S+8NaCN+O_(2) 2H_(2) O to 4NA[Ag(CN_(2))]+4NaOH+2S`

Discussion

9.	Silver metal is extracted by cyanide process from silver glance (Ag_(2)S). The ore is concentrated through forth floatation process. The concentrated ore is then leached and solution is treated with reducing agent to get spongy silver which is then purified by fusion with an oxidsing agent followed by electrolysis. Which of the following statement about electrolytic refining of silver is not true
Answer» Anode consists of impure silver Cathode of impure silver Electrolytic SOLUTION consists of `AgNO_(3)` and nitric acid Electrolytic consists of `AgNO_(3)` and hydrochloric acid Solution : Electrolyticsolution cannot consist of `AgNO_(3) and HCl` because Ag will GET precipitarted as AGCL in the solution.

Discussion

10.	Silver metal crystallizes with a face centred cubic lattice. The length of the unit cell is found to be 4.077 xx 10^(-8) cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108 u, N_A = 6.02 xx 10^(23) mol^(-1))
Answer» Solution :(i) `d = (Z xx M)/(a^3 xx N_A)""...(1)` According to the given DATA: `a =4.077 xx 10^(-8)` cm, Z=4, M = 108 G `mol^(-1)`, `N_A=6.02 xx 10^(23) g mol^(-1)` SUBSTITUTING these values in the EXPRESSION (1), we get `d=(4 xx 108 g mol^(-1))/((4.077 xx 10^(-8) cm)^3 xx 6.022 xx 10^(23) mol^(-1))` = 10.58 g `cm^(-3)` `R=(a)/(2sqrt2)=(4.077 xx 10^(-8) cm)/(2sqrt2) = 1.44 xx 10^(-8) cm`

Discussion

11.	Silver metal crystallises with a face centred cubic lattice. The length of unit cell is formed to be 4.077 xx 10^(-8) cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108 u, N_(A) = 6.023 xx 10^(23) mol^(-1)).
Answer» Solution :GIVEN : `a = 4.077 xx 10^(-8)` cm, Z = 4, M = 108 g `MOL^(-1)` `N_(A) = 6.023 xx 10^(23)` `d = (Z xx M)/(a^(3) xx N_(A))` `= (4 xx 108)/((4.077 xx 10^(-8))^(3) xx 6.023 xx 10^(23))` `= 6.22 g//cm^(3)` For fcc `r = (a)/(2 SQRT(2))` `= (4.077 xx 10^(-8) cm)/(2 sqrt(2))` `= 1.44 xx 10^(-8)` cm

Discussion

12.	Silver, mercury (ous) and lead are grouped together in the scheme of qualitative analysis because they form
Answer» soluble nitrates carbonates which DISSOLVE in DILUTE nitricacid insoluble chlorides All of above Solution :Insoluble chlorides are formed by `Hg_(2)^(2+), PB^(2+)` and `Ag^(+)`

Discussion

13.	Silver, mercury (ous) and lead are grouped together in a scheme of qualitative analysis because they form
Answer» soluble NITRATES CARBONATES which DISSOLVE in dilute nitric acid insoluble CHLORIDES all of these Solution :Insoluble chlorides are formed by `Hg_(2)^(2+) , Pb^(2+)` and `AG^(+)`

Discussion

14.	Silver, mercury and lead have been placed in same group of qualitative analysis, because they form:
Answer» Carbonates SOLUBLE in DILUTE `HNO_3` Nitrates Insolule chlorides Same TYOE of coloured compounds Answer :C

Discussion

15.	Silver may be obtained by __________.
Answer» reduction of `Ag_(2)S` by (Zn+HCl) melting `Ag_(2)S` with NaCl `Ag_(2)S` TREATED with NACN then by adding zinc POWDER `Ag_(2)S` when treated with ZnS Answer :C

Discussion

16.	Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm^2 by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited.
Answer» Solution :CALCULATIONS of MASS of Ag deposited: The electrode reaction is `Ag^(+)+e^(-) to Ag` The QUANTITY of ELECTRICITY passed `= Current times time` `=0.5 (amp.) times 2 times 60 times 60 (sec)` =3600 C. From the electrode reaction, it is clear that 96500C of electricity deposit Ag=108g 360 C of electricity will deposit Ag=`108/96500 times 3600` =4.03g Calculation of thickness: Let the thickness of deposited be x cm. Mass= Volume `times` density = Area `times` thickness `times` density (Volume= Area` times` thickness) `4.03g=900(cm^2) times x (cm) times 10.5 (G cm^-3)` `x=4.03/(900 times 10.5)=4.26 times 10^-4 cm`

Discussion

17.	Silver iodide is used for producing artificial rain because AgI
Answer» is EASY to spary at high altitude is easy to synthesize has CRYSTAL structure SIMILAR to ICE is INSOLUBLE in water, Answer :C

Discussion

18.	Silver is removed electrically from 200 ml of a 0.1 N solution of AgnO_(3) by a current of 0.1 ampere. How long will it take to remove half of the silver from the solution.
Answer» 16 sec 96.5 sec 100 sec 10 sec Answer :B

Discussion

19.	Silver is removed electrolytically from 200 mL of a 0.1 N solution of AgNO_3by a current of 0.1 ampere. Hoe long will it take to remove half of the silver from the solution:
Answer» 10 sec 16 sec 100 sec 9650 sec Answer :D

Discussion

20.	Silver is monovalent and has atomic mass of 108. Copper is divalent and has an atomic mass of 63.6. The same electric current is passed for the same length of time through a silver couloment and a copper coulometer. If 27.0 g of silver is deposited, then the corresponding amount of copper deposited is
Answer» 63.60 g 31.80 g 15.90 g 7.95 g Answer :D

Discussion

21.	Silver is extracted from argentiferous lead by :
Answer» MOND PROCESS Parkes process HABER process Bergius process ANSWER :B

Discussion

22.	Silver is extracted by
Answer» CUPELLATION METHOD Parke's PROCESS Pattinson's process All Solution :SILVER is extracted by all process. These three processes can be USED for the removal of Ag from Pb

Discussion

23.	Silver is exfracted by:
Answer» CUPELLATION method Parke.s process Pattinson.s process All Answer :D

Discussion

24.	Silver is electro-deposited on a metallic vessel of surface area 800 cm^(2) by passing current 0.2 ampere for 3 hours. Calculate the thickness of silver deposited. Given the density of silver as 10.47 g/cc (Atomic mass of Ag=107.92amu)
Answer» Solution :Quantity of electricity passed`=0.2xx3xx60xx60C=2160C` `At^(+)+e^(-)toAg` 96500 C deposit `Ag=107.92g` 2160 C will deposit `Ag=(107.92)/(96500)xx2160g=2.4156g` VOLUME DEPOSITED`=(Mass)/(Density)=(2.4156)/(10.47)c c=0.2307c c` THICKNESS deposited`=("Volume")/("Area")=(0.2307)/(800)=2.88xx10^(-4)cm`

Discussion

25.	Silver is a soft metal . It is hardened by alloying it with small amount of :
Answer» ANSWER :D

Discussion

26.	Silver iodide is used to produce artificial rain because :
Answer» It is easily prepared Its structure is ice- like It can easily be sprayed at HIGH altitude It is INSOLUBLE in rain water ANSWER :B

Discussion

27.	Silver halides are used in photography because they are :
Answer» Photosensitive SOLUBLE in HYPO SOLUTION Soluble in `NH_4OH` Insoluble in acids Answer :A

Discussion

28.	Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).
Answer» Solution : Since the lattice is ccp, the number of Ag atoms per unit cell (z) = 4 Molar mass of Ag `= 107.9 "g mol"^(-1)` Edge length of (a) = 408.6 PM Volume of cube = `(408.6)^3 xx 10^(-30) "CM"^3` `d = (Z xx M)/(N_A xx a^3)` `therefore d= (4 xx (107.9 "g mol"^(-1)))/((6.022 xx 10^(23) "atomsmol"^(-1)) xx (408.6)^3 xx 10^(-30) "cm"^3)` `therefore d = 10.5 "g cm"^(-3).`

Discussion

29.	Silver from silver nitrate is deposited by copper, because :
Answer» `E^@(CU^(2+)//Cu) lt E^@(Ag^+//Ag)` `E^@(Cu^(2+)//Cu)gt E^@(Ag^+//Ag)` `E^@(Cu^(2+)//Cu)=E^@(Ag^+//Ag)` None Answer :A

Discussion

30.	Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9u).
Answer» Solution : Edge length = 408.6 pm = `4.086 xx 10^(-8)` CM M=107.9 U, For ccp, Z = 4, `d =(ZM)/(N_A a^3)G cm^(-3)=(4 xx 107.9)/(6.023 xx 10^(23) xx (4.086)^3 xx 10^(-24)cm^3)= 10.5 g cm^(-3)`

Discussion

31.	Silver crystallizes in CCP lattice. The edge length of its unit cell is 408.6 pm. Calculate density of silver (atomic mass of silver is 107.9)
Answer» SOLUTION :here, `Z=4, a=408.6xx10^(-12)m` `:.d=(ZxxM)/(a^3xxN_A)=(4xx108)/((408.6xx10^(-12))^3xx6.023xx10^(23))=10.5g//cm^3`

Discussion

32.	Silver forms ccp lattice and x-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver. (Atomic mass of Ag = 107.9 u)
Answer» Solution :Since it is, cop the number of atoms PER UNIT cell, Z = 4 Molar MASS of silver,`M=108 g mol^(-1)` Edge length of the unit cell, `a=408.6` pm. `=408.6xx10^(-10)cm`. DENSITY of the unit cell `d=(ZM)/(N_(A)a^(3))=(4xx108 gmol^(-1))/(6.022xx10^(23)mol^(-1)xx(408.6xx10^(-10)cm)^(3))` `d=10.51 gmol^(-3)`.

Discussion

33.	Silver dissolves in the solution of an alkali cyanide in the presence of oxygen, forming:
Answer» `[Ag(CN)_2]^-` `[Ag(CN)_4]^3-` `[Ag(CN)_2]^(-2)` `[Ag(CN)_3]^(2-)` Answer :A

Discussion

34.	Silver dissolves in the solution of an alkali cyanide in the presence of oxygen to form:
Answer» `[AG(CN)_2]^-` `[Ag(CN)_4]^3-` `[Ag(CN)_2]^(-2)` `[Ag(CN)_3]^(2-)` Answer :A

Discussion

35.	Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 xx 10^(-8) cm and its density is 10.5 g cm^(-3). Calculate on this basis the atomic mass of silver. [N_(A)= 6.02 xx 10^(23) mol^(-1)]
Answer» Solution :APPLY the RELATION : `=(dxxa^(3)xxN_(A))/(z)` `"Given :d = 10.5 g cm"^(-3), a=4.077xx10^(-8)cm` In fcc LATTICE, z= 4 Substituting the values in the above equation, we get `M=(10.5xx(4.077)^(3)xx10^(-24)xx6.02xx10^(23))/(4)=107.09` Thus, the ATOMIC mass of silver = 107.09u.

Discussion

36.	Silver crystallizes in face-centred cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms).
Answer» SOLUTION :a=400 PM For a face-centred cubic unit CELL, radius (r) = `(a)/(2sqrt2)=(400)/(2xx1.41)` pm = 141.8 pm

Discussion

37.	Silver crystallizes in a face centered cubic structure. If the edge length is 4.077 xx 10^(-8) cm and density is 10.5 g//cm^(3), calculate the atomic mass of silver.
Answer» Solution :`d = (ZM)/(a^(3)N_(A))` `M = (d XX a^(3) xx N_(A))/(Z)` `= (10.5 xx (4.077)^(3) xx 10^(-24) xx 6.022 xx 10^(23))/(4)` `= (428.50)/(4)` The atomic mass of silver (M) = 107.12 u

Discussion

38.	Silver crystallizes in ſec lattice. If edge length of the cell is 4.077 xx 10^(-8) cm and density is 10.5 g cm^(-3). Calculate the atomic mass of silver.
Answer» Solution :`"Density" rho= (Z xx M)/(a^(3) xx N_(0))` or`M= (rho xx a^(3) xx N_(0))/(Z)` Z= 4(fcc lattice) `rho= 10.5 g cm^(-3), N_(0)= 6.022 xx 10^(23), a= 4.077 xx 10^(-8)` cm `:. M= ((10.5 xx 4.077 xx 10^(-8))^(3) xx 6.022 xx 10^(23))/(4)=107.12 "g MOL"^(-1)`

Discussion

39.	Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. Calculate the radius of silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms.)
Answer» SOLUTION :For fcc UNIT cell `R = (a)/(2 SQRT(2))` Given, a = 400 PM `r = 400//2 sqrt(2)` pm r = 141.4 pm

Discussion

40.	Silver crystallises in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms)
Answer» Solution :For fcc unit CELL the relation between EDGE length and RADIUS is GIVEN by `r=(a)/(2SQRT2)` Given a = 400 pm `therefore""r=(400)/(2sqrt2) or r =" 141.4 pm."`

Discussion

41.	Silver crystallises in fcc lattice. If edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 g cm^(-3), calculate the atomic mass of silver.
Answer» Solution :The relation between atomc MASS, density, edge length and z is given by `M=(d xx a^(3)xxN_(A))/(z)` Substituting the VALUES, we get `M=((10.5 g CM^(-3))xx(4.077xx10^(-8)cm)^(3)xx6.022xx10^(23)mol^(-1))/(4)="107.09 g mol"^(-1)`.

Discussion

42.	Silver crystallises in fcc lattice. If edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 "g cm"^(-3), calculate the atomic mass of silver.
Answer» Solution :Edge length `(a) = 4.07 xx 10^(-8)` CM `therefore` Volume of unit CELL `= a^3 = (4.07)^3 xx 10^(-24) "cm"^3` Density `(d) = 10.5 "g cm"^3` Atomic Mass (M) = ? NUMBER of ATOMS PER unit cell (Z) = 4 `therefore M= (d xx N_A xx a^3)/(Z)` `therefore M = (10.5 xx 6.022 xx 10^23 xx (4.07)^3 xx 10(-24))/(4)` `therefore` M= 107.12 g/mol .

Discussion

43.	silvercrystallises in face- centred cubicstructure . Theedgelength of unitcell isfoundto be408.7pm. Calculatedensityofsilver(Ag= 108 g mol^(-1))
Answer» SOLUTION :Density=10.51 `G CM^(-3)`

Discussion

44.	Silver crystallies in face centred cubic (fcc) unit cell. If the radius of silver atom is 145 pm, what is the length of each side of the unit cell ?
Answer» Solution :For fcc CELL Radius `= (sqrt(3))/(4)a` a = 316.5 pm Radius `= (sqrt(3))/(4) XX 316.5` pm `= (1.732)/(4) xx 316.5` = 334.87 pm

Discussion

45.	Silver containing lead as impurity is purified by :
Answer» POLING CUPELLATION Lavigation Distillation ANSWER :B

Discussion

46.	Silver containing lead as an impurity is removed by
Answer» poling CUPELLATION lavigation distillation Solution :Silver CONTAINING LEAD is PURIFIED by cupellation.

Discussion

47.	Silver containing lead as an impurity is not purified by
Answer» Poling Cupellation LEVIGATION Distillation ANSWER :B

Discussion

48.	Silver chloride is soluble in methylamine due to the formation of :
Answer» `[AG(CH_(3)NH_(2))_(4)]CL` `[Ag(CH_(3)NH_(2))_(2)]Cl` `[Ag(CH_(3)NH_(2))_(3)]Cl` `[Ag(CH_(3)NH_(2))]Cl` Answer :B

Discussion

49.	Silver chloride is soluble in methylamine due to the formation of:
Answer» `Ag(CH_(3)NH_(2))Cl` `Ag+CH_(3)Cl+NH_(4)Cl` `[Ag(CH_(3)NH_(2))_(2)]Cl` `AgOH.` Answer :C

Discussion

50.	Silver chloride is prepared by (i) dissolving 0.5 g of silver wire in nitric acid and adding excess of hydrochloric acid to silver nitrate formed. The silver chloride precipitated is separated, washed and dried. The weight of silver chloride is 0.66 g. (ii) heating 1 g of silver metal in a current of dry chlorine gas till the metal is completely converted into its chloride. It is found to weight 1.32 g. Illustrate the law of constant composition by the above data.
Answer» SOLUTION :`%" of AG in AgCl in 1st case"=(0.5)/(0.66)xx100=75.76%` `%" of Cl"=24.24%` `%" of Ag in AgCl in 2ND case"=(1)/(1.32)xx100=75.76%` `5" of Cl"=24.24%`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Silver nitrate is prepared by

Silver Mirror test is given by which one of the following compounds ?

Silver nitrate is mainly used

Silver metal crystallizes with a face centred cubic lattice. The length of the unit cell is found to be 4.077 xx 10^(-8) cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108 u, N_A = 6.02 xx 10^(23) mol^(-1))

Silver metal crystallises with a face centred cubic lattice. The length of unit cell is formed to be 4.077 xx 10^(-8) cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108 u, N_(A) = 6.023 xx 10^(23) mol^(-1)).

Silver, mercury (ous) and lead are grouped together in the scheme of qualitative analysis because they form

Silver, mercury (ous) and lead are grouped together in a scheme of qualitative analysis because they form

Silver, mercury and lead have been placed in same group of qualitative analysis, because they form:

Silver may be obtained by __________.

Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm^2 by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited.

Silver iodide is used for producing artificial rain because AgI

Silver is removed electrically from 200 ml of a 0.1 N solution of AgnO_(3) by a current of 0.1 ampere. How long will it take to remove half of the silver from the solution.

Silver is removed electrolytically from 200 mL of a 0.1 N solution of AgNO_3by a current of 0.1 ampere. Hoe long will it take to remove half of the silver from the solution:

Silver is extracted from argentiferous lead by :

Silver is extracted by

Silver is exfracted by:

Silver is electro-deposited on a metallic vessel of surface area 800 cm^(2) by passing current 0.2 ampere for 3 hours. Calculate the thickness of silver deposited. Given the density of silver as 10.47 g/cc (Atomic mass of Ag=107.92amu)

Silver is a soft metal . It is hardened by alloying it with small amount of :

Silver iodide is used to produce artificial rain because :

Silver halides are used in photography because they are :

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).

Silver from silver nitrate is deposited by copper, because :

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9u).

Silver crystallizes in CCP lattice. The edge length of its unit cell is 408.6 pm. Calculate density of silver (atomic mass of silver is 107.9)

Silver forms ccp lattice and x-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver. (Atomic mass of Ag = 107.9 u)

Silver dissolves in the solution of an alkali cyanide in the presence of oxygen, forming:

Silver dissolves in the solution of an alkali cyanide in the presence of oxygen to form:

Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 xx 10^(-8) cm and its density is 10.5 g cm^(-3). Calculate on this basis the atomic mass of silver. [N_(A)= 6.02 xx 10^(23) mol^(-1)]

Silver crystallizes in face-centred cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms).

Silver crystallizes in a face centered cubic structure. If the edge length is 4.077 xx 10^(-8) cm and density is 10.5 g//cm^(3), calculate the atomic mass of silver.

Silver crystallizes in ſec lattice. If edge length of the cell is 4.077 xx 10^(-8) cm and density is 10.5 g cm^(-3). Calculate the atomic mass of silver.

Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. Calculate the radius of silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms.)

Silver crystallises in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms)

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 g cm^(-3), calculate the atomic mass of silver.

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 "g cm"^(-3), calculate the atomic mass of silver.

silvercrystallises in face- centred cubicstructure . Theedgelength of unitcell isfoundto be408.7pm. Calculatedensityofsilver(Ag= 108 g mol^(-1))

Silver crystallies in face centred cubic (fcc) unit cell. If the radius of silver atom is 145 pm, what is the length of each side of the unit cell ?

Silver containing lead as impurity is purified by :

Silver containing lead as an impurity is removed by

Silver containing lead as an impurity is not purified by

Silver chloride is soluble in methylamine due to the formation of :

Silver chloride is soluble in methylamine due to the formation of: