94420 + Interview Questions in Chemistry in Class 12 Page 1775 InterviewSolution

88701.	A metal which has no effect on asolution of mercury chloride is :
Answer» Zn Al Fe Ag Answer :D

Discussion

88702.	A metal weighing 0.43 g was dissolved in 50 mL of 1 N H_(2)SO_(4). The unreacted H_(2)SO_(4) required 14.2 mL of 1 N NaOH for neutralisation. Calculate the equivalent weight of the metal?
Answer» Solution :`H_(2)SO_(4)" LEFT unreacted "-=NaOH" reacted= 14.2 mL of 1 N"` `therefore"1 N "H_(2)SO_(4)" reacted with the metal "=(50-14.2)mL=35.8mL` `"35.8 mL of 1 N "H_(2)SO_(4)="35.8 MILLIEQUIVALENTS"` If E is the EQUIVALENT wt. of the metal, milliequivalents of the metal reacted `"=(0.43)/(E)xx1000=(430)/(E)` As metal and the acid will react in equivalent amounts, `(430)/(E)=35.8 or E=(430)/(35.8)=12.01`

Discussion

88703.	A metal surface ejects electrons when hit by blue light but not when hit by green light. When the electrons be ejected if the surface is hit by yellow light?
Answer» Yes No Yes, if the beam USED is QUITE intense Yes,if the beam is allowed to fall on the surface for a longer time. Answer :B

Discussion

88704.	A metal sulphide A gives gas B with rotten egg smell and a colourless sulphate C, when treated with dil H_2SO_4. B reacts with K_2Cr_2O_7//H^+ to form D, a white grey element. D burns in oxygen to yield E a colourless gas. If B is added to E, it gives D and a colourless liquid, which turns anhydrous CuSO_4 blue. C gives precipitate with NH_(3(aq)) or NaOH, which dissolves in excess of NaOH
Answer» `CuSO_4` `BaSO_4` `ZnSO_4` `Na_(2)ZnO_2` Solution :`underset((A))(ZNS )+ H_2SO_4 rarr underset((B))(H_2S) + underset((C))(ZnSO_4)` `H_2S + K_2Cr_2O_7 rarr underset((D))(S+Cr_2O_3)` `S+O_2 rarr underset((E)) SO_2` `H_2S + SO_2 rarr S + H_2O` `CuSO_4 + H_2O rarr underset((BLUE))(CuSO_4.) 5H_2O` `ZnSO_4 + NaOH rarr underset((white))(Zn(OH)_2darr) overset(NaOH) rarr underset("(SOLUBLE)")(Na_2ZnO_2)`

Discussion

88705.	A metal sulphide (A) gives (B) gas with rotten egg and (C) a colourless sulphates, when treated with dil. H_(2)SO_(4) (B) reacts with K_(2)Cr_(2)O_(7)//H^(+) to form (D) a white grey element. (D) burns in oxygen to yield (E) a colourless gas. If (B) is added in (E), it gives (D) and colourless liquid, which turns anhydrous CuSO_(4) blue. (C) gives a precipitate with NH_(3(aq)) or NaOH which dissolves in excess of NaOH. E is
Answer» `H_(2)S` `SO_(3)` `SO_(2)` NONE of these Solution : `UNDERSET((A))(ZnS)+DIL. H_(2)SO_(4) to underset((B))(H_(2)S)+underset((C))(ZnSO_(4))` , `H_(2)S OVERSET(K_(2)Cr_(2)O_(7)H^(+))to underset((D))(S) overset(O_(2))to underset((E))(SO_(2))`

Discussion

88706.	A metal sulphide (A) gives (B) gas with rotten egg and (C) a colourless sulphates, when treated with dil. H_(2)SO_(4) (B) reacts with K_(2)Cr_(2)O_(7)//H^(+) to form (D) a white grey element. (D) burns in oxygen to yield (E) a colourless gas. If (B) is added in (E), it gives (D) and colourless liquid, which turns anhydrous CuSO_(4) blue. (C) gives a precipitate with NH_(3(aq)) or NaOH which dissolves in excess of NaOH. C is
Answer» `CuSO_(4)` `BaSO_(4)` `ZnSO_(4)` `Na_(2)ZnO_(2)` Solution :Due to formation of `ZnSO_(4)`

Discussion

88707.	A metal sulphide (A) gives (B) gas with rotten egg and (C) a colourless sulphates, when treated with dil. H_(2)SO_(4) (B) reacts with K_(2)Cr_(2)O_(7)//H^(+) to form (D) a white grey element. (D) burns in oxygen to yield (E) a colourless gas. If (B) is added in (E), it gives (D) and colourless liquid, which turns anhydrous CuSO_(4) blue. (C) gives a precipitate with NH_(3(aq)) or NaOH which dissolves in excess of NaOH. B is
Answer» `SO_(2)` `SO_(3)` `H_(2)S` `CS_(2)` Solution :Due to formation of `H_(2)S` gas

Discussion

88708.	A metal solution when treated with dimethyl glyoxime and NH_(4) OH gives a rose red complex. The metal is
Answer» NI V Co Mn Answer :A

Discussion

88709.	A metal salt solution gives a yellow precipitate with silver nitrate.The precipitate dissolves in dilute nitric acid as well as in dilute ammonia solution.The solution contains :
Answer» bromide ions iodide ions phosphate ions chromate ions Solution :`Ag_(3)PO_(4)` is yellow precipitate which is soluble both in dilute ammonia solution and dilute `HNO_(3)` `HPO_(4)^(2-) + 3Ag^(+) to Ag_(3)PO_(4) darr +H^(+)` `Ag_(3)PO_(4)+2H^(+)toH_(2)PO_(4)^(-) darr+3Ag^(+),Ag_(3)PO_(4)darr +6NH_(3) to 3[Ag(NH_(3))_(2)]^(+)+PO_(4)^(3-)` PALE yellow precipitate of `AgBr` is not soluble in dilute `HNO_(3)`,bright yellow precipitate of `Agl` is not soluble in both ,`Ag_(2)CrO_(4)` is obtained as red precipitate.

Discussion

88710.	Ametal salt solutionforms a yellowprecipitatewith potassiumchromate in acetic acid , a whiteprecipitate with dil sulphuric acid butgives no precipitate with sodiumchloride oriodate .The whiteprecipitateabtained when sodium carbonate is added to the metal saltsolution consiste of
Answer» LEAD carbonate basic lead carbonate Bariumcarbonate Strontiumcarbonate Solution :`BaCrO_(*4)` is YELLOW ,`BaSO_(4)` is WHITE

Discussion

88711.	A metal rod dipped into a solution containing its own ions at unit activity at 298 K is called
Answer» SINGLE electrode half cell metal couple standard electrode Answer :D

Discussion

88712.	A metal oxide is yellow when hot and white when cold. The metal oxide is
Answer» ZnO CuO PbO all of these SOLUTION :`ZnO` GIVES yellow COLOUR on heating and white on COOLING .

Discussion

88713.	A metal present in insulin is
Answer» aluminium ZINC iron copper Solution :METAL PRESENT in INSULIN is zinc.

Discussion

88714.	A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 3.15g of the oxide have yielded 1.05 g of the metal. We may deduce that
Answer» The EQ. weight of the metal is 8. The ATOMIC weight of the metal is 8 The atomic weight of the metal is 4 The eq. weight of the metal is 4 Answer :D

Discussion

88715.	A metal oxide is reduced by heating it in a stream of hydrogen it is found that after completereduction, 3.15 g of the oxide have yielded 1.05 g of the metal. We may conclude that:
Answer» ATOMIC MASS of the METAL is 4 atomic mass of the metal is 8 equivalent mass of the metal is 4 equivalent mass of the metal is 8 Answer :C

Discussion

88716.	A metal oxide has the formula X_(2)O_(3). It can be reduced by hydrogen to give free metal and water. 0.1596g of metal oxide requires 6 mg of hydrogen for complete reduction . The atomic mass of metal in amu is:
Answer» 15.58 155.8 5.58 55.8 Solution :`X_(2)O_(3)+3H_(2)rarr2X+3H_(2)O` 1MOL 3 MOL (2a+48)g 6g 0.006 g `H_(2)` is required by 0.1596 g oxide `THEREFORE 6g H_(2)` will be required by 159.6 g oxide `therefore 2a+48=159.6` a=55.8

Discussion

88717.	A metal oxide has the formula M_(2)O_(3). It can be reduced by H_(2) to free metal and water. 0.1596g of M_(2)O_(3) required 6 mg of H_(2) for complete reduction. The atomic mass of the metal is:
Answer» 27.9 79.8 55.8 159.8 Solution :`M_(2)O_(3)+3H_(2)rarr2M+3H_(2)O` `(2x+48)G 6G` x=Atomic mass of METAL `because` 0.006 g `H_(2)` reduces 0.1596g `M_(2)_(3)`. `therefore 6g H_(2)` will reduce `(0.1596)/(0.006)xx6 g M_(2)O_(3)=159.6M_(2)O_(3)` `2x+48=159.6` `2x=11.6` `x=55.8`.

Discussion

88718.	A metal oxide has 40% oxygen. Thepercentage of chlorine in its chloride is
Answer» `90%` `88.75%` `74.7%` `35%` SOLUTION :N/A

Discussion

88719.	A metal oxide has a formula Z_(2)O_(3). It can be reduced by hydrogen to give free metal and water. 0.01596g of the metal requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal is
Answer» 27.9 159.6 7.98 55.8 Answer :C

Discussion

88720.	A metal obtained by hydrometallurgy is :
Answer» Ag Fe Na Mg Answer :A

Discussion

88721.	A metal object is to be coated with a nickel layer 0.3 mm thick. The surface areaof the object is 100 cm^2 . The density of nickel is 9.0 g//cm^3. How long will it take to pass a 3-amp current if the current yield is 90%?
Answer» SOLUTION :9 H 8 min 25 s

Discussion

88722.	A metal nitride on hydrolysis with steam gives
Answer» `Cl_(2)` `NH_(3)` `O_(2)` `CO_(2)` ANSWER :2

Discussion

88723.	A metal nitrate solution reacts with dilute hydrochloric acid to give a white precipitate which is soluble in concentrated potassium chloride.White precipitate on passing of hydrogen sulphide gas is converted into black precipitate.The black precipitate on boiling with hydrogen peroxide (3%) is converted again to a white precipitate which is soluble in ammonium acetate.The cation of the metal nitrate is :
Answer» `Pb^(2+)` `Ag^(+)` `HG^(2+)` `BI^(3+)` Solution :`Pb^(2+)+2Cl^(-)toPbCl_2darr ("white"), PbCl_2+ 2Cl^(-)to[PbCl_4]^(2-)` (COLOURLESS soluble complex ) `PbCl_2darr +H_2StoPbSdarr`(black)+2HCl `PbS+4H_2O_2toPbSO_4`(white)+`4H_2O` `PbSO_4darr +2CH_3COONH_4to(NH_4)_2SO_4+(CH_3COO)_2Pb`

Discussion

88724.	A metal nitrate reacts with KI to give a black precipitate which on addition of excess of Kl is converted into orange colour solution.The cation of the metal nitrate is :
Answer» `HG^(2+)` `Bi^(3+)` `Pb^(2+)` `Cu^(+)` SOLUTION :N//A

Discussion

88725.	A metal nitrate reacts with KI to give a black precipitate which on addition of excess of KI is converted into orange colour solution. The cation of the metal nitrate is
Answer» `Hg^(2+)` `BI^(3+)` `Pb^(2+)` `CU^(+)` Solution :`Bi(NO_(3))_(3)(AQ)+3KI(aq)tounderset(BiI_(3)(s))+3KNO_(3)(aq),BiI_(3)(s)+KI(aq)to underset("Orange")(K[BiI_(4)])`

Discussion

88726.	A metal nitrate reacts with KI to give a black precipitate, which on addition of excess of KI converts to orange colour solution. The cation of metal nitrate is :
Answer» `Hg^(2+) ` `Bi^(3+) ` `PB^(2+)` `Cu^(+)` SOLUTION :`Bi(NO_3)_3 (AQ) + 3KI to BiI_3(s) + 3KNO_3 (aq)` `BiI_3 + KI to underset("SOLUBLE")(K[BiI_4])`

Discussion

88727.	A metalnitratereacts with Ki to give a balckprecipitatewhich on additionof excess of KIcovered to an orange colour solution .The cationof metalnitrate in
Answer» `Hg^(2+)` `Bi^(3+)` `Pb^(2+)` `CU^(o+)` Solution :`Bi^(3+)3KI rarrunderset(Black)(BiI_(3) darr)+ 3K^(o+)` `BiI_(3) +KI RARR underset("orange solution")(K[BiI_(4)])`

Discussion

88728.	A metal nitrate reacts with KI to give a black precipitate which addition of excess of KI converts into orange colour solution. The cation of metal nitrate is:
Answer» `Hg^(2+)` `Bi^(3+)` `PB^(2+)` `Cu^(+)` Answer :B

Discussion

88729.	A metal M which is not affected by strong acids like conc. HNO_(3), conc, H_(2)SO_(4) and conc. Solution of alkalies like NaOH, KOH. It formes MCl_3 which is used for toning in photography. The metal M is
Answer» Ag Hg Au Cu Answer :C

Discussion

88730.	A metal M^(n+) having d^(4) valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Delta_(0)gtP. (i) Write the electronic configuration of d^(4) ion. (ii) What type of hybridisation will M^(n+) ion has ? (iii) Name the type of isomerism exhibited by this complex.
Answer» Solution :(i) `t_(2g)^(4)` (ii) `d^(2)sp^(3)` (one d-electron will pair up making two empty d-orbital available for hybridisation). (iii) Optical isomerism (SIMILAR to `[Co(EN)_(3)]^(3+)` or `[CR(o x)_(3)]^(3-)`.

Discussion

88731.	A metal M with specific heat (0.16) have chlorine 68.3% then the formula of the compound is similar to
Answer» MCl `MCl_(2)` `MCl_(3)` `MCl_(4)` Solution :MASS of CHLORINE in the metal chloride = 68.3% Mass of metal = 100-68.3 = 31.7 Equivalent mass of the metal `=("mass of metal")/("mass of chlorine") xx 35.5` `=31.7/68.3 xx 35.5 = 16.47` According to Dulong and Petit.s LAW, Approx, atomic mass of the metal `=6.4/(0.16) = 40` VALENCY `=("Approximation atomic mass")/("Equivalent mass") = 40/16.47 = 2.42 =2` Hence, the formula of the compound = `MCl_(2)`

Discussion

88732.	A metal M which is not affected by strong acids like conc. HNO_3, conc. H_2SO_4 and conc. solution of alkalies like NaOH, KOH forms MCl_(3), which is used for toning inphotography, The metal M is :
Answer» AG HG Au Cu Answer :C

Discussion

88733.	A metal M (specific heat 0.16) forms a metal chloride with a 65% chorine present in it. The formula of the metal chloride will be
Answer» MCl `MCl_(2)` `MCl_(3)` `MCl_(4)` Solution :According to Dulong-Petits law: Approx. at mass `XX` SP. HEAT = 6.4 (actually `6.4 +- 0.6`) `THEREFORE` At. Mass (approx) `=6.4/(0.16 = 40` Suppose, formula of the chloride `=MCl_(x)` `therefore` CHLORINE in it `=(35.5x xx 100)/(40 + 35.5 x)%` So, according to the question `(35.5x)/(40 + 35.5x) = 65/100 rArr 40/(35.5x) + 1= 100/65` `rArr 35.5 x = (40 xx 13)/7` `rArr x = 2.09 =2` `therefore` Formula of the metal chloride `=MCl_(2)`

Discussion

88734.	A metal M readily forms water soluble sulphate MSO_(4), water insoluble hydroxide M(OH)_(2) and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. The metal is :
Answer» Be Mg Ca Sr Answer :A

Discussion

88735.	A metal M reacts with N_(2) to give a compound 'A'(M_(3)N). 'A' on heating at high temperature gives back 'M' and 'A' on reacting with H_(2)O gives a gas 'B'.'B' turns CuSO_(4) solution blue on passing through it A and B can be
Answer» `AL and NH_3` `Li and NH_3` `NA and NH_3` `MG and NH_3` ANSWER :B

Discussion

88736.	A metal (M) produces a gas (N) on reaction with alkalies like NaOH and KOH. Same gas is produced when the metal reacts with dilute sulphuric acid. Gas (N) reacts with another toxic gas (P) to form methanol at high temperature and pressure . (N) also reacts with metals like (Q) to form electrovalent hydrides. M, N, P and Q respectively are
Answer» `ZN, H_(2), CO, Na` `Na, H_(2), Cl_(2), Ca` `Al, H_(2), H_(2)S, B` `Mg, H_(2), NO_(2), Al` Answer :A

Discussion

88737.	A metal "M" reacts with nitrogen gas to afford "M_3N". "M_3N" on heating at high temperature gives back "M" and on reaction with water produces gas "B". Gas "B" reacts with aqueous CuSO_4 to form deep blue compound. "M" and"B" are respectively......
Answer» Na and `NH_3` Li and `NH_3` Ba and `N_2` AL and `N_2` Solution :`6LI + N_(2) to 2L_(3)N overset(Delta)to 6Li + N_(2)` `2Li_(3)N to 6LiOH + 2NH_(3) overset(Cu^(2+)) to UNDERSET("DEEP blue")([Cu(NH_(3))_(4)]^(2+))`

Discussion

88738.	A metal (M) produces a gas (N) on reactions with alkalies like NaOH and KOH. Same gas is produced when the metal reacts with dilute sulphuric acid. Gas (N) reacts with another toxic gas (P) to form methanol at high temperature and pressure. (N) also reacts with metals like (Q) to form electrovalent hydrides, M,N,P and Q respectivelyare
Answer» <P>`Zn, H_(2), CO, Na` `Na, H_(2), Cl_(2), Ca` `Al, H_(2), H_(2)S, B` `Mg, H_(2), NO_(2), Al` Solution :`underset((M))(Zn)+2NaOHrarrNa_(2)ZnO_(2)+underset((N))(H_(2))` `Zn+H_(2)SO_(4)rarrZnSO_(4)+H_(2)` `underset((P))(CO)+2H_(2)overset(Delta,200atm)underset("catalyst")rarr CH_(3)OH` `underset((Q))(2Na)+H_(2)rarr 2NaH`

Discussion

88739.	A metal M of molar mass 96 reacts with fluorine to form a salt that can be represeneted as MF_(x). In order to determiner x a 9.18 a sample of the salt is dissolved in 100g ofwater and its K_(b)(water 0.512 kg/mol. Assuming complete dissociation of salf.
Answer» ANSWER :`x=4`

Discussion

88740.	A metal M of equivalent mass E forms an oxide of molecular formula M_(x)O_(y). The atomic mass of the metal is given by the correct equation:
Answer» 2E(x/y) xyE E/y y/E Solution :Let atomic mass of METAL M is 'a' Mass of metal =`axx x` Mass of oxygen=`16xxy` EQUIVALENT mass of element`=("Mass of element")/("Mass of oxygen")XX8` `E=(ax)/(16y)xx8` `a=2E((y)/(x))`

Discussion

88741.	A metal M of equivalent mass E forms an oxide of molecular formula M_(x)O_(y). The atomic mass of the metal is given by the correct equation
Answer» `2E(y//X)` `xy//E` `E//y` `y//E` Solution :If m is the atomic mass of the metal M, then 16 y g of oxygen combine with x m g of the metal. `therefore"8 g of oxygen will combine with metal"` `=(xm)/(16Y)XX8=(xm)/(2y)` ` therefore (xm)/(2y)=E"or"m=2E(y//x)`

Discussion

88742.	A metal M of atomic mass 54.94 has a density of 7.42 g/cc. Calculate the apparent volume occupied by one atom of the metal.
Answer» ANSWER :`1.23xx10^(-23)CC`

Discussion

88743.	A metal M of at. wt. 24 forms an oxide having 40% by wt. of O_2 .The probability formulae of oxide is
Answer» `MO_2` `M_2 O` MO `M_2 O_3` ANSWER :C

Discussion

88744.	A metal 'M' is crystallised in F.C.C lattice. The number of unit cells in it having 2.4 xx 10^(24) atoms
Answer» N N/2 2n and n 4N Answer :A

Discussion

88745.	A metal M readily forms water soluble sulphate, and water insoluble hydroxide M(OH)_(2). Its oxide MO is amphoteric, hard and having high melting point. The alkaline earth metal M must be :
Answer» `Be` `MG` `CA` `SR` ANSWER :A

Discussion

88746.	A metal M forms chlorides in tis +2 and +4 oxidation state. Which of the following statement about these chlorides is correct?
Answer» `MCl_(2)` more ionic than `MCL_(4)` `MCl_(2)` more easily HYDROLYSIS than `MCL_(4)` `MCl_(2)` more volatile than `MCL_(4)` `MCl_(2)` more SOLUBLE in anhydrous ethanol than `MCL_(4)` Solution :Metal atoms on the lower oxidition state forms the ionic bond and in the higher oxidation state the convent bond because higher oxidation state means small size and great polazing power and hence GREATER the covalent CHARACTER. Hence, `MCl_(2)` is more ionic than `MCl_(4)`.

Discussion

88747.	A metal ,M forms chlorides in its +2 and +4 oxidatio states . Which of the following statements about these chlorides is correct ?
Answer» `MCl_(2)` is more volatile than `MCl_(4)` `MCl_(2)` is more SOLUBLE in anhydrous ethanol than `MCl_(4)` `MCl_(2)` is more ionic than `MCl_(4)` `MCl_(2)` is more easily hydrolysed than `MCl_(2)` Answer :C

Discussion

88748.	A metal M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?
Answer» `MCl_(2)` is more soluble is ANHYDROUS ETHANOL than `MCl_(4)` `MCl_(2)` is more ionic than `MCl_(4)` `MCl_(2)` is more easily hydrolysed than `MCl_(4).` `MCl_(2)` is more volatile than `MCl_(4)`. Solution :Because of Fazan.s rule `MCl_(2)` would be more ionic than `MCl_(4)`.

Discussion

88749.	A metal M forms an alum which contains the element potassium and is isomorphous with ordinary alum(K_2SO_4. Al_2(SO_4)_3 24H_2O) . If the alum of M contains 10.42% of M, then atomic weight of M is (at. mass of K= 39, O=16, S=32 and H=1):
Answer» 52 104 156 208 Solution :Since the alum is isomorphous with `K_2SO_4` `Al_2(SO_4)_3.24H_2O ` its FORMULA is `K_2SO_4`. `M_2(SO_4)_3 . 24 H_2O` Molecular weight = 894 + 2M where M is the atomic weight of the metal Now ` (2M)/(894 + 2M) xx 100 = 10.42 ` SOLVING M = 52

Discussion

88750.	A metal (M) burns with dazzling brilliance in air to give a white powder.The white powder reacts with water to form a white precipitate and a colourless gas with a characteristic smell.The metal (M) decomposes hot water but not cold water, liberating the inflammable hydrogen gas. The metal (M) is:
Answer» K Ca Mg Rb Solution :HYDROGEN is LIBERATED when Mg reacts with hot water. Rb, K and Ca gives hydrogen gas even with COLD water. `Mg+O_2toMgO , Mg+N_2toMg_3N_2` `Mg_3N_2+6H_2O to 3MG(OH)_2 + 2NH_3 uarr` `Mg+2H_2O("hot")toMg(OH)_2+H_2`

Discussion

Explore topic-wise InterviewSolutions in .

A metal which has no effect on asolution of mercury chloride is :

A metal weighing 0.43 g was dissolved in 50 mL of 1 N H_(2)SO_(4). The unreacted H_(2)SO_(4) required 14.2 mL of 1 N NaOH for neutralisation. Calculate the equivalent weight of the metal?

A metal surface ejects electrons when hit by blue light but not when hit by green light. When the electrons be ejected if the surface is hit by yellow light?

A metal solution when treated with dimethyl glyoxime and NH_(4) OH gives a rose red complex. The metal is

A metal salt solution gives a yellow precipitate with silver nitrate.The precipitate dissolves in dilute nitric acid as well as in dilute ammonia solution.The solution contains :

Ametal salt solutionforms a yellowprecipitatewith potassiumchromate in acetic acid , a whiteprecipitate with dil sulphuric acid butgives no precipitate with sodiumchloride oriodate .The whiteprecipitateabtained when sodium carbonate is added to the metal saltsolution consiste of

A metal rod dipped into a solution containing its own ions at unit activity at 298 K is called

A metal oxide is yellow when hot and white when cold. The metal oxide is

A metal present in insulin is

A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 3.15g of the oxide have yielded 1.05 g of the metal. We may deduce that

A metal oxide is reduced by heating it in a stream of hydrogen it is found that after completereduction, 3.15 g of the oxide have yielded 1.05 g of the metal. We may conclude that:

A metal oxide has the formula X_(2)O_(3). It can be reduced by hydrogen to give free metal and water. 0.1596g of metal oxide requires 6 mg of hydrogen for complete reduction . The atomic mass of metal in amu is:

A metal oxide has the formula M_(2)O_(3). It can be reduced by H_(2) to free metal and water. 0.1596g of M_(2)O_(3) required 6 mg of H_(2) for complete reduction. The atomic mass of the metal is:

A metal oxide has 40% oxygen. Thepercentage of chlorine in its chloride is

A metal oxide has a formula Z_(2)O_(3). It can be reduced by hydrogen to give free metal and water. 0.01596g of the metal requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal is

A metal obtained by hydrometallurgy is :

A metal object is to be coated with a nickel layer 0.3 mm thick. The surface areaof the object is 100 cm^2 . The density of nickel is 9.0 g//cm^3. How long will it take to pass a 3-amp current if the current yield is 90%?

A metal nitride on hydrolysis with steam gives

A metal nitrate reacts with KI to give a black precipitate which on addition of excess of Kl is converted into orange colour solution.The cation of the metal nitrate is :

A metal nitrate reacts with KI to give a black precipitate which on addition of excess of KI is converted into orange colour solution. The cation of the metal nitrate is

A metal nitrate reacts with KI to give a black precipitate, which on addition of excess of KI converts to orange colour solution. The cation of metal nitrate is :

A metalnitratereacts with Ki to give a balckprecipitatewhich on additionof excess of KIcovered to an orange colour solution .The cationof metalnitrate in

A metal nitrate reacts with KI to give a black precipitate which addition of excess of KI converts into orange colour solution. The cation of metal nitrate is:

A metal M which is not affected by strong acids like conc. HNO_(3), conc, H_(2)SO_(4) and conc. Solution of alkalies like NaOH, KOH. It formes MCl_3 which is used for toning in photography. The metal M is

A metal M with specific heat (0.16) have chlorine 68.3% then the formula of the compound is similar to

A metal M which is not affected by strong acids like conc. HNO​_3​, conc. H_2​SO​_4 and conc. solution of alkalies like NaOH, KOH forms MCl_(3)​, which is used for toning inphotography, The metal M is :

A metal M (specific heat 0.16) forms a metal chloride with a 65% chorine present in it. The formula of the metal chloride will be

A metal M readily forms water soluble sulphate MSO_(4), water insoluble hydroxide M(OH)_(2) and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. The metal is :

A metal M reacts with N_(2) to give a compound 'A'(M_(3)N). 'A' on heating at high temperature gives back 'M' and 'A' on reacting with H_(2)O gives a gas 'B'.'B' turns CuSO_(4) solution blue on passing through it A and B can be

A metal "M" reacts with nitrogen gas to afford "M_3N". "M_3N" on heating at high temperature gives back "M" and on reaction with water produces gas "B". Gas "B" reacts with aqueous CuSO_4 to form deep blue compound. "M" and"B" are respectively......

A metal M of molar mass 96 reacts with fluorine to form a salt that can be represeneted as MF_(x). In order to determiner x a 9.18 a sample of the salt is dissolved in 100g ofwater and its K_(b)(water 0.512 kg/mol. Assuming complete dissociation of salf.

A metal M of equivalent mass E forms an oxide of molecular formula M_(x)O_(y). The atomic mass of the metal is given by the correct equation:

A metal M of equivalent mass E forms an oxide of molecular formula M_(x)O_(y). The atomic mass of the metal is given by the correct equation

A metal M of atomic mass 54.94 has a density of 7.42 g/cc. Calculate the apparent volume occupied by one atom of the metal.

A metal M of at. wt. 24 forms an oxide having 40% by wt. of O_2 .The probability formulae of oxide is

A metal 'M' is crystallised in F.C.C lattice. The number of unit cells in it having 2.4 xx 10^(24) atoms

A metal M readily forms water soluble sulphate, and water insoluble hydroxide M(OH)_(2). Its oxide MO is amphoteric, hard and having high melting point. The alkaline earth metal M must be :

A metal M forms chlorides in tis +2 and +4 oxidation state. Which of the following statement about these chlorides is correct?

A metal ,M forms chlorides in its +2 and +4 oxidatio states . Which of the following statements about these chlorides is correct ?

A metal M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

A metal M forms an alum which contains the element potassium and is isomorphous with ordinary alum(K_2SO_4. Al_2(SO_4)_3 24H_2O) . If the alum of M contains 10.42% of M, then atomic weight of M is (at. mass of K= 39, O=16, S=32 and H=1):

A metal (M) burns with dazzling brilliance in air to give a white powder.The white powder reacts with water to form a white precipitate and a colourless gas with a characteristic smell.The metal (M) decomposes hot water but not cold water, liberating the inflammable hydrogen gas. The metal (M) is:

A metal M which is not affected by strong acids like conc. HNO_3, conc. H_2SO_4 and conc. solution of alkalies like NaOH, KOH forms MCl_(3), which is used for toning inphotography, The metal M is :