Question Answers
Sahay Sir > Question Answers
‘Softening of lead’ means the (a) Conversion of PbS into Pb (b) Removal to tin from common solder (c) Removal of impurities from Pb (d) Addition of tin to lead
17
Nov
‘Softening of lead’ means the (a) Conversion of PbS into Pb (b) Removal to tin from common solder (c) Removal of impurities from Pb (d) Addition of tin to lead 'Softening of lead' means the (a) Conversion of PbS into Pb (b) Removal to tin from common solder (c) Removal of impurities from Pb (d) [...]
In which of the following the inert pair effect is most prominent (a) C (b) Si (c) Ge (d) Pb
17
Nov
In which of the following the inert pair effect is most prominent (a) C (b) Si (c) Ge (d) Pb In which of the following the inert pair effect is most prominent (a) C (b) Si (c) Ge (d) Pb November 17, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Which of the following has least tendency to undergo catenation (a) C (b) Si (c) Ge (d) Sn
17
Nov
Which of the following has least tendency to undergo catenation (a) C (b) Si (c) Ge (d) Sn Which of the following has least tendency to undergo catenation (a) C (b) Si (c) Ge (d) Sn November 17, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Diamond is harder than graphite because (a) Graphite is planer (b) Diamond has free electron (c) Graphite is sp3 hybridised
17
Nov
Diamond is harder than graphite because (a) Graphite is planer (b) Diamond has free electron (c) Graphite is sp3 hybridised Diamond is harder than graphite because (a) Graphite is planer (b) Diamond has free electron (c) Graphite is sp3 hybridised November 17, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Red lead is (a) Pb3O3 (b) PbO (c) PbO2 (d) Pb4O3
17
Nov
Red lead is (a) Pb3O3 (b) PbO (c) PbO2 (d) Pb4O3 Red lead is (a) Pb3O3 (b) PbO (c) PbO2 (d) Pb4O3 November 17, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Identify the correct order of solubility of Na2S,CaS and ZnS in aqueous medium.
17
Nov
Identify the correct order of solubility of Na2S,CaS and ZnS in aqueous medium. CaS and ZnS in aqueous medium. Identify the correct order of solubility of Na2S November 17, 2020 Category: Chapter 7 - Equilibrium , Chemistry , JEE Adv and Mains Chemistry 41 Years Solved Video Solutions 1979 - 2019 ,
A solution which is 10^−3M each in Mn^2+, Fe^2+,Zn^2+, and Hg^2+ it treated with 10^−16M sulphide ion. If the Ksp of MnS, FeS, ZnS and HgS are 10^−15,10^−23,10^−20,and 10^−54, respectively, which one will precipitate first?
17
Nov
A solution which is 10^−3M each in Mn^2+, Fe^2+,Zn^2+, and Hg^2+ it treated with 10^−16M sulphide ion. If the Ksp of MnS, FeS, ZnS and HgS are 10^−15,10^−23,10^−20,and 10^−54, respectively, which one will precipitate first? 10^−20 10^−23 A solution which is 10^−3M each in Mn^2+ and 10^−54 and Hg^2+ it treated with 10^−16M sulphide ion. [...]
HX is a weak acid ( Ka=10^−5 ).lt forms a salt Na (0.1M) on reacting with caustic soda. The degree of hydrolysis of Na X is
17
Nov
HX is a weak acid ( Ka=10^−5 ).lt forms a salt Na (0.1M) on reacting with caustic soda. The degree of hydrolysis of Na X is HX is a weak acid ( Ka=10^−5 ).lt forms a salt Na (0.1M) on reacting with caustic soda. The degree of hydrolysis of Na X is November 17, 2020 [...]
CH3NH2 (0.1 mole, kb = 5 x 10^-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , resulting hydrogen ion concentration is
17
Nov
CH3NH2 (0.1 mole, kb = 5 x 10^-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , resulting hydrogen ion concentration is CH3NH2 (0.1 mole kb = 5 x 10^-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre resulting [...]
2.5 ml of 2/5 M weak monoacidic base (Kp=1×10^−12 at 25^∘ C) is titrated with 2/15 M HCI in water at 25^∘ C. The concentration of H ‘ at equivalence point is (Ksp=1×10^−14 at 25^∘C)
17
Nov
2.5 ml of 2/5 M weak monoacidic base (Kp=1×10^−12 at 25^∘ C) is titrated with 2/15 M HCI in water at 25^∘ C. The concentration of H ‘ at equivalence point is (Ksp=1×10^−14 at 25^∘C) 2.5 ml of 2/5 M weak monoacidic base (Kp=1×10^−12 at 25^∘ C) is titrated with 2/15 M HCI in water [...]