Write the chemical equations to represent the following reactions.
(a) The oxidation of `HCI (aq)` to `Cl_(2) (g)` by `PbO_(2)`.
(b) The disproportionation of `SnO` to `Sn` and `SnO_(2)`.
(a) `H_(3)PO_(3)+2AgNO_(3) +H_(2)O to 2Ag +2HNO_(3) +H_(3)PO_(4)` (b) `3Cl_(2)+6NaOH("Conc.") to 5 NaCl+NaClO_(3)+3H_(2)O` (c ) `2XeF_(2) +2H_(2)O to 2Xe +O_(2)+4HF`.
2 Answers 1 views(i) `Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+) to 2Cr^(3+)+6Fe^(3+)+7H_(2)O` (ii) `2MnO_(4)^(-)+16H^(+)+10I^(-) to 2Mn^(2+)+5I_(2)+8H_(2)O`. (iii) `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10CO_(2)+8H_(2)O`.
2 Answers 1 views(a) `({:(K_(2)Cr_(2)O_(7)+2H_(2)SO_(4)to KHSO_(4)+2CrO_(3)+H_(2)O),(NaCl+H_(2)SO_(4)to NaHSO_(4)+HCLxx4),(2CrO_(3)+4HCl to 2CrO_(2)Cl_(2)+2H_(2)O):})/(K_(2)Cr_(2)O_(7)+4NaCl+6H_(2)SO_(4) to 2CrO_(2)Cl_(2)+4NaHSO_(4)+2KHSO_(4)+3H_(2)O)`
2 Answers 1 viewsCorrect Answer - a `"Molarity (M)"=("Mass of HCI//Molar mass")/("Volame of solution in litrese")` Mass of HCI solution = 200+5.5=205.5 g Density of solution = 0.79 g `mL^(-1)` `"Volume of solution"=((205.5g))/((0.79gmL^(-1)))` =260L...
2 Answers 1 viewsCorrect Answer - c `M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=((0.5M)xx(750mL)+(250 mL))/((750 mL+250 mL))` 0.875 M.
2 Answers 2 viewsCorrect Answer - b If the two solutions containing the same solute are mixed, the molarity of the resulting solution is given as : `M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/V_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `=((3.0M)xx(25mL)+(4.0M)xx(75mL))/((25+75)mL)` =3.75 M
2 Answers 1 viewsCorrect Answer - c For solution, consult Example 2,14 (Text part).
2 Answers 1 viewsCorrect Answer - `-0.08` volt First determine `E^(@)` for `Hg_(2)Cl_(2) rarr 2Hg^(2+) +2Cl^(-) +2e^(-)` electrode. It comes to -1.44 volt. This electrode is now coupled with `Cl_(2)+2e^(-) rarr 2Cl^(-)` electrode.
2 Answers 1 viewsIn combustion of methane, Carbon exhibits variation from -4 to +4. The reaction is as follows : CH4 + 2O2 → CO2 + 2H2O In CH4, The oxidation state of carbon is -4 while in...
2 Answers 1 viewsHoffmann-bromamide degradation reaction : It is given by the primary amine having one carbon less. `underset(1^(@)"amide")(R-CON)H_(2) + Br_(2) + 4KOH to underset((1^(@)"amine"))(RNH_(2)) + K_(2)Co_(3) + 2KBr + 2H_(2)O` (ii) Carbylamine...
2 Answers 1 views