Advertisements
Advertisements
प्रश्न
Multiple Choice Question
The metal which reacts with cold water to produce hydrogen is
विकल्प
magnesium
aluminium
calcium
iron
उत्तर
calcium
APPEARS IN
संबंधित प्रश्न
Match the statements in Column A with those in Column B.
| Column A | Column B |
|
1. A metal which reacts with cold water to from hydrogen. |
1. Reduction |
| 2. A gas which is inflammable and non-supporter of combustion | 2. Hydrogenation |
|
3. A process in which vanspati ghee is prepared from vegitable oils |
3. oxidation |
|
4. The removal ofhydrogen or addition of oxygen |
4. sodium |
| 5. The addition of hydrogen or removal of oxygen | 5. Hydrogen |
Which metal is preferred for the preparation of hydrogen?
from acid?
Why nitric acid is not used in the preparation of hydrogen?
Hydrogen may be prepared in the laboratory by the action of a metal on an acid.
Which of the acids dilute sulphuric, concentrated sulphuric, dilute nitric acid and concentrated nitric acid would you choose? Explain why you would not use the acids you reject.
‘Hydrogen is obtained by electrolysis of acidified water’.
Answer the following pertaining to the preparation of hydrogen by electrolysis,
The meaning of the term ‘electrolysis’ and ‘electrolyte’.
Select the correct options for the following statement:
The metal which reacts with steam liberating hydrogen & the reaction is reversible.
On heating which of the following substances i.e. copper carbonate, zinc carbonate, washing soda, copper sulphate, zinc nitrate, copper nitrate, lead nitrate, ammonium chloride and ammonium dichromate – relate to the reaction given below.
A white substance which leaves an amphoteric oxide as a residue [whose colour varies in the heated and in the cold state] and evolves a gas which turns lime water milky.
Complete and balance the equation:
[General method]
Reactions of metals with steam
Aluminium - 2Al + 3H2O → ______+ _______ [g]
State why hydrogen is not prepared in the laboratory by the action of sodium with cold water.
Give a balanced equation for the following conversion.
\[\ce{KAlO2<-KOH->K2ZnO2}\]
