Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
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2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
Answer:
Metals
1) Sodium : It is used in the formation of table salt i. e. sodium chloride.
2) Copper : It is used in the wires of electricity due to better conduction.
3) Silver : It is used in wires and for making jewelry.
4) Aurum: It is used for making jewelry.
Non metals
1) Chlorine : It is used in the production of sodium chloride salt which is used in foods.
2) Flourine : It is used in making toothpaste.
3) Hydrogen : It is used in the formation of acids.
4) Oxygen : It is widely used in the production of steel and plastic.
Explanation:
Thus, those metals which remain unaffected by moisture, oxygen and carbon dioxide of the air can occur native or free. In other words, the unreactive metals occur in nature in free state because of their low reactivity towards chemical reagents. ... Metals usually occur in combination with nonmetallic elements.
