Answer:
The swirling yellow solid formed is lead iodide (PbI₂).
Explanation:
- The reaction of potassium iodide (KI) with lead nitrate (Pb(NO₃)₂) will produce lead iodide (PbI₂) and potassium nitrate (KNO₃) according to the equation:
2KI + Pb(NO₃)₂ → PbI₂↓ + 2KNO₃
- Lead iodide (PbI₂) is a yellow swirling precipitate that is formed from the reaction.
Answer: The given statement is true.
Explanation: If this reaction would have occurred, then this reaction would be considered as displacement reaction.
Displacement reactions are the reaction in which more reactive element displaces the less reactive element in a chemical reaction. This is based on the reactivity of elements.
Reactivity of elements is the tendency of the elements to gain or loose electrons. The reactivity decreases down the group in a periodic table.
In the given reaction, Iodine and chlorine are the elements of the same group in the periodic table and iodine lies below chlorine in the group. So, the reactivity of iodine is less than the reactivity of chlorine.
Hence, in the given reaction, iodine will not replace chlorine because it lies below in the periodic table.
Answer:
OPTION (A) : Testing a rock sample for gold content
Explanation:
For testing a rock sample of gold content you will need a Chemist. To test the material, the sample is rubbed on black stone which will leave a mark on the stone. This mark is tested by applying aqua fortis i.e nitric acid on the mark. If the mark gets dissolve then the material is not gold. If the mark sustain the it is further tested by applying aqua regia i.e nitric acid and hydrochloric acid which will prove the sample is of gold if it gets dissolve on using hydrochloric acid. The purity of the sample can be checked by differing the concentration of the aqua regia and comparing it with the gold material of the known purity.
