To solve this question, we need to understand the chemical reaction that occurs when Lead Sulphide (PbS) reacts with dilute nitric acid (HNO3).
The chemical reaction between lead sulphide and dilute nitric acid can be represented as follows:
\(PbS + 4HNO_3 \rightarrow Pb(NO_3)_2 + 2H_2O + 2NO_2 + S\)
Breaking down the reaction, we can observe the formation of the following products:
From the given options, we can analyze and conclude:
Based on the analysis, the correct answer is:
Nitrous oxide (N2O) is not formed in this reaction.
The reaction between Lead Sulphide (PbS) and dilute nitric acid (\(HNO_3\)) proceeds as follows:
\(PbS + 2HNO_3 \rightarrow Pb(NO_3)_2 + NO + S + H_2O\)
Key Points:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: