When nitrogen gas (\( \text{N}_2 \)) reacts with calcium carbide (\( \text{CaC}_2 \)), the product formed is calcium cyanide (\( \text{CaCN}_2 \)) along with carbon. This reaction is commonly known as the formation of calcium cyanide, and it occurs under high temperatures. Let's break down the reaction and the process involved:
The chemical equation for the reaction is as follows:
\(\text{CaC}_2 + \text{N}_2 \rightarrow \text{CaCN}_2 + \text{C}\)
In the reaction, the calcium carbide (\( \text{CaC}_2 \)) reacts with nitrogen gas (\( \text{N}_2 \)) under high temperature conditions to form calcium cyanide (\( \text{CaCN}_2 \)) and carbon. The reaction proceeds as follows:
\(\text{CaC}_2 + \text{N}_2 \rightarrow \text{CaCN}_2 + \text{C}\)
Calcium cyanide has several important industrial uses:
In summary, when calcium carbide (\( \text{CaC}_2 \)) reacts with nitrogen gas (\( \text{N}_2 \)), it produces calcium cyanide (\( \text{CaCN}_2 \)) and carbon. This reaction plays an important role in industrial chemistry, especially in the production of materials used in gold extraction and chemical synthesis.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :