\(V_1\), Volume of \(0.2\ g\) \(H_2\) at \(200\ K\)
\(=\frac { 0.2 \times R \times 200}{2 \times P}\)
\(V_2\), Volume of \(3.0\ g\) of gas A at \(300\ K\)
\(= \frac {0.3 \times R \times 300}{M \times P}\)
Given that, \(V_1 = V_2\)
\(⇒ \frac {0.2 \times R \times 200 }{ 2 \times P} = \frac {0.3 \times R \times 300}{M \times P}\)
\(M = 45\) g mol-1
So, the answer is \(45\).
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
Read More: Some Basic Concepts of Chemistry
There are two ways of classifying the matter:
Matter can exist in three physical states:
Based upon the composition, matter can be divided into two main types: