\(V_1\), 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 \(45\).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
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: