
At STP, 1 mole of a gas occupies 22.7 liters. The given volume of O2 is 2.8375 liters. The number of moles is calculated as:
\[ n_{\text{O}_2} = \frac{\text{Volume of O}_2}{22.7} \]
Substitute the given values:
\[ n_{\text{O}_2} = \frac{2.8375}{22.7} = 0.125 \, \text{moles} \]
The number of molecules of O2 is given by:
\[ \text{Molecules of O}_2 = n_{\text{O}_2} \cdot N_A \]
Where \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \) (Avogadro’s number).
Substitute the values:
\[ \text{Molecules of O}_2 = 0.125 \cdot 6.022 \times 10^{23} = 7.527 \times 10^{22} \, \text{molecules} \]
At STP, one mole of any gas occupies 22.4 liters. For the given volume of oxygen gas (2.8375 liters):
The number of moles is calculated as:
\[ \text{Number of moles of O}_2 = \frac{2.8375 \, \text{L}}{22.4 \, \text{L/mol}} = 0.125 \, \text{mol} \]
Using Avogadro’s number (\( 6.022 \times 10^{23} \, \text{molecules/mol} \)):
\[ \text{Number of molecules of O}_2 = 0.125 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \]
\[ \text{Number of molecules of O}_2 = 7.527 \times 10^{22} \, \text{molecules} \]
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

The matter is made up of very tiny particles and these particles are so small that we cannot see them with naked eyes.
The three states of matter are as follows: