D = m / V
0.736 = 225.0 / V
V = 225.0 / 0.736
V = 305.7 cm³
Based upon Max Planck's theory of black-body radiation, Einstein theorized that the energy in each quantum of light was equal to the frequency multiplied by a constant, later called Planck's constant. A photon above a threshold frequency has the required energy to eject a single electron, creating the observed effect.
They are too small to see with the naked eye
Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,

where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:


The relation between the equilibrium constant and standard Gibbs free energy is:

where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:


Therefore, the value of equilibrium constant for this reaction at 262.0 K is 