1. <span>Count Sergei Lvovich Levitsky invented the Bellows camera. The correct option among all the options that are given in the question is the last option or option "D".
2. The substance with which the first photograph was made from, was Bitumen. The correct option among all the options given in the question is the first option or option "A".
3. Kodak DCS 100 was the first commercially available digital camera. The correct option among all the options that are given in the question is the first option or option "A".</span>
False
Although we use many of their ideas to describe atoms today, such as the existence of a tiny, dense nucleus in an atom (proposed by Rutherford), or the notion that all atoms of an element are identical (proposed by Dalton), some of their ideas have been rejected by the modern theory of the atom.
For example, Thompson came up with the plum pudding model to describe an atom, which resembled a sphere of positive charge with electrons embedded in it. We know now, however, that atoms are mostly empty space with a tiny, dense nucleus.
Another example is Dalton's atomic theory, which stated that atoms are indivisible particles. However, this was disproved by the discovery of subatomic particles.
Answer:
Light can be reflected, transmitted, or absorbed. You have read that EM waves can interact with a material medium in the same ways that mechanical waves do.
Explanation:
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 