The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell.
The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
It would be 23.46, since the next number is a 5 so we can round up.
Significant figures are:
ANY number that is not 0
Any 0 between two nonzero digits
Any 0 following a decimal (i.e 1.50)
Answer : The value of equilibrium constant for this reaction at 328.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 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 = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 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 328.0 K is
The atomic mass is the average mass of all the isotopes. The atomic number is the number of protons in the nucleus of an atom. In an uncharged atom the atomic number is also equal to the number of electrons.
Hope this helps!
your answer would be C hope this helps
