B - Atomic number. Dmitri Mendeleev organised the table according to atomic weight, however this caused problems with elements such as iodine and tellurium, Iodine has a higher mass, but a lower atomic number. And to make iodine in the same group as similar elements (halogens), Mendeleev had to break his own rules and put it before tellurium in the table. Moseley fixed this problem by ordering the elements according to atomic (proton) number.
Assuming that both cases describe hydrogen‑like atoms with one electron, More energy is emitted or absorbed for case 2. The correct option is D.
<h3>What is emitting of energy, by electron?</h3>
The energy of the electron decreases as it changes levels, and emission of photons happens in the atom.
With the electron moving from a higher to a lower energy level, the photon is emitted. The photon's energy is the same as the energy lost by an electron moving to a lower energy level.
Thus, the correct option is D, More energy is emitted or absorbed for case 2.
Learn more about emitting of energy
brainly.com/question/14350185
#SPJ4
Answer:
A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:
Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n
According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:
Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
