Answer:
See explanation
Explanation:
Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.
In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.
Answer:
Using the formula cards again, add the coefficient of 2 in front of the formula and have them recalculate the number of each element and the total number of atoms in each element.
Explanation:
A frequency of 60 MHz is close to the lower end of the old VHF-TV band.
c = f λ ...... where c is the speed of light, f is the frequency and λ is the wavelength
λ = c / f = 3.00x10^8 m/s / 6.0x10^7 1/s
λ = 5.0 m
