In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outer most orbital. For example, it would be far easier to take electrons away from the larger element of Ca (Calcium) than it would be from one where the electrons are held tighter to the nucleus, like Cl (Chlorine). Hope this helped a little not the exact answer though :)
Answer:
2n^2 electrons
Explanation:
for example, the first level contain 2*(1^2) = 2 electrons, the second level contain 2*(2^2) = 8 electrons, etc
Answer:
mass ratio of A/B is 2:1
Explanation:
Since the mass of box A = 10g
mass of box B = 5g
Mass of box C = mass of box A + mass of box
A ratio compares two quantities. To find the ratio of the two boxes:
Ratio of A to B =
Ratio of A to B = = 2
The mass ratio is 2:1 i.e box A has twice the mass of B
A3B9 represents a molecular formula. The representation of the empirical formula for this compound is AB3. This is so because the empirical formula is the simplest ratio of the atoms present in the molecule. You get AB3 when you divide the subscripts of A3B9, this is 3 and 9, by the greatest common factor, which is 3. 3/3 = 1 and 9/3 = 3, so the subscripts for the empirical formula are 1 and 3, which is what AB3 represents. <span>Answer: AB3.</span>
There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital
here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.