Solution of the schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. each function is characterized by 3 quantum numbers: n, l, and ml. if the value of n = 1 the quantum number l can have values from to . the total number of orbitals possible at the n = 1 energy level is . if the value of l = 1 the quantum number ml can have values from to . the total number of orbitals possible at the l = 1 sublevel is .

If n=1, l can only have the value l=0 and the total number of orbitals is 0. If l=1, ml can be -1,0 or 1 and the total number of orbitals at l=1 is 3.

Explanation:

When solving the radial part of Schrödinger equation, one needs to expand in power series that lead to Laguerre polynomials of the form , being n and l the quantum numbers. As is known, in the Laguerre polynomials the subindex must be greater than or equal to 0, which implies . So if n=1, the only possible value of l is l=0.

Likewise, when solving the angular part, one gets the spherical harmonics can be . Therefore the sublevel l=| has 3 possible orbitals.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%