What is a possible quantum number set for an electron in the 2s orbital of a beryllium (be) atom?

Answer is: 2,0,0,±1/2.

1) n = 1. The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

2) l = 0. The azimuthal quantum number is a quantum number for an atomic orbital that determines its orbital angular momentum and describes the shape of the orbital.

3) ml = 0. Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.

Magnetic quantum number (ml) specifies the orientation in space of an orbital of a given energy and shape . Magnetic quantum number divides the subshell into individual orbitals which hold the electrons, there are 2l+1 orbitals in each subshell.  

4) The spin quantum number, ms, is the spin of the electron; ms = +1/2 or -1/2.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml