Sulfur is a group six element in period 3 with atomic number 16 and an electronic configuration of 2:8:6. Therefore, to attain a stable configuration it requires to gain two electrons forming an ion with a charge of -2. The negative charge is due to the gaining of electrons.
Answer:

Explanation:
If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
Look up the molar mass of carbon.
Set up a ratio using the molar mass.

Since we are converting 3.06 grams to moles, we multiply by that value.

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.
The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................
Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
Answer:
At -13
, the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

where
and
are initial and final pressure respectively.
and
are initial and final volume respectively.
and
are initial and final temperature in kelvin scale respectively.
Here
,
,
,
and
Hence 



So at -13
, the gas would occupy 1.30L at 210.0 kPa.