IF 3.25 mol of argon gas occupies a volume of 100. L at a particular temperature and pressure, wihat volume does 14.15 moi of argon occupy under the same conditions?

435.38 L

Explanation:

From the question given above, the following data were obtained:

Initial mole (n₁) = 3.25 mole

Initial volume (V₁) = 100 L

Final mole (n₂) = 14.15 mole

Final volume (V₂) =?

The final volume occupied by the gas can be obtained as follow:

V₁/n₁ = V₂/n₂

100 / 3.25 = V₂ / 14.15

Cross multiply

3.25 × V₂ = 100 × 14.15

3.25 × V₂ = 1415

Divide both side by 3.25

V₂ = 1415 / 3.25

V₂ = 435.38 L

Thus, the final volume of the gas is 435.38 L

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.