What is the volume of 6.9 mol oxygen (o2) gas at 233 k and a pressure of 4.0 atm? (the universal gas constant is 0.0821 l•atm/mol•k.)

Volume of Oxygen O2 would be 33 L.

Solution : Given information - Mole (n) = 6.9 mol

Pressure (P) = 4.0 atm

Temperature (T) = 233 K

Universal gas constant (R) = 0.0821 L.atm/mol.K

Volume(V) is unknown , which we need to find.

Volume can be calculated by using the Ideal gas equation which is :

PV = nRT  (Ideal gas equation)

On rearranging the above formula we will get :

Where , P = pressure , V = volume , n = mole , R = universal gas constant , T = temperature.

We will plugin the values of P , n , R and T in the ideal gas equation and then calculate Volume (V)

V = 32.998 L or 33 L

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.