A 50.5 L flask contains 3.25 mol of an unknown gas at 288.6 K, what is the pressure of
this gas?

1.52atm is the pressure of the gas

Explanation:

To solve this question we must use the general gas law:

PV = nRT

Where P is pressure in atm = Our incognite

V is volume = 50.5L

n are moles of gas = 3.25moles

R is gas constat = 0.082atmL/molK

And T is absolute temperature = 288.6K

To solve pressure:

P = nRT / V

P = 3.25mol*0.082atmL/molK*288.6K / 50.5L

P = 1.52atm is the pressure of the gas

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.

b. examine and describe cells