58. What is the temperature if 30.1 g of N, in a 1.00-L container is at a pressure of 25.2 atm? DUE TONIGHT. LINKS DONT WORK SO PLS SHOW WORK. THANK YOU SO MUCH

14°C

Explanation:

In order to use the ideal gas law, you need the number of moles of N2 present.

30.1 g x (mol / 28.02 g) = 1.07 mol N2

Use the ideal gas law: PV = nRT

Now isolate for T(temperature) and plug in your other values:

T = PV / nR

T = (25.2 atm)(1.00 L) / (1.07 mol)(0.08206)

T = 287 K

T(°C) = 287 K - 273 K = 14°C

Therefore, the temperature is 14°C

answer; /// i do believe that the correct answer is (a) boiling

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.