Question

14

of 18

7 points

How many liters are occupied by 18.0g of oxygen gas (O2)at STP (standard air

temperature and pressure)? ONLY ENTER THE NUMBER OF YOUR

ANSWER. ROUND TO THE HUNDREDTHS PLACE IN YOUR ANSWER. DO

NOT LABEL



Submit

The volume of oxygen gas at STP is 12.60 L

Explanation:

Step 1: data given

Mass of oxygen = 18.0 grams

Molar mass of oxygen = 32.0 g/mol

STp = 1atm and 273 K

Step 2: Calculate moles of oxygen

Moles O2 = mass O2 / molar mass O2

Moles O2 = 18.0 grams / 32.0 g/mol

Moles O2 = 0.5625 moles

Step 3: Calculate volume of the gas

p*V = n*R*T

⇒with p = the pressure at STP = 1.00 atm

⇒with V = the volume of the oxygen gas = TO BE DETERMINED

⇒with n = the number of moles of oxygen gas = 0.5625 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature at STP = 273 K

V = (n*R*T)/p

V = (0.5625*0.08206 * 273) / 1 atm

V = 12.60 L

The volume of oxygen gas at STP is 12.60 L

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.