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

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

1 a

2 d

3 a

4 c

explanation: