# Chemistry: An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container, if the gas has a pressure of the 845mmHg at 22degrees Celsius? ​

There are 29.4 grams of oxygen in the container

Explanation:

Step 1: Data given

Volume = 20.0 L

Pressure = 845 mmHg

Temperature = 22.0 °C

Molar mass of O2 = 32 g/mol

Step 2: Ideal gas law

p*V = n*R*T

⇒ p = the pressure of the gas = 845 mmHg = 1.11184

⇒ V = the volume of the gas = 20.0 L

⇒ n = the number of moles = TO BE DETERMINED

⇒R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C + 273 = 295 Kelvin

n = (p*V)/(R*T)

n = (1.11184*20.0)/(0.08206*295)

n = 0.9186 moles

Step 3: Calculate mass of NO2

Mass of O2 = Moles O2 * Molar mass O2

Mass of O2 = 0.9186 moles * 32 g/mol

Mass of O2 = 29.4 grams

There are 29.4 grams of oxygen in the container

## Related Questions

Butane, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete combustion of butane is 2c4h10(g)+13o2(g)?8co2(g)+10h2o(l) at 1.00 atm and 23 ?c, what is the volume of carbon dioxide formed by the combustion of 3.20 g of butane?

Molar mass of butane = 12g/mol*4+1g/mol*10=58g/mol

Mass of 2 moles of butane=2mol*58g/mol=116g

2c4h10(g)+13o2(g)->8co2(g)+10h2o(l)

116g                        8 moles

3.20g                           x

116g butane/8moles CO2=3.20g butane/x

x=3.20g butane*8moles CO2/116g butane=0.2207moles CO2

T=23°C=296K

PV=nRT

V=nRT/P=0.2207moles*0.082(atm*dm³/(K/mol))*296K/(1atm)=5.36dm³

Which of these pairs of elements is most likely to be part of a polyatomic ion? Cr and O
K and F
Li and Br
Mg and O

Cr and O are most likely to be part of a polyatomic ion.

Cr belongs to the transition metal series in the periodic table and has vacant d-orbitals. As a result it can exist in more than one oxidation state, the most common being +6, +3 and +2. Oxygen exists in -2 oxidation state. Hence, Cr and O can combine to form different polyatomic ions like: CrO₄²⁻ and Cr₂O₇²⁻ (Cr in +6 state)

K, Li and Mg exist in only one oxidation state +1 (K and Li) and +2(Mg). F and Br exhibit an oxidation state of -1. Therefore, they are not likely to form polyatomic ions.

Ans (a) Cr and O

Cr and O I believe. I feel like I've seen it before as a polyatomic ion

True or false: a mixture is always made up of a combination of elements