Question 2(multiple choice worth 5 points) (06.04 hc) during a laboratory experiment, 36.12 grams of al2o3 was formed when o2 reacted with aluminum metal at 280.0 k and 1.4 atm. what was the volume of o2 used during the experiment? 3o2 + 4al → 2al2o3 8.70 liters 9.22 liters 10.13 liters 12.81 liters

8.70 liters.

Explanation:

3O₂ + 4Al → 2Al₂O₃,

It is clear that 3.0 moles of O₂ react with 4.0 moles of Al to produce 2.0 Al₂O₃.

n = mass/molar mass = (36.12 g)/(101.96 g/mol) = 0.3543 mol.

using cross multiplication:

3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.

??? mol of O₂ produces → 0.3543 mol of Al₂O₃.

∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃ = (3.0 mol)(0.3543 mol)/(2.0 mol) = 0.5314 mol.

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.4 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 0.5314 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 280 K).

∴ V = nRT/P = (0.5314 mol)(0.0821 L.atm/mol.K)(280 K)/(1.4 atm) = 8.726 L ≅ 8.70 L.

So, the right choice is: 8.70 liters.

8.70 liters

Explanation:

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.