1) Convert 12.9 liters of Oxygen to mol at the given conditions:
PV = nRT ⇒ n = PV/RT
n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]
n = 0.636 mol of O2
2) use the stoichiometry derived from the balanced chemical equation
1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2
x = 0.636 / 3 mol O2 = 0.212 mol O2.
Answer: 0.212 mol O2
The empirical and molecular formula would be
and
respectively.
<h3>Empirical formula</h3>
The compound contains N and O.
N O
0.608/14 = 0.0434 1.388/16 = 0.0867
Divide by the smallest.
N = 1 O = 2
Thus, the empirical formula would be 
To get the molecular formula:
Empirical formula mass = 14 + (16x2) = 46
n = molar mass/empirical formula mass
= 92/46 = 2
Thus, the molecular formula would be 
More on empirical formulas can be found here: brainly.com/question/14044066
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<span>Calculating mass using Atomic Mass Units (amu):. Ex. Calculate the mass, in amu, of a sample of aluminum that contains 75 atoms. 1 Al atom = 26.98 amu.</span>
Answer: The density of the material is 2.66 g/mL and it is likely this is made of Aluminum
Explanation:
The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) =
. Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:

This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.