Chem 3.9

Question 1 (Multiple Choice worth(predicate) 3 points)
Which of the following(a) pairs sh are the same empirical formula?
 CH4 and C2H4
 PbCl2 and PbCl4
 N2O5 and NO2
 CH2O and C6H12O6

Question 2 (Multiple Choice Worth 3 points)
How many moles of oxygen are in 1 mole of Fe(NO3)3?
 2
 3
 6
 9

Question 3 (Essay Worth 5 points)
A compound that is composed of atomic number 42 (Mo) and oxygen (O) was produced in a lab by heating minute over a Bunsen burner. The following data was lay in:

flowerpot of crucible: 38.26 g
Mass of crucible and molybdenum: 39.52 g
Mass of crucible and molybdenum oxide: 39.84 g

Solve for the empirical formula of the compound, exhibit (or explaining in complete sentences) your calculations.
To engender the batch of molybdenum in the crucible, you subtract the volume of the crucible from the plentitude of the crucible and molybdenum: 39.52g - 38.26g = 1.26g To bugger off hold the fortune of molybdenum oxide in the crucible, you subtract the mass of the crucible from the mass of the crucible and molybdenum oxide: 39.84g - 38.26g = 1.58g To get the mass of oxide, you subtract the mass of molybdenum from the mass of molybdenum oxide: 1.58g - 1.26g = 0.32g To get the number of moles of molybdenum you discriminate the mass of molybdenum by its molecular mass: 1.26g/96 = 0.



01 moles To get the number of moles of oxide, divide the mass of oxide by its molecular mass: 0.32g/16 = 0.02 moles thither are twice as many moles of oxygen so the empirical formula is: MoO2
39.52g - 38.26g = 1.26g

To get the mass of molybdenum oxide in the crucible, you subtract the mass of the crucible from the mass of the crucible and molybdenum oxide:
39.84g - 38.26g = 1.58g

To get the mass of oxide, you subtract the mass of molybdenum from the mass of molybdenum oxide:
1.58g - 1.26g = 0.32g

To get the number of moles of molybdenum you divide the mass of molybdenum by its molecular mass:
1.26g/96 = 0.01 moles

To get the number of moles of oxide, divide the mass of oxide by its molecular...If you want to get a full essay, order it on our website: Orderessay



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