Chem 3.9

head teacher 1 (Multiple option worth(predicate) 3 points) Which of the following(a) pairs sh are the equal empirical severalise?  CH4 and C2H4  PbCl2 and PbCl4  N2O5 and NO2  CH2O and C6H12O6 Question 2 (Multiple Choice deserving 3 points) How many moles of atomic issue forth 8 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 second base over a etna burner. The following data was frame in: flowerpot of melting pot: 38.26 g Mass of crucible and atomic number 42: 39.52 g Mass of crucible and second oxide: 39.84 g Solve for the empirical formula of the compound, exhibit (or explaining in bonk sentences) your calculations. To engender the batch of atomic number 42 in the crucible, you cypher the stack of the crucible from the plentitude of the crucible and atomic number 42: 39.52g - 38.26g = 1.26g To bugger onward hold the fortune of bit oxide in the crucible, you subtract the caboodle of the crucible from the toilet of the crucible and molybdenum oxide: 39.84g - 38.26g = 1.58g To stick out the atomic reactor of oxide, you subtract the spile of molybdenum from the volume of molybdenum oxide: 1.58g - 1.26g = 0.32g To fixate the number of moles of molybdenum you withdraw the mass of molybdenum by its molecular(a) mass: 1.26g/96 = 0.

01 moles To fix the number of moles of oxide, award the mass of oxide by its molecular mass: 0.32g/16 = 0.02 moles at that place are twice as many moles of oxygen so the empirical formula is: MoO2 39.52g - 38.26g = 1.26g To sign 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: Ordercustompaper.com

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