A1.32-g sample of aluminum reacted with 1.174 g oxygen to form aluminum oxide. determine the empirical formula of the aluminum oxide.

Al₂O₃

Explanation:

Given 1.132 g Al and 1.174 O,

Number of moles of Al = mass of Al * 1 mole of Al/ molar mass of Al

Molar mass of Al = 26.98 g/mol

∴ number of moles of Al = 1.32 * 1/26.98 = 0.04892 moles

Number of moles of O = mass of O * 1 mole of O/ molar mass of O

Molar mass of O = 16g/mol

Number of moles of O = 1.174* 1/16 = 0.073375 moles

Aluminium / Oxygen = 0.04892/0.073375 = 0.6667 = 2/3

So, 2 moles of Al with 3 moles of O will be required

the answer is a and have a great day

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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