Answer:
The answer to your question is: 70.8%
Explanation:
Data
Al₂O₃ = 60 g
C = 30 g
CO = gas
Al = 22.5 g
MW Al₂O₃ = 102 g
MW C = 12 g
MW Al = 54 g
Reaction
Al₂O₃ + 3C ⇒ 3 CO + 2 Al
Limiting reactant
102 g of Al₂O₃ -------------- 54 g Al
60 g -------------- x
x = 31.8 g
36 g of C ------------------ 54 g of Al
30 g of C ------------------ x
x = 45 g of Al
Limiting reactant = Al₂O₃
Percent yield = 
Percent yield = 70.75 %
Answer:
11
Explanation:
electronic configuration = 1s^2, 2s^2, 2p^6, 3s^1
therefore total electrons = 2+2+6+1 = 11
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Answer:
I would use calorimetric to determine the specific heat and I would measure the mass of a sample
Explanation:
I would use calorimetry to determine the specific heat.
I would measure the mass of a sample of the substance.
I would heat the substance to a known temperature.
I would place the heated substance into a coffee-cup calorimeter containing a known mass of water with a known initial temperature.
I would wait for the temperature to equilibrate, then calculate temperature change.
I would use the temperature change of water to determine the amount of energy absorbed.
I would use the amount of energy lost by substance, mass, and temperature change to calculate specific heat.
The volume of one mole of any gas at Standard Temperature and Pressure (1 atm and 0 degrees Celsius [273K]) is 22.4 L.
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
