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# 2015 AP Chemistry free response 7

Energy necessary to recycle aluminum from aluminum oxide. From 2015 AP Chemistry free response 7.

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• In part (a) at , how do you know if it is appropriate to use the 'molar quantity' rather than the molecular mass of Aluminium? (as in q = mass x specific heat capacity x temp change)? Is this only because we are given the MOLAR heat capacity? • Hi! A question about part B: Don't we have to take into account the phase and temperature difference between Al (s) (for extracting) and Al (l) (for purifying @ 933K) if we look at energy requirement? • Actually you wouldn't worry about the difference of energy between the Al(extracted from Al2O3) and the Al(purified) because the purpose of the two processes is to get some aluminum (for making a cool sword of aluminum, for example), even though they're in different phases. What you could think about is how to get benefits from the already melted Aluminum to raise the temperature of other things (Although I don't know if and when it really would occurs). I hope this cleared out your doubts :) .
• Hi! So, The heat of fusion is the heat that is actually needed to both melt AND solidify?
Am I right? And also, for part a, why does the answer has to be in 2 sig.fig? Thanks. • Correct, the heat of fusion is the amount of heat required to melt a substance and the amount of heat given off by a substance solidifying.

The correct number of sigfigs is 2 because your final answer can't have more than the minimum number you started with. To see why this is true for 15,000 + 10,700, it may help to think about what 15,000 really means. Sigfigs are related to rounding — 15,000 means the real answer might be any amount from about 14,500 - 15,400. If we add 10,700 (really this is also a range) to each of those numbers we get 25,200 - 26,100. If we take the average of those we get 25,650. Rounded to 2 sigfigs that gives us 26,000, which is better than 25,650 because it conveys that we know the real answer is 20,000 + around 6,000 ...
• Why do we not multiply in the formula by Al in grams? Why do we multiply using moles instead? • Sal used the heat equation to calculate the heat required to increase the temperature of one mol of Al from 298K to 933K. Since the specific heat capacity is in units of J/(mol)(K), he use this form of the heat equation: q = n(C)(deltaT) ; where n = number of moles. If the specific heat capacity had been given in J/gK, then he would have converted one mole to grams using the molar mass. And he would have used this form of the heat equation:
q = mass(C)(deltaT)
• dont you think that the heat of fusion of Al is the answer to part A?..because fusion means Solid to Liquid. So isn't 10.7KJ is the heat required to melt 1 mole of Al to evaporate impurities?...plz help • Yes, but it is only a part of the answer. Just like you explained, we will eventually have to melt the Al (transform it from solid state to liquid state). But at what temperature does this physical transformation take place? It takes place at 933 K. The Al at our disposition is only at 298 K, so we need to reach the fusion temperature, which we will achieve by heating the Al. So the final answer for part A will have two parts: first, the heating of the Al from 298 K to 933 K; second, since we now have reached the necessary temperature for fusion, the actual fusion of Al.
• at we divide by two, i thought we had to multiply. help please? • hello , i can not see what formula do you use to calculate the needed heat to rise till 635K, and the other formula too , where we calculate the melting heat.
Thx. • Why don't you just write the whole answer at ? Why round?   