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This is regarding a lab, that's why it refers to "you"
-You are expected to use a 35cm^2 piece of aluminium from a pop can. What is the average thickness of such beverage cans if you have 1.00 gram of metal from this piece?
Please explain if you can, I'm not sure which steps to take.

2007-03-26 16:43:15 · 3 answers · asked by universitystudent 2 in Science & Mathematics Chemistry

3 answers

I hope you mean that the piece is 1.00 grams rather than you have 1 g "from" the piece. Sloppy wording.
If yu assume that you have pure Al (not a particullarly good assumption) then the density of AL is expressed in g/cc
If you have 35 cm² then doing a dimensional analysis:
1.0g divided by density is g per (g/cc) or cc (cubic cm).
given 35cm² and the calculated volume of Al (in cm^3)
you will get the average thickness: vol/area=width

2007-03-26 16:55:26 · answer #1 · answered by Anonymous · 0 0

You need to know the density of aluminum (about 2.7 g/cc) Then you have the volume of the 1gram piece in cc. The thickness is then the volume divided by 35 cm^2.

2007-03-26 16:49:31 · answer #2 · answered by cattbarf 7 · 0 0

Enthalpy of reaction = Enthalpy of formation of goods MINUS Enthalpy of formation of the reactants dHf = enthalpy of formation (Greek delta and subscript are actually not showing will right here) First compute the products, remembering to multiply the stoichiometric coefficients (numbers) used to stability the equation. dHf [ products ] = 6 dHf [ CO2(g) ] + 6 dHf [ H2O(l)] dHf [ products ] = 6 x ( -393.5 kJ/mol ) + 6 x ( -285.8 kJ / mol ) dHf [ products ] = -2361.0 kJ/mol + ( -1714.8 kJ / mol ) dHf [ products ] = -4075.8 kJ / mol 2nd, compute the enthalpy of formation of the reactants. observe that Oxygen gas has an enthalpy of formation of 0 kJ/mol. via definition, all aspects have 0 kJ/mol as their enthalpy of formation dHf [ reactants ] = dHf [ C6H12(l) ] + 9 dHf [ O2(g) ] dHf [ reactants ] = -151.9 kJ/mol + 9 x 0 kJ / mol dHf [ reactants ] = -151.9 kJ/mol very final step, compute the enthalpy of reaction from the enthalpies of formation. dHreaction = dHf [ products ] - dHf [reactants ] dHreaction = -4075.8 kJ/mol - ( -151.9 kJ/mol) dHreaction = -3923.9 kJ/mol As an particularly final observe, be careful with stoichiometric indices, and the indications of the enthalpies. a basic mistake is to forget approximately the stoichiometric coefficients, or to multiply the indications incorrect.

2016-11-23 18:20:52 · answer #3 · answered by ? 4 · 0 0

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