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chem experts help please
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| makija |
ok i need some help on this one ... anyone that can come up with a solution and answer i would appriciate it alot ;)
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k so this is a friend in a problem ... anyone that is good in chemistry help out please
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In a chem lab i have a graduated cylinder with 20.0 mL of water.
The Mass of the cylinder and water together is 107.470 g
Then had to continue adding an unknown metal in the water in increments of 10.00 g until i have added a total mass of 50.00 g. and After each addition, recorded the mass and volume.
so it looked like this
10.00 mL - 27.41 mL / 127.470 g
10.00 mL - 31.11 mL / 137.470 g
10.00 mL - 34.81 mL / 147.470 g
10.00 mL - 38.52 mL / 157.470 g
10.00 mL - 42.22 mL / 167.470 g
10.00 mL - 45.93 mL / 177.470 g
10.00 mL - 49.63 mL / 187.470 g
The questions :
The final volume of water in the graduated cylinder after addition of the unknown metal (in mL or cm3):
(1) The net volume of the unknown metal (in mL or cm3):
(2) mass of the unknown metal (in g):
(3) density of the unknown metal (in g/cm3):
(4) Calculate and record here the average density of the unknown metal. Express your results in g/cm3.
(5) Identify your coded unknown metal considering the data below. The following is a chart of the literature values for the densities of some standard metals.
Aluminum 2.702 g/cm3
Copper 8.960 g/cm3
Iron 7.865 g/cm3
Lead 11.34 g/cm3
Magnesium 1.738 g/cm3
Silver 10.50 g/cm3
Zinc 7.149 g/cm3
6. Calculate the percentage error of your experimental value in comparison with the accepted literature value:
%error = [(expt. value) - (lit. value)] / (lit. value) x 100% |
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| UWM |
I'll be back with an answer for you in 5 minutes.
p.s This problem is really easy.
(1) 29.63 cm3
(2) 80 g
(3) 80g/29.63cm3
(4) 2.7g/cm3
(5) Aluminum
(6) .07% |
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| makija |
| thx man haha you got msn or something i might need some more help in the future ;) |
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| UWM |
Would help more but I'm in the midst of studying for something a bit more challenging atm :) (Biochem 620 - Advanced Eukaryotic Molecular Biology).
I'll PM you my msn later so if you have any ?s you can fire them away ... after 10:00 (GMT-6) tomorrow :D |
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| SidMl |
| quote: | Originally posted by UWM
I'll be back with an answer for you in 5 minutes.
p.s This problem is really easy.
(1) 29.63 cm3 = final value - initial value = 49.63 - 20
(2) 80 g = final value - initial value = 187.47 - 107.47
(3) 80g/29.63cm3 = mass / volume
(4) 2.7g/cm3 = how is this different from #3?
(5) Aluminum = closest one on the table
(6) .07% |
UWM FTW.
I wasn't even going to look at the problem. But yeah, it's pretty easy. UWM's answers are correct but I added stuff to show how he got them. |
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| UWM |
| I didn't really know what the difference was between #3 and #4, so I just put those answers there. It seemed like the same question to me. |
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