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May 22 2013 09:10pm
Instruction
for the following, report answers with correct sig figs. scientific notation is optional

1) a lead brick has a weight of 168oz(dry). convert to units of grams
2)the same brick occupies a volume of 14.3 fluid oz. convert to units of cm^3
3)using the data above, calculate the density of the lead brick in units of g/cm^3

please explain mainly im trying to understand so must put explanation to get fg
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May 22 2013 09:33pm
1.)

1oz is 28.3495g, to find how many 168oz in gram, you just mutiply it. So, 168oz x 28.3495 = 4762.416 grams. But the limiting sig fig is 168 (it only has 3, where as the 28.3495 has 6). .---> so the final answer has to have 3 sigfg. Which equals to 4,760 grams.

2) Same logic as problem one. You find that 1oz has 29.57 cm3, so you just multiply 168 x 29.57cm3 .. which equals 4967.76, reduce it to 3 sigif, becomes 4980cm3. (you pick 496 as the three, but have to round up because of the 7).

3 Density is mass divide by volume. 4760/4980 = .9558232932 g/cm3 . Reduce it to 3 sigfig, becomes .956g/cm3. (round up the 5 because of the 8).

Good luck hope it helps!
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May 22 2013 09:39pm
Quote (XylaMints @ May 22 2013 11:33pm)
1.)

1oz is 28.3495g, to find how many 168oz in gram, you just mutiply it. So, 168oz x 28.3495 = 4762.416 grams. But the limiting sig fig is 168 (it only has 3, where as the 28.3495 has 6). .---> so the final answer has to have 3 sigfg. Which equals to 4,760 grams.

2)  Same logic as problem one. You find that 1oz has 29.57 cm3, so you just multiply 168 x 29.57cm3 .. which equals 4967.76, reduce it to 3 sigif, becomes 4980cm3. (you pick 496 as the three, but have to round up because of the 7).

3 Density is mass divide by volume.  4760/4980 = .9558232932 g/cm3 . Reduce it to 3 sigfig, becomes .956g/cm3. (round up the 5 because of the 8).

Good luck hope it helps!


2) Same logic as problem one. You find that 1oz has 29.57 cm3, so you just multiply 168 x 29.57cm3 .. which equals 4967.76, reduce it to 3 sigif, becomes 4980cm3. (you pick 496 as the three, but have to round up because of the 7).
for that u said it equals to 4867.76 and u said it needs to be round up wouldnt that make it 487? not 4980?
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May 22 2013 09:43pm
adding more questions

4.using ur rounded density from question 1(c), calculate i) the volume of a lead rod that has a mass of 1986.4g, and ii) the mass of a lead block that has a volume of 112.8cm^3

5. a graduated cylinder contains 3.55ML of water. a piece of non-reactive metal weighing 3865mg is placed in the cylinder, and the water level rises to 4.30ML. is this metal also lead? Explain.

This post was edited by Huntyoudown on May 22 2013 09:44pm
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May 22 2013 11:01pm
Quote (Huntyoudown @ May 22 2013 08:39pm)
2) Same logic as problem one. You find that 1oz has 29.57 cm3, so you just multiply 168 x 29.57cm3 .. which equals 4967.76, reduce it to 3 sigif, becomes 4980cm3. (you pick 496 as the three, but have to round up because of the 7).
for that u said it equals to 4867.76 and u said it needs to be round up wouldnt that make it 487? not 4980?


yea my bad on that one. So 4967.76 becomes 4970.
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May 22 2013 11:12pm
Quote (Huntyoudown @ May 22 2013 08:43pm)
adding more questions

4.using ur rounded density from question 1(c), calculate i) the volume of a lead rod that has a mass of 1986.4g, and ii) the mass of a lead block that has a volume of 112.8cm^3

5. a graduated cylinder contains 3.55ML of water. a piece of non-reactive metal weighing 3865mg is placed in the cylinder, and the water level rises to 4.30ML. is this metal also lead? Explain.



Answer for 4:

Density =(mass/v) ----> So you know the density to be .956g/cm3, set this equal to Mass/Volume -----> .956= 1986.4g/v , solve for v= 2077.8cm3, reduce this to 3 sigif, and it equals 2080cm3.

ii). Density = mass/v, we are given density and v, find m ---> .956=m/112.8cm3, solve for m, ---> m =107.8368g, reduce this to 3 sigif, 108g.

5. The difference in water level means the total volume that the metal displace, so the total volume is the final water level minus the initial water level. (4.30-3.55 = 0.75ml). We can find the density of the metal by divinding its mass by its volume ----> 3865mg/.75 = 5153mg/cm3 , which is equal to 5.153g/cm3. The density of lead is .956g/cm3, this does not match it, therefore it is not lead.
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May 23 2013 03:49pm
By the way, 4970 is 4 significant figures. If you want to use proper scientific notation and 3 sig figs, you have to write it as 4.97 * 10^3
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May 23 2013 05:22pm
Quote (russian @ May 23 2013 03:49pm)
By the way, 4970 is 4 significant figures. If you want to use proper scientific notation and 3 sig figs, you have to write it as 4.97 * 10^3


i believe how it is written its 3 sig figs. 4970. with a decimal would be 4 i think. its been awhile since ive had to worry about sig figs
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May 23 2013 06:43pm
Quote (cialda @ May 23 2013 04:22pm)
i believe how it is written its 3 sig figs. 4970. with a decimal would be 4 i think. its been awhile since ive had to worry about sig figs


http://www.edu.pe.ca/gray/class_pages/krcutcliffe/physics521/sigfigs/sigfigRULES.htm
says it's 4 sig figs
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May 23 2013 07:15pm
Quote (russian @ May 23 2013 06:43pm)


http://www.sciencegeek.net/Chemistry/taters/Unit0SigFigCounting.htm

says its 3 or 4. its like saying 530,000 mph has 6 six figs even though you can only measure it to "5" and the 3 is an estimate - in reality its only 2 for such said case <-- this is at least what i have been taught.

this is different than being able to measure the speed being 530,000, where the last bolded number is an estimate and the zero in the tens place is the last digit you measured.

This post was edited by cialda on May 23 2013 07:16pm
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