Anybody can solve this problem? please help me.

[exderwall]

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Gen.Chemistry problem.

A roll of aluminum wire is used to make fishing nails. If the nails are 5.00 cm long and 0.300 cm in diameter and the roll of wire has a mass of 225 kg, how many nails can be made from one roll?

I got stucked with this one.
Please help.

 

[doc3232]

Gen.Chemistry problem.

A roll of aluminum wire is used to make fishing nails. If the nails are 5.00 cm long and 0.300 cm in diameter and the roll of wire has a mass of 225 kg, how many nails can be made from one roll?

I got stucked with this one.
Please help.

Not enough information. You either forgot to tell us something or the question is wrong.

 

[billw]

You need to look up the density of aluminum, calculate the volume of each nail, calculate the mass of each nail, and then divide into the total mass of wire.

 

[doc3232]

You need to look up the density of aluminum, calculate the volume of each nail, calculate the mass of each nail, and then divide into the total mass of wire.

I was thinking it was a DAT question...you just want help for your homework.

 

[klutzy1987]

You also have to somehow calculate the volume of the nail without knowing howit changes at the head of the nail. There is nowhere near enough info. However if you could calculate the volume of the block of aluminum by getting the density of alluminum (2.7g/cm³) and dividing the mass by the density to get the volume which is 83333.33 cm³. Then you would calculate the volume of the nail which you cant because the volume would change at the head. Lets assume that it doesnt change though, then the volume of the nail would be (.15^2*pi)*5cm. This equals .1125 Then divide 83333.33 by .1125 to get 740740.72. So you should be able to get 740740 nails. Question is very flawed though.

 

[doc toothache]

You also have to somehow calculate the volume of the nail without knowing howit changes at the head of the nail. There is nowhere near enough info. However if you could calculate the volume of the block of aluminum by getting the density of alluminum (2.7g/cm³) and dividing the mass by the density to get the volume which is 83333.33 cm³. Then you would calculate the volume of the nail which you cant because the volume would change at the head. Lets assume that it doesnt change though, then the volume of the nail would be (.15^2*pi)*5cm. This equals .1125 Then divide 83333.33 by .1125 to get 740740.72. So you should be able to get 740740 nails. Question is very flawed though.

Reading too much into the problem.