How Many Meters of 34-Gauge Wire Can Be Made from 5.01 lb of Covellite?

AI Thread Summary
The discussion revolves around calculating the length of 34-gauge wire that can be produced from 5.01 lb of covellite, which contains 66% copper by mass. The participants clarify that to find the volume of copper, one must first convert the mass of covellite to grams and then determine the mass of copper. Using the density of copper, they can calculate the volume available for wire production. The volume of the wire, treated as a cylinder, is expressed as V = πr²L, where L is the length of the wire. The conversation highlights the importance of correctly interpreting variables in formulas, particularly distinguishing between height and length.
metalmagik
Messages
130
Reaction score
0

Homework Statement


Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10^-3 in) can be produced from the copper in 5.01 lb of covellite, an ore of copper that is 66% copper by mass? (Hint: Treat the wire as a cylinder. d of copper = 8.95 g/cm^3)


Homework Equations


V = (pi)(r^2)(h)


The Attempt at a Solution


So I used the formula for Volume of a cylinder, V = (pi)(r^2)(h) and plugged in half of the given diameter for r and then the whole diameter for h. And I got 1.968 x 10^-7 cubic inches. How should I go about doing the rest of the problem? It's really confusing me. Help please!
 
Physics news on Phys.org
Think of it this way. You are given a certain mass of covellite (5.01 lb = 2273 grams), of which 66% is copper (1500 grams). You know the density of copper, so you can calculate the volume of copper available, since

d = \frac{m}{V}

Once you calculate the volume, you want to form the copper into the shape of a cylinder (wire), and the volume of the cylinder = volume of copper available. For a cylinder, the volume can be written

V = pi r^{2} L

where L is the length we wish to find. Take V from the first part and solve this equation for L.
 
Ah man I totally get it now. I wasn't thinking of the variable in the Volume formula as length, my book labels it as h, which I interpreted for height.

Thanks so much edziura! That really helped me to clarify my thinking, all this stuff with mass is so difficult, its all over my homework which is due tomorrow, and my professor hasn't even gotten past dimensional analysis in lecture! Thanks again.
 
Your welcome.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top