Calculating Al Foil Thickness in mm

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To calculate the thickness of a sheet of aluminum foil with an area of 1.000 ft² and a mass of 3.636 g, the density of aluminum (2.699 g/cm³) is used in the formula d = m/v. The volume (v) can be derived from the mass divided by the density. The initial attempt at a solution was incorrect, indicating a misunderstanding of the units and calculations involved. Clarification on the variables used in the density equation was sought, emphasizing the need for accurate unit conversion to find the thickness in millimeters. The discussion highlights the importance of correctly applying formulas and understanding unit conversions in physics problems.
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Homework Statement



A Sheet of aluminum(Al) foil has a total area of 1.000ft^2 and a mass of 3.636g.

What is the thickness of the foil in millimeters (density of Al=2.699g/cm^3)?

Homework Equations



d=m/v



The Attempt at a Solution



1.347=totally wrong... help please
 
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1.347 is probably the right answer in some system of units :-p

Also, what is the d in your "relevant equations"? Is it the d for density? Or is it the d that is sometimes used for small distances (such as, say, thickness?) And what is v? How do you calculate it from the given data?

Your "attempt" is not so much an attempt from which we can clearly locate some error, as some numerical guess, without units or any other relevant information.
 
thank you
 
You're welcome.
I take it this means you solved it?
 
Thread 'Confusion regarding a chemical kinetics problem'

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...
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