I have an aluminum cylinder; length =10cm and radius =.25cm density of aluminum =2.70g/cm3 1Al atom =4.48*10-23g How many Al Atoms are in the cylinder? Would you find the volume first? then do mass = d*v? and divide that by the mass of 1 Al atom?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I have an aluminum cylinder; length =10cm and radius =.25cm density of aluminum =2.70g/cm3 1Al atom =4.48*10-23g How many Al Atoms are in the cylinder? Would you find the volume first? then do mass = d*v? and divide that by the mass of 1 Al atom?

Chemistry
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

@toxicsugar22 do you know?
@3-sfera doing what i explained in the question i got 1.18*10^23 atoms of Al, do you know if i did this correctly
let me do some math

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

no my result is 1,2x10^27
1 Attachment
i did Vcylinder = pi r^2(h) pi(.25^2)(10) = 1.96cm^3 then mass = d*v m= 2.70*1.96 = 5.292 then 5.292 / 4.48*10^-23 = 1.18*10^23
in the question it says taht 1 atom of Al = 4.48 * 10^-23 g
my solution is with the molar mass, your volume is incorrect i think. the volume is 1,96*10^3 cm^3
so my number of atoms is correct too? 1.18*10^23
uhm actually not :).. maybe just some different approximation..maybe the mass on your book is wromg
Well by the book, the answer is correct then?
@aaronq can you please help?
the answer with your atom mass is 1.8*10^26
sorry for late i'm a little bit busy
i don't understand how you got that.
oh flutter wait.. the radius is 0.25
*flutter
ahha i wanna write f*u*c*k but it shows flutter
by the way, i misread the radius
your answer is correct i think

Not the answer you are looking for?

Search for more explanations.

Ask your own question