Asked by Krista
an aluminum cylinder is 10.0 cm in length and has a radius of .25 cm. if the mass of a single atom of aluminum is 4.48X10^-23 calculate the number of aluminum atoms present in the cylinder. the density of aluminum is 2.70 g/cm^3
Please answer quickly.
Please answer quickly.
Answers
Answered by
Steve
g/cm^3 * cm^3 * atoms/g = atoms
or,
g/cm^3 * cm^3 ÷ g/atom = atoms
2.70 * (pi * .25^2 * 10) / (4.48*10^-23) = 1.18 * 10^23
or,
g/cm^3 * cm^3 ÷ g/atom = atoms
2.70 * (pi * .25^2 * 10) / (4.48*10^-23) = 1.18 * 10^23
Answered by
muhammed
An aluminum cylinder is 10.0 cm in length and has a radius of 0.25 cm.
If the mass of a single Al atom is 4.48 × 10−23g, calculate the number of
Al atoms present in the cylinder. The density of aluminum is 2.70 g/cm
If the mass of a single Al atom is 4.48 × 10−23g, calculate the number of
Al atoms present in the cylinder. The density of aluminum is 2.70 g/cm
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.