# Math Help - Chemistry Question

1. ## Chemistry Question

Assume that an atom of neon is spherical with a volume V of 4.85 x 10^-24 m^3. If the equation relating radius r and volume V of a sphere is

V = 4/3 pie r^3 (sorry, i don't know how to do the pie symbol on the computer)

determine the atomic radius of a neon atom.

Not sure if I'm right, but I have:

V = 4.85 x 10^-24m^3 = 4/3 pie r^3

0.0485m^3 = 4/3 pie r^3

Is this right? And if so, do I solve it any further? If I'm supposed to, how?

Cheers

2. $V = \frac{4}{3} \pi r^3$

$\frac{3V}{4\pi} = r^3$

$r = \sqrt[3]{\frac{3V}{4\pi}}$

btw, this is pie ...

... and this is pi.

3. ## Chemistry Question

lol, I it was late and I had food on the brain .

So, where it says 3V, would you have

3 x 4.85x10^-24
_________________________________
4pi

?

r = 1.050068691x10^-8

Don't know if I'm right, or if I'm completely confused?

4. Originally Posted by AJShaw
lol, I it was late and I had food on the brain .

So, where it says 3V, would you have

3 x 4.85x10^-24
_________________________________
4pi

?

r = 1.050068691x10^-8

Don't know if I'm right, or if I'm completely confused?
You would never round it off that much anyway. As many digits are given in the question will do - in this case 3sf.

Where skeeter put 3V you put in the value of V given then evaluate the expression.

$\sqrt[3]{\frac{3 \times 4.85 \times 10^{-24}}{4\pi}} = 1.05 \times 10^{-8} \text { m } = 10.5 \text{ nm }$

According to Data this is about 2 orders of magnitude too large but the calculated answer holds given the volume named

5. I get it now. Thanks heaps