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At room temperature, an oxygen molecule, with mass of #5.31 x 10^-26 kg#, typically has a KE of about #6.21 x 10-21# J. How fast is the molecule moving?

Answer 1

Adrian Ayala

#484ms^(-1)#

#E_k=(mv^2)/2=6.21*10^(-21)J#

#6.21*10^(-21)=(5.31*10^(-26)v^2)/2#

#v=sqrt((2(6.21*10^(-21)))/(5.31*10^(-26)))=483.630339~~484ms^(-1)#

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Answer 2

Lily Adams

To find the speed of the oxygen molecule, we can use the kinetic energy formula:

KE = (1/2)mv^2

Where: KE = kinetic energy (6.21 x 10^-21 J) m = mass of the molecule (5.31 x 10^-26 kg) v = speed of the molecule (what we're solving for)

Rearranging the formula to solve for v:

v = sqrt((2*KE)/m)

Substitute the given values:

v = sqrt((2 * 6.21 x 10^-21 J) / (5.31 x 10^-26 kg))

Calculate:

v ≈ sqrt((2 * 6.21 x 10^-21) / (5.31 x 10^-26)) ≈ sqrt(2 * (6.21 / 5.31) * 10^5) ≈ sqrt(2 * 1.17 * 10^5) ≈ sqrt(2.34 * 10^5) ≈ 483.29 m/s

So, the oxygen molecule is moving at approximately 483.29 meters per second.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.