Question

Determine the mean root square speed of CO2 molecules that have the average kinetic energy of 4.21 x 10^-21 in m/s

Answer 1

@ParthKohli can you help?

Answer 2

Oh, never mind.

Answer 3

For one CO2 molecule, the kinetic energy is \(\frac{1}{2} \times m \times v^2.\)

Do you know the mass of a CO2 molecule?

Answer 4

yes 44.01

Answer 5

Just take that as 44.

Answer 6

okay

Answer 7

But you need to convert it to kilograms. Can you do that?

Answer 8

.044?

Answer 9

OK, nope...

First of all, do you know how you would convert from atomic mass units to grams?

Answer 10

no

Answer 11

will it be 44 times avagadros number?

Answer 12

to change it into kg

Answer 13

Alright, look:

44 g = Avogadro's No. * 44 u
So 44 u = (44/Avogadro's No.) grams.

Answer 14

I got 7.3

Answer 15

* 10^(-25).

Answer 16

well 7.306 x 10^-23

Answer 17

(44/6*10^23*10^3) kg
= 44/(6*10^26) kg
= 7.3 * 10^(-26) kg

I've been making so many errors. Ugh. Anyway...

Answer 18

\[v^2 = \frac{\rm KE}{\frac{1}{2}m}\]

Answer 19

so that answer is the m in the equation?

Answer 20

No, it is \(v\)

Answer 21

okay

Answer 22

v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 57600 m^2/s^2

v = 240 m/s

is that correct?

Answer 23

v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 57600 m^2/s^2

v = 240 m/s

is that correct?

Answer 24

v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 57600 m^2/s^2

v = 240 m/s

is that correct?

Answer 25

v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 57600 m^2/s^2

v = 240 m/s

is that correct?