At a temperature of 408K, which gad will have the highest velocity?

1 - NO2 at 339 K

2 - Ne at 371 K

3 - H2 at 371 K

4 - H2 at 425 K

Explanation:

Kinetic Energy is directly related to temperature; the higher the temperature the higher the kinetic energy. Kinetic energy is also equal to 12m⋅v2, so if we want a high velocity we want high temp and low mass. So let's list out approximate masses:

m(H2)≈2

m(NO2)≈46

m(Ne)≈20

So we have NO2 at 339 K, the lowest temperature out of the mix, and the highest mass out of the mix, so this is moving the slowest.

In contrast, we have H2 at 425 K, the highest temperature out of the mix, and the lowest mass out of the mix, so this is moving the fastest.

Now we have Ne and H2 at 371 K, since they are at the same temperature they have the same kinetic energy. But H2 is lighter than Ne so it must be faster. To quantify this mathematically, let's assume (this is wrong but just as an assumption for an example) KE at 371 K is 100:

100=12⋅m⋅v2

200=m⋅v2

√200m=v

So H2 is about v=10 and Ne is about v=√10≈3

So the order to recap is:

1 - NO2 at 339 K

2 - Ne at 371 K

3 - H2 at 371 K

4 - H2 at 425 K

Hope that makes it clearer!

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g