What would happen to the rate of a reaction with rate law rate = k no2 h2 if the concentration of h2 were halved?

The question is not formatted correctly. The proper question should be

What would happen to the rate of a reaction with rate law

rate=k [NO]2[H2]

if the concentration of H2 were halved?

Rate would be halved.

Explanation:-

Let originally [NO]= 1 and [H2]=1.

Then original rate= k [1]2 [1] = k.

Now if [H2] is halved then new [H2] = 1/2

While [NO] remains same.

Then

new rate = k [1]2 [1/2] = k/2

Thus the rate is halved.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

i'm pretty sure it is letter c.) car