A balloon occupies a volume of 44 L at 200K, what will the temperature of the balloon become if it is changed to 500K?

110L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 44L

Initial temperature (T1) = 200K

Final temperature (T2) = 500K

Final volume (V2) =..?

The new volume of the balloon can be obtained by using Charles' law equation as shown below:

V1/T1 = V2/T2

44/200 = V2/500

Cross multiply

200 x V2 = 44 x 500

Divide both side by 200

V2 = 44 x 500/200

V2 = 110L

Therefore, the new volume of the balloon is 110L

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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons