A gas in a sealed container has a pressure of 100 kPa at a temperature of 27 C. If the pressure in the container is increased to 20 kPa, what is the new temperature?

32.4 C

Explanation: Gay-Lussac's Law or Third Gas Law states that for a constant volume, the pressure is directly proportional to absolute temperature: P alpha T; also stated as P/T = K, where K is a constant, and similarly, P1/T1 = P2/T2 .

Here P1= 100 kPa, P2=120 kPa (If pressure increases by 20 kPA), T1=27 C [Given]

Hence, T2= (T1*P2)/P1 = (27*120)/100=32.4 C [C is degree in Celcius]

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

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d