Temperature depends on the average kinetic energy of the particles in a sample. how does this with explaining why a plasma-based light does not have a high temperature?

no particles in the plasma have high kinetic energy.
the fastest particles have the greatest mass.
most particles in the plasma are heavy and slow.
the fastest particles transfer most of their energy through collisions.

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚c inside a freezer for a night.  when he took it out, its new volume was 322 milliliters, but its pressure was the same.  if the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer? the temperature of the freezer is  kelvins.

Actual ingredients of lab (the cookies i am actually making) 1/2 cup sugar 1/2 cup brown sugar 1 1/3 stick margarine 1 egg 1/2 tsp salt 1 tsp vanilla 1/2 tsp baking soda 1 1/2 cup flour 1 1/3 cup chocolate chip can you answer the questions below ? discussion 1. suppose you are given the following amounts of ingredients: 1 dozen eggs 24 tsp. of vanilla 1 lb. (82 tsp.) of salt 1 lb. (84 tsp.) of baking soda 3 cups of chocolate chips 5 lb. (11 cups) of sugar 2 lb. (4 cups) of brown sugar 1 lb. (4 sticks) of margarine a. for each ingredient, calculate how many cookies could be prepared if all of that ingredient were consumed. (for example, the recipe shows that using 1 egg- with the right amounts of the other ingredients- yields 24 cookies. how many cookies can you make if the recipe is increased proportionately for 12 eggs? ) b. to determine the limiting reactant for the new ingredients list, identify which ingredient will result in the fewest number of cookies. c. what is the maximum number of cookies that can be produced from the new amounts of ingredients?

Consider the reduction reactions and their equilibrium constants. cu+(aq)+e−↽−−⇀cu(s)pb2+(aq)+2e−↽−−⇀pb(s)fe3+(aq)+3e−↽−−⇀fe(=6.2×108=4.0×10−5=9.3×10−3 cu + ( aq ) + e − ↽ − − ⇀ cu ( s ) k =6.2× 10 8 pb 2 + ( aq ) +2 e − ↽ − − ⇀ pb ( s ) k =4.0× 10 − 5 fe 3 + ( aq ) +3 e − ↽ − − ⇀ fe ( s ) k =9.3× 10 − 3 arrange these ions from strongest to weakest oxidizing agent.