The picture shows two containers filled with a gas, both initially at room temperature. A Heat
B Room Temperature
Which statement is correct?

The average kinetic energy of the gas particles is greater in container A because its particles move faster.

The average kinetic energy of the gas particles is greater in container B because it has a lower temperature.

The gas particles in both containers have the same average kinetic energy because they have the same volume.

The gas particles in both containers have the same average kinetic energy because they have equal number of particles​

Schrodinger and heisenberg developed an alternate theory about atomic nature that contradicted some of bohr's model of the atom. how do changes resulting from new technology and evidence affect the reputation of the atomic theory?

Part b: study of equilibrium on solubility: mg(oh)2(s) ⇌ mg2+(aq) + 2 oh–(aq) cloudy clear (pink) 7. a. b. 8. a. b. 9. 10. 11. 12. when adding concentrated hydrochloric acid, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 7a. you should indicate which ion was added to or removed from the equilibrium mixture. when adding edta, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 8a. you should indicate which ion was added to or removed from the equilibrium mixture. upon heating in which direction is the equilibrium shifting? upon cooling in which direction is the equilibrium shifting? is the forward reaction a. endothermic explain your answers to questions 9, 10, and 11. (l) left (r) right (l) left (r) right b. exothermic

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.