Acid rain is caused by human activity. true or false?

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow.part ao2-(aq)+h2o(l)< => express your answer as part of a chemical equation. identify all of the phases in your answer.o2-(aq)+h2o(l) < => oh-(aq)+oh-(aq)part bpredict whether the equilibrium lies to the left or to the right of the equation in previous part.h2o is a stronger acid than oh–, so the equilibrium lies to the right.h2o is a weaker acid than oh–, so the equilibrium lies to the left.h2o is a stronger acid than oh–, so the equilibrium lies to the left.h2o is a weaker acid than oh–, so the equilibrium lies to the right.part cch3cooh(aq)+hs? (aq) < => express your answer as part of a chemical equation. identify all of the phases in your answer.ch3cooh(aq)+hs-(aq) < => h2s(aq)+c2h3o2-(aq)h2s(aq)+c2h3o2-(aq)part dpredict whether the equilibrium lies to the left or to the right of the equation in previous part.ch3cooh is a weaker acid than h2s, so the equilibrium lies to the right.ch3cooh is a weaker acid than h2s, so the equilibrium lies to the left.ch3cooh is a stronger acid than h2s, so the equilibrium lies to the right.ch3cooh is a stronger acid than h2s, so the equilibrium lies to the left.part eno2-(aq)+h2o(l) < => express your answer as part of a chemical equation. identify all of the phases in your answer.no2-(aq)+h2o(l) < => part fpredict whether the equilibrium lies to the left or to the right of the equation in previous part.hno2 is a stronger acid than h2o, so the equilibrium lies to the right.hno2 is a weaker acid than h2o, so the equilibrium lies to the left.hno2 is a stronger acid than h2o, so the equilibrium lies to the left.hno2 is a weaker acid than h2o, so the equilibrium lies to the right.

Experiment: effect of solution concentration on reaction rate you have learned that as the concentration of reactants increases, there will most likely be a greater number of collisions, and hence increase the rate of a reaction. in this experiment, you will see a demonstration of this, with a twist. there will be three reactions going on in this experiment. objectives determine how solution concentration can affect the rate of a reaction. the first reaction will be a reaction of the iodide ion (i-1) with hydrogen peroxide (h2o2) in an acidic solution. this reaction produces a slightly orange solution. in our experiment, we will add some orange food coloring to make this solution more orange. 2 h+ (aq) + 2 i- (aq) + h2o2 (aq) ⟶ i2 (aq) + 2 h2o (l) the next reaction will be between the iodine and starch i2 + starch ⟶ i2-starch complex (blue-black) so, when starch is added to the iodine solution made from the first reaction, the solution will turn black immediately, so it is difficult to find the rate of reaction. in order to be able to time this reaction, you will slow it down with another reaction. adding ascorbic acid will react with the iodine, reducing the concentration of the iodine available to react with the starch. c6h8o6 (aq) + i2 (aq) ⟶ 2i- (aq) + c6h6o6 (aq) + 2 h+ (aq) when the ascorbic acid is used up, the remaining iodine molecules can react with the starch and form the black color. the more ascorbic acid you add, the slower the reaction to form the iodine-starch complex will be. use your data and observations to complete the assignment. analysis and conclusions submit your data and the answers to these questions in the essay box below. what was your hypothesis? plot your data as drops of ascorbic acid vs. time. as the concentration of ascorbic acid was increased, did the rate of the formation of the iodine-starch complex increase or decrease? explain your answer in terms of the chemical reactions involved. was your hypothesis correct? make a general rule about the effects of concentration of reactants on reaction rates. for practice, the molecular formula for ascorbic acid is c6h8o6, and you used 6 g in this experiment, calculate the molarity of the ascorbic acid. now calculate the concentration in moles per drop (assume 1 ml = 20 drops).

Mercury turns to vapor at 629.88 k how much heat is lost 175 g of mercury vapor at 650 current condenses to a liquid at 297 ca mercury turns to weber at 629.88 kelvin how much he is lost 175 g of mercury vapor and 650 coming condensers to liquidate 297 kevin