Which combination below will be a buffer solution? O HNO3 and NaNO3
O NaBr and NaOH
O HC2H2O2 and NaC2H302
O HCl and CH-
O All of the above are true.

Which of the following observations indicates that there is a small, dense, positively charged part in the center of an atom? some uncharged particles are scattered by a gold foil. all uncharged particles are attracted towards a gold foil. all positively charged particles pass straight through a gold foil. some positively charged particles bounce back from a gold foil.

These questions are based on the attached photo. the experiment is about burning magnesium metal with oxygen. 1. write the balanced chemical equation for the reaction you are performing. 2. calculate the mass of magnesium metal used in each trial. o trial 1: o trial 2: 3. calculate the actual yield of magnesium oxide for each trial. o trial 1: o trial 2: 4. magnesium is the limiting reactant in this experiment. calculate the theoretical yield of mgo for each trial. o trial 1: o trial 2: 5. determine the percent yield of mgo for your experiment for each trial. o trial 1: o trial 2: 6. determine the average percent yield of mgo for the two trials. your company currently uses a process with a similar cost of materials that has an average percent yield of 91 percent. if the average percent yield of this process is higher than that, this could save the company money. what is your recommendation to the company? support your recommendation using your data, calculations, and understanding of stoichiometry gathered from this lab.

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes