what is the molecular formula of something with an empirical formula CO that has a molar mass of 56g/mol ?

150.0 grams of asf3 were reacted with 180.0 g of ccl4 to produce ascl3 and ccl2f2. if the actual yield of ccl2f2 was 127 g, what is the percent yield?

Astudent made a graph plotting the progress of a reaction over time. the student forgot to label the y-axis of the graph. a graph is shown with two graph lines. one graph line starts at a higher position on the y axis and slopes downwards towards the right. the other graph line starts at a lower position on the y axis and slopes upwards towards the right. the two graph lines stop short of intersecting each other and continue as separate lines which gradually become straight and parallel to the x axis. a vertical line is shown at a point where the two graph lines finally became parallel to the x axis. this vertical line is labeled equilibrium. the title on the x axis is time and an arrow pointing towards the right is shown above time. the title on the y axis is left blank. what best explains the label that the student should use on the y-axis? amount, because as the amount of product decreases, the amount of reactant increases over time. reaction rate, because forward and backward reaction become equal at equilibrium. amount, because the amounts of reactants and products become constant after equilibrium is reached. reaction rate, as the rate of forward reaction increases and rate of backward reaction decreases over time.

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