Cu(s)+4HNO3(aq)Cu(NO3)2(aq)+2H2O(+2NO2(g) Let's say we started with 0.0015molCu(s). In the reaction equation, note that 4HNO3 react with 1Cu. This means you can use the relation as a conversion factor: 1Cu4HNO3 or 4HNO31Cu Start with the 0.0015molCu(s), use the conversion factor above to calculate the moles HNO3. How many moles of HNO3 would react with your moles of Cu(s) ? Use dimensional analysis to solve. mole HNO3 would be needed to react with 0.0015molCu(s). Your laboratory instructor added 1.0mL of 16MHNO3(aq) in this reaction. Calculate the moles of HNO3 present. (No clue where to start with this one? See the video on the Sway website https://sway.office.com/OyhCyMfBnG5azH86?ref=Link for help in the Post Lab Questions part at the end of the Sway). moles HNO3 were added to the beaker. Part 3 (1.3 points) Compare your answers for parts 1 and 2 above. Which is greater, the amount of nitric needed to react (part 1) or the amount added (part 2)? Choose one: Part 1 (the amount of nitric needed to react) was larger. Part 2 (the amount of nitric acid actually added) was larger