The molar mass for PCL3 is 137.33 g/mol
Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
Answer: 207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu
2.855 + 49.639 + 45.741 + 108.979 = <em><u>207.217amu</u></em>
Explanation:
You would have started with 5 grams of reactants