1) The U.S. Bureau of Alcohol, Tobacco and Firearms has ruled that nitrocellulose containing more tha 12.6% nitrogen is classified as a high explosive. Calculate the mass percent of nitrogen in tri- nitrocellulose, made up of the monomer, C6H7O₂(NO3)3, and of di-nitrocellulose, C6H8O3(NO3)2. Percent nitrogen in C6H₂O₂(NO3)3 Percent nitrogen in C6H8O3(NO3)2 2) Complete and balance the combustion reaction, To save a copy of this document, click Duplicate. C6H10O5 + 3) Complete and balance the combustion reaction, below, for the tri-nitrocellulose monomer. _C6H7O₂(NO3)3 + CO₂ + N₂ + CO₂ 4) Tri-nitrocellulose can burn explosively (hence the ATF ruling). Why? + H₂O H₂O

Lighters typically contain butane (C4H10), which has a molar mass of 58.12 g/mol. Consider that there may have been a mixture of hydrocarbons present in the lighter. If the experimentally obtained molar mass of this mixture was 56.1 g/mol, what other hydrocarbons may have been present? Choices include: CH4 (methane), C2H6 (ethane), C3H8 (propane), C5H12 (pentane), C6H14 (hexane), C7H16 (heptane), or C8H18 (octane). Explain.

2. If 2.500 g of Na2CO3 is used in the reaction, calculate the theoretical yield of NaCl.

This is part of a continuing problem with the previous carbohydrate sample. Use the previous data in this part. 1) A 12.915 g carbohydrate sample composed of CHO was analyzed by combustion analysis from which was obtained 18.942 g CO2 and 7,74 g H2O. The molar mass of the sample was found to be 455.32 g/mole by independent analysis. How many repeat units are in this molecule? Input the numeric value here. Hint: This is a whole number value. Hint: The repeat units are determined by the ratio of the empirical formula molar mass to the molar mass of the sample; the empirical formula is determined by the ratio of moles of the compound's elements. 2) What is the molecular formula? Include this for your notes for personal practice. Table molar mass (g/mol): C 12.00, O 16.00, H 1.008.