CHEM 115- Midterm Exam Guide - Comprehensive Notes for the exam ( 12 pages long!)
Document Summary
Mass n2= 500. kg = 5. 00 x 105 g. Mol n2= 5. 00 x 105 g / 28 g/mol = 1. 785 x 104 mol. Mass h2= 1. 00. kg= 1. 00 x 105 g. Mol h2= 1. 00 x 105 g / 2. 016 g/mol = 4. 960 x 104 mol. 1. 785 x 104 mol n2 would require 5. 355 x 104 mol h2 so, Sometimes you"ll need to find the excess in mass of other reagent, in this case n2. Remember % yield= actual yield x 100 theoretical yield. Slide 39 is a practice question, the answer will be posted. M= amount of solute (in mols) volume of solute. Slide 43 is a practice question, the answer will be posted. Remember the number of moles doesn"t change unless more solvent is added to solution. Slide 45 is a practice question, the answer will be posted. 150l pb(no3)2 x . 175 mol pb(no3)2 x 2mol kcl x 1 l kcl = . 350l kcl solution.
