CHEM 1127Q Study Guide - Midterm Guide: Chemical Formula, Molar Mass

• School: UConn
• Department: Chemistry
• Course: CHEM 1127Q
• Professor: Cady

Document Summary

0. 1253 moles/l x 0. 265l = 0. 0332 moles fecl3. 0. 1253 moles/l fecl3 x v = 1. 000 moles fecl3. V = 7. 981 l: so to get 1. 000 moles fecl3 you need, the stock solution of 0. 1253 moles/l, add in 7. 981 l. What we need (ca2+) does not match what we have (ca3(po4)2) --> make them the same. 1. 023 moles/l ca3(po4)2 x 3ca2+/1 ca3(po4)2 = 3. 069 moles/l ca2: amount of calcium in the calcium phosphate. 3. 069 m ca2+ x v = 0. 001 moles ca2+ V = 0. 0003 l ---> 3 x 10-4 l. You have a 0. 13 m sample of cl-. How much of this stock solution do you need to make 2. 5l of a. Concentration problem: m1 v1 = m2 v2 (0. 13m cl-) (v1) = (0. 05m cl-) (2. 5l) (0. 13m)(v1) = 0. 125 moles cl- You have a 0. 13 m sample of mgcl2. M1 v1 = m2 v2 (0. 13m mgcl2) (v1) = (0. 05m cl-) (2. 5l)