Description
1,,seclect conductor or nonconductor
melted aluminum iodide
solid sodium chloride
oxygen dissolved in pure water
pure liquid water
2,,How many mL of a 8.00 M solution of nitric acid are required to prepare 350 mL of a 0.150 M solution?
Explanation & Answer
1. melted aluminum iodide .............conductor
solid sodium chloride.......................nonconductor
oxygen dissolved in pure water.......very very weak conductor (nonconductor)
pure liquid water...............................very very weak conductor (nonconductor)
2. use dilution eqn (c1)(v1) = (c2)(v2)
solving for v2...
v2 = (c1)(v1)/(c2)
v2 = (0.150 M)(350 mL) / (8.00 M)
= 6.6 mL (with sigfigs in mind)
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