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Words: 1458
Pages: 5
(approximately 235 words/page)
Pages: 5
(approximately 235 words/page)
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LECTURE 6 NOTES, CHM 101, SEC. 01
SOLUTION CONCENTRATIONS
THE ACTUAL WEIGHING OF REACTANTS OFTEN PROVES IMPRACTICAL OR INCONVENIENT. THIS IS ESPECIALLY TRUE IF REACTANTS ARE GASES, LIQUIDS, OR VERY REACTIVE.
HOW CAN WE DELIVER KNOWN WEIGHTS OF REACTANTS WITHOUT WEIGHING THEM FIRST?
CONSIDER: HNO3 + NaOH = NaNO3 + H2O
NEITHER NITRIC ACID, NOR SODIUM HYDROXIDE CAN BE WEIGHED EASILY. WHY?
HOWEVER, SOLUTIONS OF KNOWN AMOUNT OF SOLUTE PER UNIT VOLUME CAN BE DELIVERED IN KNOWN VOLUME TO GIVE
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showed first 75 words of 1458 total
showed last 75 words of 1458 total
reagent and all the carbon dioxide is not used up. The number of moles of the carbonate will be the same as the number of moles carbon dioxide used, 0.075. 83.a Calculate the molarity of a solution prepared from 6.00 mol HCl in 2.50 L solution. M = moles solute/liters solution = 6.00 moles/2.50 L = 2.4 moles/liter 87. How many mL of a 0.215 M solution are required to contain 0.0867 mol NaBr? Note M x VL = moles = 0.0867 = 0.215 x VL VL = 0.0867/.215 = 0.403 liters = 403 mL
reagent and all the carbon dioxide is not used up. The number of moles of the carbonate will be the same as the number of moles carbon dioxide used, 0.075. 83.a Calculate the molarity of a solution prepared from 6.00 mol HCl in 2.50 L solution. M = moles solute/liters solution = 6.00 moles/2.50 L = 2.4 moles/liter 87. How many mL of a 0.215 M solution are required to contain 0.0867 mol NaBr? Note M x VL = moles = 0.0867 = 0.215 x VL VL = 0.0867/.215 = 0.403 liters = 403 mL