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OpenStudy (anonymous):
How many moles of KCl are required to prepare 1.50 L of 5.4 M KCl? Can someone explain the steps for this problem?
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OpenStudy (anonymous):
The answer choices are : 0.28 mol KCl 3.6 mol KCl 6.9 mol KCl 8.1 mol KC
OpenStudy (anonymous):
I'm just not sure how to find the answer. Can someone walk me through it please?
OpenStudy (anonymous):
Use the formula mol=Concentration*volume/1000, in which the volume is in cm^3, the concentration is M or mol/dm^3. 1L = 1dm3. Just make the conversion to cm^3 and get the answer.
OpenStudy (danjs):
|dw:1419297938614:dw|
OpenStudy (danjs):
Mol KCL
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