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OpenStudy (anonymous):
How many potassium hydroxide (KOH) formula units are present in 6.89 mol of KOH?:A. 4.1498 × 1024 B. 4.1498 × 1023C. 8.299 × 1024 D. 8.299 × 1023
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OpenStudy (anonymous):
PLEASE SOMEONE>>>HELP!!!!!...How many potassium hydroxide (KOH) formula units are present in 6.89 mol of KOH?:A. 4.1498 × 1024 B. 4.1498 × 1023C. 8.299 × 1024 D. 8.299 × 1023
OpenStudy (anonymous):
n=6,89 mol n=N/L --> N= n*L (L is Avogadro's constant) and answer is A
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