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Chemistry 18 Online
OpenStudy (anonymous):

Please help!! How many moles of NaCl are required to prepare 2.90 L of 1.8 M NaCl? 0.62 mol NaCl 1.6 mol NaCl 4.7 mol NaCl 5.2 mol NaCl

OpenStudy (anonymous):

Umm, I'm not too sure, but I'm thinking it's either A. 0.62 or C. 4.7... probs ask @aaronq, @.Sam. or @abb0t to check it though :)

sam (.sam.):

Use \[n=[M]V\] Where n = number of moles [M] = concentration V = Volume

OpenStudy (anonymous):

lol so i guess i was totally off :P @.Sam. it would be D. 5.2 then huh? :)

sam (.sam.):

yep

OpenStudy (anonymous):

haha thanks for teaching me as well!! haha :P

OpenStudy (anonymous):

Thanks so much!

OpenStudy (anonymous):

np:) @.Sam. is awesome haha :P i totally got the formula wrong at first lol... my bad! ;/ hehe i guess we just need to memorize that formula like @.Sam. does :P

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