asked 43.6k views
5 votes
How many moles of AgNO3 are present in 1.50 L of a 0.050 M solution?

asked
User Christin
by
8.1k points

2 Answers

0 votes
c = n / V

n = c × V
n = 0,05 × 1,5 = 0,075[mol]

:-) ;-)
answered
User Darren Willows
by
7.2k points
1 vote
0.075 moles of AgNO3 is the answer ....
answered
User Brett Allred
by
8.4k points
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