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What volume in ML of 12.0M HCL is needed to contain 3.00 moles of HCL?
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2 votes
What volume in ML of 12.0M HCL is needed to contain 3.00 moles of HCL?

User Tobias Otto
by
4.7k points

1 Answer

6 votes

Answer:

V= 250ml

Step-by-step explanation:

From n= CV

3= 12×V

V= 0.25L= 250ml

User Eric Auld
by
4.9k points
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