G.15 part (b)

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Sophia B 3G
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Joined: Fri Sep 24, 2021 6:13 am

G.15 part (b)

Postby Sophia B 3G » Wed Oct 20, 2021 4:54 pm

Can someone explain G.15 part (b)? I got part (a) right but I did not get the right answer to part (b).
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KPINTO 1B
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Re: G.15 part (b)

Postby KPINTO 1B » Wed Oct 20, 2021 6:16 pm

So, this is a M1 x L1 =M2 x L2 problem. I would multiply the desired molarity and liters together to get the mols. Then divide that mols value by the 2.5M of the reagent bottle, and you'll get .012 L or 12 mL needed of that reagent bottle added to 48mL of water.
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Sophia B 3G
Posts: 103
Joined: Fri Sep 24, 2021 6:13 am

Re: G.15 part (b)

Postby Sophia B 3G » Sat Oct 23, 2021 5:27 pm

805781531 wrote:So, this is a M1 x L1 =M2 x L2 problem. I would multiply the desired molarity and liters together to get the mols. Then divide that mols value by the 2.5M of the reagent bottle, and you'll get .012 L or 12 mL needed of that reagent bottle added to 48mL of water.

This really helped! Thank you!
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