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blondinia [14]
3 years ago
8

Im hvaing a hard time getting the right answer

Chemistry
1 answer:
Tanya [424]3 years ago
6 0

Answer:

V=23.9mL

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

n=2.49g*\frac{1mol}{125.55 g}=0.0198mol

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

M=\frac{n}{V}\\\\V=\frac{n}{M}

By plugging in the moles and molarity, we obtain:

V=\frac{0.0198mol}{0.830mol/L}=0.0239L

Which in mL is:

V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL

Best regards!

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A. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution<br> IMO)
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Volume of solution  = 1L

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    number of moles of HgCl₂  = molarity of solution x volume

                                                   =   1 x 1

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