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Phantasy [73]
3 years ago
11

What volume of a 2.75 m solution of naoh is required to make 500.0 ml of a 1.27 m solution of naoh?

Chemistry
2 answers:
Alex_Xolod [135]3 years ago
3 0

The volume of 2.75 M solution of NaOH required to make 500.0 ml of a 1.27M solution of NaOH is 230.9 Ml


calculation


by use of M1V1=M2V2 volume formula where


M1=2.75 m

V1=?

M2= 500 Ml

V2= 1.27 M


make V1 the subject of the formula V1= M2V2/M1


V1 is therefore= 500Ml x 1.27 M/ 2.75 M = 230.9 Ml

Hitman42 [59]3 years ago
3 0

Answer:

V_1=230.9mL

Explanation:

Hello,

This is about a dilution exercise which could be mathematically solved by taking into account that during the dilution process the moles are conserved:

n_1=n_2

Now, as the moles could be expressed in term of molarity an volume one defines:

M_1V_1=M_2V_2

Then, we're asked to report V_1 as it is the volume we need to take from the inital 2.75M solution of NaOH, thus:

V_1=\frac{M_2V_2}{M_1} =\frac{1.27M*500.0mL}{2.75M}\\V_1=230.9mL

Best regards.

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Adopting the number of avogrado 6.02 * 10²³ / mol 
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2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)
defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

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Lead solution = 0.000013 M

Volume = 100 L

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Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

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Molarity = \frac{number of moles}{volume}

Solve for number of moles

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Substitution

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                        x = (0.0013 x 207.2) / 1

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