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denis23 [38]
3 years ago
11

A 51.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 208 mL. A 104 mL portion of that solution is diluted by a

dding 169 mL of water. What is the final concentration?Assume the volumes are additive.
Chemistry
1 answer:
S_A_V [24]3 years ago
6 0

Answer:

The final concentration is 0.226 M.

Explanation:

In this problem the dilution has occurred two times. So the symbol initial concentration, concentration after first dilution and the concentration after second dilution is given as C1, C2, and C3 respectively. Same can be done for the volume i.e, V1, V2, and V3 will be the initial volume, the volume taken after first dilution, and the volume after second dilution.

So, let's use the dilution formula for the first dilution

C_1 \times V_1 = C_2 \times V_2

C_2 = \frac{C_1\times V_1}{V_2}\\\\C_2 = \frac{1.50\times 51.0}{208}\\\\C_2 = 0.368 M

Now, for the second dilution

C_2 \times V_2 = C_3 \times V_3\\\\C_3 = \frac {C_2 \times V_2}{V_3}\\\\C_3 = \frac {0.368 \times 104}{169}\\\\C_3 = 0.226 M

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What is the maximum mass of b4c that can be formed from 2.00 moles of boron(iii) oxide?
anzhelika [568]

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<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.

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3 0
2 years ago
A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat
shutvik [7]
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150                0.0394
0.250                0.109
0.350                0.214
0.500                0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
3 0
3 years ago
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