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Sagot :
Answer:
1) 5.0 mL, 4.0 mL, 3.0 mL and 2.0 mL respectively.
2) 3.0 mL
Explanation:
Hello there!
1) In this case, by considering that the equation we use for dilutions contain the initial and final concentrations and volumes:
[tex]C_1V_1=C_2V_2[/tex]
For the first four solutions, we compute the volume of the stock one (V1) as shown below:
[tex]V_1=\frac{C_2V_2}{C_1}[/tex]
Thus, we obtain:
[tex]V_1^1=\frac{3.5x10^{-5}M*10.00mL}{7.0x10^{-5}M} =5.0mL\\\\V_1^2=\frac{2.8x10^{-5}M*10.00mL}{7.0x10^{-5}M} =4.0mL\\\\V_1^3=\frac{2.1x10^{-5}M*10.00mL}{7.0x10^{-5}M} =3.0mL\\\\V_1^4=\frac{1.4x10^{-5}M*10.00mL}{7.0x10^{-5}M} =2.0mL[/tex]
2) In this case, for a final concentration of 2.1x10-5 M and a volume of 10.00 mL, the volume of the stock solution would be:
[tex]V_1=\frac{2.1x10^{-5}M*10.00mL}{7.0x10^{-5}M} =3.0mL[/tex]
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