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Zepler [3.9K]
2 years ago
5

PLEASE HELP!

Chemistry
1 answer:
White raven [17]2 years ago
4 0

Benzene be B

Toluene be T

We have made then one lettered to make things easier

  • \sf P^0_B=95.1mmHg,P^0_T=28.4mmHg
  • \sf x_B=x_T=0.5

\\ \rm\Rrightarrow \dfrac{P^0_B-P_B}{P^0_B}=x_B

\\ \rm\Rrightarrow 95.1-P_B=95.(0.5)

\\ \rm\Rrightarrow 95.1-P_B=47.55

\\ \rm\Rrightarrow P_B=47.55mmHg

And

\\ \rm\Rrightarrow \dfrac{P^0_T-P_T}{P^0_T}=x_T

\\ \rm\Rrightarrow 28.4-P_T=28.4(0.5)

\\ \rm\Rrightarrow P_T=14.2mmHg

Total vapour pressure

  • 28.4+95.1
  • 123.6mmHg
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Explanation:

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15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno
timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

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Concentration of iodate ions:

[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M

Solubility product of barium iodate,K_{sp}=4.01\times 10^{-9}

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K_i=[Ba^{2+}][IO_2^{-}]^2

K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}

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As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

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