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saw5 [17]
3 years ago
5

why is H2SO4 considered as strong acid? What colour does it give with phenolphthalein and methyl orange?​

Chemistry
2 answers:
Oksanka [162]3 years ago
4 0
He said that i was like crying and he
natulia [17]3 years ago
4 0
This requires a bit of background knowledge of indicators and how they work. An indicator like methyl orange is just a weak acid that forms an equilibrium (and doesn't fully dissociate). The acid and the dissociated negative ion differ in color, and so when you add an acid or base (dissociated H+ or OH- ions), the equilibrium shifts and you get the color change.

Now, for titration, we use the point the color changes to find our equivalence point. The point in methyl orange that changes color is when there are equal amounts of the non-dissociated acid and the negative ion. If you are familiar with the Henderson-Hasselbach equations, this happens when the pH of the solution is equal to the pKa of the solution, which for methyl orange is 3.7. The human eye notices the color change in the range 3.1–4.4.

Now, for a titration, we need to see when the equivalence point has been reached. This means the indicator should change color at the equivalence point. For a strong acid strong base reaction, the pH at equivalence is 7, which is outside of the pH range of methyl orange, which means when equivalence point has been reached, methyl orange won't change color, and we won't know. That's why we can use methyl orange as the indicator for the titration.

I assumed some previous knowledge of acids and bases, and equilibria. If there's something you aren't sure about, feel free to comment and I'll try my best to explain it more thoroughly.

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At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare
OleMash [197]

Answer:

P_{CCl_4}=52.43torr

Explanation:

Hello there!

In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576

Therefore, the partial pressure of chloroform turns out to be:

P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr

Regards!

3 0
3 years ago
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liubo4ka [24]

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a is the anwser

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4 0
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Harman [31]

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