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Inessa05 [86]
3 years ago
7

When titrating a strong monoprotic acid and koh at 25°c, the ph will be less than 7 at the equivalence point. ph will be greater

than 7 at the equivalence point. titration will require more moles of base than acid to reach the equivalence point. ph will be equal to 7 at the equivalence point. titration will require more moles of acid than base to reach the equivalence point?
Chemistry
1 answer:
iVinArrow [24]3 years ago
8 0
By definition titraion of a monoprotic acid with means that the equivalence point implies netrality of the solution, which is pH = 7.

So, the answer is that pH will be equal to 7 at the equivalence.

Given that the acid is monoprotic and KOH has one OH- radical per molecule of KOH, the titration will require the same number of moles of acid than base to reach the equivalence point, as you can see in this equation, representing the monoprotic acid as HA:

 HA + KOH = K(+) + A(-) + H2O => 1 mol HA per 1 mol KOH.
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For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for
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Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
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3 0
3 years ago
What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus
Nastasia [14]

Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

Explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

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Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

Best regards!

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LenaWriter [7]

Answer:

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Explanation:

5 0
2 years ago
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PV=nRT

P=nRT/V

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6 0
3 years ago
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