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IRINA_888 [86]
3 years ago
8

n the laboratory, two forms of sodium phosphate will be available (the monobasic monohydrate NaH2PO4·H2O, F.W. = 137.99 g/mol, a

nd the dibasic Na2HPO4, F.W. = 141.96 g/mol). Explain the most efficient way of making this phosphate buffer by using only one of the compounds, and specify which compound you will use.
Chemistry
1 answer:
const2013 [10]3 years ago
6 0

Answer:

The compound you will use is the Dibasic phosphate

Explanation:

Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,

In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.

To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.

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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

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The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

A=\pi ^{1/3} (6V)^{2/3}

The area of the 10.0 cm³-sphere is:

A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}

The area of each 1.25 cm³-sphere is:

A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of  8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0
3 years ago
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