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3 years ago

13. Explain two of the methods of separating mixtures that are used to make wheat flour. Be specific

Chemistry

1 answer:

kirill [66]3 years ago

8 0

I need this answer too, can someone please help us , or did you ever find the right answer?!

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A solution of 0.207 M aspartic acid, the charge neutral form of the amino acid, is titrated with 0.0690 M NaOH . The pKa values

Softa [21]

Answer:

pH = 6,951

Explanation:

The neutral form of aspartic acid is in its isolectric point. For aspartic acid the isoelectric point is the average of pka1 and pka2, thus:

$pH =\frac{1}{2} (3,900+1,990)$ = 2,945

The addition of stoichiometric amounts of NaOH at the first equivalence point will increase the pH at average of pka2 and pka3 thus:

$pH = \frac{1}{2} (3,900+10,002)$ = 6,951

Thus, <em>pH at the first equivalence point is 6,951</em>

I hope it helps!

5 0

3 years ago

Which group on the periodic table contains only metals?

Snezhnost [94]

Answer: GROUP 2 is the only group to include only metals. Group one includes the most metalic metals but hydrogen is usually put into group one and obviously hydrogen is not a metal. Group 2 includes Br, Mg, Ca, Sr, Ba,Ra.

Explanation:

5 0

3 years ago

How does a rock from the mountain become a pebble on the shore

Semmy [17]

From erosion i think good luck :D

7 0

3 years ago

Read 2 more answers

65. Write an equation for the combustion of octane.

siniylev [52]

Answer:

2C8H18+25O2→16CO2+18H2O.

4 0

2 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago