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3 years ago

Can arsenic be broken down by a chemical change

1 answer:

7 0

Arsenic is a pure element. Elements (atoms) are the simplest building blocs excluding smaller elementary particles such as electrons, neutrons, quarks, protons etc.

This means that Arsenic can not be broken down any further by chemical change, only nuclear decay can cause the breakdown of large elements into smaller ones. Chemical change would be the combination of multiple elements into a compound or the decomposition of a molecule/compound into pure elements or smaller compounds.

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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:

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

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3 years ago

Plant cells and animal cells have many of the same organelles, but only plant cells have

Ierofanga [76]

Answer: chlorophyll

Explanation:

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3 years ago

How is a wound healing a chemical change? please explain

masya89 [10]

Answer:

its because fibrinogen is a chemical substance that builds up at the wound and gets hard by the action of air to prevent bleeding

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Judging by the electron domain geometry, which type of hybrid orbitals does this molecule have? s sp sp2 sp3

Aleksandr-060686 [28]

The molecule have TETRAHEDRAL HYBRID ORBITAL.
SP3 hybridization involves the mixing of one orbital of S sub level and three orbitals of P sub level of the valence shell. All the orbitals possess equivalent energies and shapes. The SP3 orbital has 25% S character and 75% P character. S and P refers to the s and p sub shells.<span />

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Name of these two Alkanes

fgiga [73]

3-ethyl-5,6-dimethyloctane

3-ethyl-2,3-dimethylpentane

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3 years ago