13500 mL convert to milliliters of mercury

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is 16.84 nm. What is the total wi

nadezda [96]

Answer:

u can just multiply 1

Explanation:

krkrfmfmmfdkd

3 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

Lists how many valence electrons are shared, gained, or lost by each atom of the 6 compounds?

Blababa [14]

<span>The number of electrons in an atom's outermost valence shell governs its bonding behavior.

In N</span>₂, three electrons are being shared by each nitrogen atom, making a total of 6 shared electrons.

In CCl₄, 4 electrons are being shared by each carbon atom and 1 electron is being shared by each chlorine atom

In SiO₂, 4 electrons are being shared by each silicon atom and 2 electrons are being shared by each oxygen atom.

In AlCl₃, 3 electrons are being shared by each aluminum atom and 1 electron is being shared by each Cl atom

In CaCl₂, 2 electrons are lost by the calcium atom and 1 electron is gained by each chlorine atom

In LiBr, 1 electron is lost by the lithium atom and 1 electron is gained by the bromine atom

6 0

3 years ago

romeo is upset that he'll no longer be able yo see juliet, but this living creature will be able to see her, be near her, even f

Lyrx [107]

Answer:

beautiful and beautiful home on

8 0

3 years ago

Read 2 more answers

Which of the following metals is most reactive?

g100num [7]

The answer is, D. Calcium