Answer:
1.8 × 10² s
Explanation:
Let's consider the reduction that occurs upon the electroplating of copper.
Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)
We will establish the following relationships:
- 1 g = 1,000 mg
- The molar mass of Cu is 63.55 g/mol
- When 1 mole of Cu is deposited, 2 moles of electrons circulate.
- The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
- 1 A = 1 C/s
The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:
Answer: 4 N-2 N=2 Nto the left. Change in motion: The box will move to the left. 7.)
Explanation: Please mark branliest.
Answer:
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not.
A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons.
Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it.
So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’.
You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not.
Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration
Co3+:1s22s22p63s23p63d6
For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split.
More specifically, you can say that
a strong field ligand will produce a more significant splitting energy, Δ a weak field ligand will produce a less significant splitting energy, Δ
Now, the spectrochemical series looks like this
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Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands.
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In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed.
This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic.
On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed.
Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.
Answer:
Explanation:
Step 1. Identify the Group that contains X
We look at the consecutive ionization energies and hunt for a big jump between them
We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.
We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.
So, X is in Group 2 and P is in Group 15.
Step 2. Identify the Compound
X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.
We must have 3 X atoms for every 2 P atoms.
The formula of the compound is 
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Cells are composed of water, inorganic ions<span>, and carbon-containing (organic) molecules. Water is the most abundant molecule in cells, accounting for 70% or more of total cell mass.</span>
