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ANTONII [103]
3 years ago
14

What is the chemical formula of the ionic compound that forms from potassium and bromine?

Chemistry
1 answer:
vekshin13 years ago
6 0

solution:

This is an example of a binary salt. A binary salt is composed of one cation (+ charged ion) and one anion (- charged ion). Cations are usually elements of first group of the periodic table and anions are usually elements of 17th group of the periodic table. All the cations of the elements of the first group of the periodic table take the charge of 1+. Remember that a molecule has no charge, so if K(potassium) is the cation and has a charge of +1, the anion must have a charge of -1. Thus, the ionic compound formed when potassium and bromine combine is KBr.

KBr is an example of this. In order for K to lose all of its outer shell electrons, you must trace your finger from K, and go left. Nothing there! K, Potassium only has one electron in its outer shell to give away. After giving the electron away, K looks externally (in terms of electrons) like Argon (a noble gas).

Now, let's look at Br. We go right, because that is the closest way to get to the noble gases, and so have a complete outer shell with no extras (called valence electrons) to react with. Br needs exactly one electron.

Therefore, pairing K and Br results in a stable state where K donates and completely gives Br an electron, making K now positively charged (lost an electron) and Br negatively charged (gained one). This electron transfer forms an ionic bond because opposite charges attract.


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Answer:

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Explanation:

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Why is an atom regarded as electrically unstable​
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Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

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Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

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Plug in all the given values in the above equation -

E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V

So, E_{cell}=2.32V

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C = 12
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