Ethanoic acid ionizes in aqueous solutions to form two ions which are
and 
<h3>Ionization of ethanoic acid</h3>
Ethanoic acid goes by the chemical formula
.
In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single 
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Answer:
6
Explanation:
This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.
<span>In order to covert a unit, you must know certain number of conversions. In here, the conversion is in unit of length. One meter is equal to 1000 meter. So if it is in cubic form, then the answer of one meter cube is also equal to 1000 cube. Then,
32 m</span>³ (1000 mm/1m)³
<span>or
</span>32 m³ (1000³ mm³/1 m³)
= 3.2 x 10¹⁰ mm³
Answer:
see explanation below
Explanation:
The question is incomplete. The missing parts are a) determine the electrophylic site. b) determine the nucleophylic site.
In order to do this, we need to write the reaction and do the mechanism. The nucleophylic site will be the site where the nucleophyle attacks to form the product. In this case the site is the carbon next to the bromine. In this place the Oxigen which is the nucleophyle goes. The electrophyle is the site where one atom substract to complete it's charges. In this case, the electrophyle is usually the hydrogen, so the site will be next to the oxygen after the nucleophyle attack.
You can see it better in the attached picture.
{ 12 mol * 27 g / mol * 0.902 j/gdeg C* ( 658 -72 ) deg C + 12 * 27 * 3.95 }
quantity of energy required to heat 12 mol of aluminium for 72 deg C to its melting point 658 C
so energy applied is 1451 057 . 33 J