Answer:
sulfur promotes oxide-reduction reactions.
Explanation:
In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.
When an electron occupies an orbital of an atom singly, rather than an electron pair such electron is said to be an unpaired electron. Each atomic orbital of an atom specified by quantum numbers can contain a maximum of two electrons having opposite spins, electron pair.
The numbers of unpaired electrons are determined by the electronic configuration of an atom. The electronic configuration of calcium, 
is:
or ![[Ar]4s^{2}](https://tex.z-dn.net/?f=%5BAr%5D4s%5E%7B2%7D)
Since, from the electronic configuration it is clear that the number of electrons in valence orbital that is 
is 2 that means the electrons are paired up in the valence orbital so, there are no unpaired electrons present in .
Hence, the number of unpaired electrons in is 0.
5.7% KCl is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g,
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius
i think it's A. cause CH is 1:1 and if you reduce C2H2, the ratio would also be 1:1
