There are 26 electrons in the nickel ion in the compound NiS.
<h3>How many electrons does the nickel ion contain in the compound NiS?</h3>
The number of electrons in a neutral Nickel atom is obtained from the atomic number of Nickel.
The atomic number of Nickel is 28, therefore, the neutral nickel atom has 28 electrons.
In the compound NiS, nickel atom loses two electrons to to form the nickel ion.
Number of electrons left = 28 - 2 = 26
Therefore, there are 26 electrons in the nickel ion in the compound NiS.
In conclusion, nickel atom loses two electrons to form nickel ion in NiS.
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Answer:
Explanation:
When two non-miscible liquids are put together, the one with the higher density will be on the bottom, while the one with the lower density will be on top.
Meaning that in this problem's case toluene would be on the top layer and water in the bottom layer.
Answer:Chewing in the mouth breaks food into smaller pieces
Explanation: all you are doing is breaking down the food in your mouth
It would be “To obey the law of conservation of mass”
Answer : The concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is 
and 
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of 
= Moles of = 
Now we have to calculate the concentration of
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of is,
