Carbon dioxide has a total of 16 valence electrons. 1. To determine the number of valence electrons of carbon dioxide (CO2), first determine the number of valence electrons of each of the elements in the molecule.
a. We have 1 carbon (C) molecule, and 2 oxygen (O) molecules.
b. The carbon molecule has 4 valence electrons and each oxygen molecule has 6 oxygen molecules.
2. Add up the valence electrons of each of the elements
4 + (2 x 6) = 16
(from C) (2 oxygen molecules, with 6 valence electrons each)
Thus, CO2 has a total of 16 valence electrons.
The number of valence electrons can be more clearly seen from the Lewis structure of the CO2 in the figure below (Source: http://chemistry.tutorvista.com/inorganic-chemistry/bonding-electrons.html). The the dots surrounding the letters represent the valence electrons.
The equilibrium constant is found by [product]/[reactant]
If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.
At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.
At equilibrium, for your reaction, it is predominantly reactants.
The best way to determine the number of atoms of arsenic in the sample will be to multiply 2.3 by Avagadro's number.
This is because Avagadro's number is the number of particles one mole of any substance has, and its value is 6.02 x 10²³
If the number of moles of a substance are known, then multiplying by Avagadro's number will give the number of particles. In this case, this is 1.38 x 10²⁴.
Answer: it would be cation, 2+
Explanation: electrons are negatively charged by 1. So if you get rid of 2 electrons it would be positive and cation is used to represent positive ions.