...a metal atom will *lose* electrons to form a *positive* cation and a nonmetal atom will *accept* electrons to form an *negative* anion.
Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
