Answer:3.5 moles * 6.02 X 10^23 particles/mole * 2 H atoms/particle of H2 = 42.14 X 10^^23 atoms
Explanation:
Answer:
lanthanum
Explanation:
lanthanum is an inner transition metal
Water is polar in that the hydrogen atoms bear a partial positive charge, whereas the oxygen atom bears a partial negative charge, due to the electronegativity of oxygen. ... Water disrupts ionic bonds and hydrogen bonds.
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change 
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:
Thus, we obtain a quadratic equation whose solution is:
Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
<em>Q = 1.08x10⁻¹⁰</em>
As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate
