To determine the element that has an electron configuration of <span>[Xe] 6s2 4f14 5d10 6p2, we must know the atomic number of Xe. From the periodic table, Xe has an atomic number of 54. We add the proceeding electrons, which is now a total of 82. The element with this atomic number is lead, Pb.</span>
We are asked to provide the net ionic equation for the reaction of HF (aq) and NaF (aq). HF is a weak acid and is in the following equilibrium:
HF (aq) ⇄ H⁺ (aq) + F⁻ (aq)
Meanwhile, NaF (aq) is an ionic compound that will dissociate completely in aqueous solutions:
NaF (aq) → Na⁺ (aq) + F⁻ (aq)
We can combine the ionic species with HF, as we are told to show F⁻ as a reactant:
HF (aq) + Na⁺ (aq) + F⁻ (aq) → HF (aq) + Na⁺ (aq) + F⁻ (aq)
We can eliminate the spectator ions, which in this case are Na⁺ ions, and this leaves us with the net ionic equation involving F⁻:
HF (aq) + F⁻ (aq) → HF (aq) + F⁻ (aq)
In this instance, the proton is just transferred between F⁻ ions and the end result is the formation of more HF, so there is no net reaction taking place.
Answer is: pH of methylamine
is 11.78.<span>
</span>
Chemical reaction: CH₃NH₂(aq)+
H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻<span>(aq).
Kb(CH</span>₃NH₂) =
4,4·10⁻⁴.<span>
c</span>₀(CH₃NH₂) = n(CH₃NH₂) ÷ V(CH₃NH₂).
c₀(CH₃NH₂) = 0.070 mol ÷ 0.8 L = 0.0875 M.
c(CH₃NH₃⁺) = c(OH⁻) = x.
c(NH₂OH) = 0.0875 M - x; equilibrium concentration of methylamine.
Kb = c(CH₃NH₃⁺) · c(OH⁻) / c(CH₃NH₂).
0.00044 = x² / (0.0875 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.00598 M.
pOH = -log(0.00598 M) = 2.22.
pH = 14 - 2.22 = 11.78.
Answer:
d i think sorry if im worng
Explanation: