Answer:
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
2. Moles of NH₃
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
(c) [NH₃]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
![K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bf%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BNi%28NH%24_%7B3%7D%24%29%24_%7B6%7D%5E%7B2%2B%7D%24%5D%7D%7D%7B%5Ctext%7B%5BNi%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BNH%24_%7B3%7D%24%5D%7D%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D)
Kf is large, so x ≪ 6.106×10⁻³. Then
![K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bf%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BNi%28NH%24_%7B3%7D%24%29%24_%7B6%7D%5E%7B2%2B%7D%24%5D%7D%7D%7B%5Ctext%7B%5BNi%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BNH%24_%7B3%7D%24%5D%7D%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5C%5C%5C%5C%5Cdfrac%7B6.106%20%5Ctimes%2010%5E%7B-3%7D%7D%7Bx%5Ctimes%200.1095%5E%7B6%7D%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5C%5C%5C%5C6.106%20%5Ctimes%2010%5E%7B-3%7D%20%3D%202.0%20%5Ctimes%2010%5E%7B8%7D%5Ctimes%200.1095%5E%7B6%7Dx%3D%20345.1x%5C%5Cx%3D%20%5Cdfrac%7B6.106%20%5Ctimes%2010%5E%7B-3%7D%7D%7B345.1%7D%20%3D%201.77%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Ni%24%5E%7B2%2B%7D%24%20at%20equilibrium%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.77%20%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
Atomic number Bi = 83
electron configuration Bi : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p3
p orbitals = 3p6, 4p6, 5p6 and 6p3
so, 6 + 6 + 6 + 3 = 21 electrons
1) The saturation point at 25°C is 35.7 g of NaCl / 100 g of water => 35.7 %
Under normal circumstances water will not accept more salt than that.
2) The solution with 1.55 mol of NaCl dissolved in 250 mL of water =>
molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
grams of NaCl = 1.55 mol * (molar mass of NaCl) = 1.55mol * 58.44 g/mol = 90.58 grams of NaCl
grams of water = 250 mL * 1 g/mL = 250 g/g.
Concentration of the solution: [90.58 g NaCl / 250 g H2O] * 100 = 36.23 %
3) Conclusion: the solution has more salt than the saturation value. This means that the solution is supersaturated.
Supersaturation is a special condition, which is unstable, but that is not part of the questions.
The answer is supersaturated,
Answer:
John Dalton
Explanation:
John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. This theory explains several concepts that are relevant in the observable world.
I feel like I know this but I just can't answer it, sorry
