Answer:
a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹
c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹
Explanation:
a. Silver iodate
Let s = the molar solubility.
AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸
E/mol·L⁻¹: s s
![K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%5BIO%24_%7B3%7D%24%24%5E%7B-%7D%24%5D%7D%20%3D%20s%5Ctimes%20s%20%3D%20%20s%5E%7B2%7D%20%3D%203.0%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%20%3D%20%5Csqrt%7B3.0%5Ctimes%2010%5E%7B-8%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%201.7%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctext%7B%20mol%2FL%7D)
b. Barium sulfate
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰
I/mol·L⁻¹: 0.02 0
C/mol·L⁻¹: +s +s
E/mol·L⁻¹: 0.02 + s s
![K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BBa%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%24%24%5E%7B2-%7D%24%5D%7D%20%3D%20%280.02%20%2B%20s%29%20%5Ctimes%20s%20%5Capprox%20%200.02s%20%3D%201.1%5Ctimes%2010%5E%7B-10%7D%5C%5Cs%20%3D%20%5Cdfrac%7B1.1%5Ctimes%2010%5E%7B-10%7D%7D%7B0.02%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%205.5%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctext%7B%20mol%2FL%7D)
c. Using ionic strength and activities
(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂
The formula for ionic strength is
![\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Csum_%7Bi%7D%20%7Bc_%7Bi%7Dz_%7Bi%7D%5E%7B2%7D%7D%5C%5C%5C%5C%5Cmu%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%28%5Ctext%7B%5BBa%24%5E%7B2%2B%7D%24%5D%7D%5Ccdot%20%282%2B%29%5E%7B2%7D%20%2B%20%5Ctext%7B%5BNO%24_%7B3%7D%24%24%5E%7B-%7D%24%5D%7D%5Ctimes%28-1%29%5E%7B2%7D%29%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%28%5Ctext%7B0.02%7D%5Ctimes%204%20%2B%20%5Ctext%7B0.04%7D%5Ctimes1%29%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%280.08%20%2B%200.04%29%5C%5C%5C%5C%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes0.12%20%3D%200.06)
(ii) Silver iodate
a. Calculate the activity coefficients of the ions

b. Calculate the solubility
AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)
![K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5Cgamma_%7BAg%5E%7B%2B%7D%7D%24%5BIO%24_%7B3%7D%24%24%5E%7B-%7D%24%5D%24%5Cgamma_%7BIO_%7B3%7D%5E%7B-%7D%7D%24%7D%20%3D%20s%5Ctimes0.75%5Ctimes%20s%20%5Ctimes%200.75%20%3D0.56s%5E%7B2%7D%3D%203.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%5E%7B2%7D%20%3D%20%5Cdfrac%7B3.0%20%5Ctimes%2010%5E%7B-8%7D%7D%7B0.56%7D%20%3D%205.3%20%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5Cs%20%3D2.3%20%5Ctimes%2010%5E%7B-4%7D%5Ctext%7B%20mol%2FL%7D)
(iii) Barium sulfate
a. Calculate the activity coefficients of the ions

b. Calculate the solubility
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq
![K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BBa%24%5E%7B2%2B%7D%24%5D%24%5Cgamma_%7B%20Ba%5E%7B2%2B%7D%7D%24%5BSO%24_%7B4%7D%24%24%5E%7B2-%7D%24%5D%24%5Cgamma_%7B%20SO_%7B4%7D%5E%7B2-%7D%7D%24%7D%20%3D%20%280.02%20%2B%20s%29%20%5Ctimes%200.32%5Ctimes%20s%5Ctimes%200.32%20%5Capprox%20%200.02%5Ctimes0.10s%5C%5C2.0%5Ctimes%2010%5E%7B-3%7Ds%20%3D%201.1%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cs%20%3D%20%5Cdfrac%7B1.1%5Ctimes%2010%5E%7B-10%7D%7D%7B2.0%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%205.5%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctext%7B%20mol%2FL%7D)
The chemical formula for sodium and sulfur is Na2S I believe...
Answer:
2.6 mole of lithium bromide have 225.81 g.
Explanation:
Given data:
Number of moles of lithium bromide = 2.6 mol
Mass of lithium bromide = ?
Solution:
Formula:
<em>Number of moles = mass/ molar mass</em>
Molar mass of lithium bromide = 86.85 g/mol
Mass = number of moles × molar mass
Mass = 2.6 mol ×86.85 g/mol
Mass = 225.81 g
Thus 2.6 mole of lithium bromide have 225.81 g.
Answer:
Option A (They happen as the Sun hits its maximum and minimum level throughout the sky) is the correct option.
Explanation:
- Solstice seems to be the moment whereas at noon this same sun achieved its maximum as well as minimum position, actually results throughout the year's shortest as well as biggest days.
- We currently believe that perhaps the solstice seems to be an apocalyptic event, exacerbated either by the tilt of the globe’s axis and therefore its direction of movement in an elliptical orbit. Globe's rotation not everyone's travel time from either the sun was what was causing winter or even just summer.
Some other options given should not be in accordance with the given context. So, the correct approach would've been option A.
The answer is D. 62 Steering Wheels
hope this helps !