Answer: Gold.
Explanation: Water, CO2, and table salt are compounds. They are composed of two or more separate elements; a mixture, where as gold is neither.
The items that are true of early nuclear science are "the first nuclear reactions were done in the 1880s" and "the first nucleus split was uranium-235." <span>The answers are letters A and D. It is impossible that nucleus was lost during the reaction as it will not follow the law of mass conservation.
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Answer: The gravitational pull on Moon B is greater than on Moon A because Moon B is closer to the new planet than Moon A.
Explanation:
The gravitational force exerted by the two objects on each other is inversely proportional to the square of the distance between the objects.

F = gravitational force or pull
G = gravitational constant on that planet
M = mass of the object-1
m = mass of object-2(Mass of Moon-A or Moon-B)
r = distance between two objects

With decease in distance 'r' the force between the object increases or vice versa.So, from this we can say that the gravitational pull on Moon-B is more than the gravitational pull on Moon-A because the Moon B is closer than the Moon-A from the new planet.
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)