Answer:
The answer to your question is 24.325
Explanation:
Data
Magnesium-24 Abundance = 78.70%
Magnesium-25 Abundance = 10.13%
Magnesium-26 Abundance = 11.17%
Process
1.- Convert the abundance to decimals
Magnesium-24 Abundance = 78.70/100 = 0.787
Magnesium-25 Abundance = 10.13/100 = 0.1013
Magnesium-26 Abundance = 11.17/100 = 0.1117
2.- Write an equation
Average atomic mass = (Atomic mass-1 x Abundance 1) + (Atomic mass 2 x
Abundance-2) + (Atomic mass 3 x Abundance 3)
3.- Substitution
Average atomic mass = (24 x 0.787) + (25 x 0.1013) + (26 x 0.1117)
4.- Simplification
Average atomic mass = 18.888 + 2.533 + 2.904
5.- Result
Average atomic mass = 24.325
The answer & explanation for this question is given in the attachment below.
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)
I got this idk if it's correct: For measuring large distances, the kilometer (103 or 1000 meters) is often used. The basic unit of volume in the metric system is the liter (l). The most common derived unit is the milliliter (ml) (10-3 or 1/1000 of a liter). The volume of a milliliter is equal to the volume of a cube 1 centimeter per side.
Answer:
If garlic repels fleas, then a dog that is given garlic every day will not get fleas. Bacterial growth may be affected by moisture levels in the air. If sugar causes cavities, then people who eat a lot of candy may be more prone to cavities.
Explanation:
