True.
<span><span>Melting points decrease down a group and increase across a period.</span>
</span>
Answer:
insulation packed around a hot water pipe
Explanation:
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)
<span>1 mole of any substance contains 6.02 x 10^23 particles.
Hence 0.25 mole of any has will contain 0.25 x 6.02 x 10^23 molecules of that gas
They cannot contain the same no. of atoms as 1 molecule of CO2 contains 3 atoms, 1 molecule H2 contains 2 molecules, and 1 molecule of NH3 contain 4 atoms
1 molecule of any gas will occupy 22.4 L at STP
hence 0.25 mol of all of these gases accupy the same volume
all of these molecules have different molar masses. thus their g.f.w cannot be same
Your answer is that they will occupy the same volume, and the same number of molecules</span>
The chemicals that cause acid rain are sulfur dioxide and nitrogen oxides . when they rise into the atmosphere they start a very acidic chemical that is later expelled into earth by the clouds