Answer:
Explanation:
1. Solubility of CaF_2
(a) Molar solubility
CaF₂ ⇌ Ca²⁺ + 2F⁻
![K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bsp%20%7D%7D%20%3D%20%5Ctext%7B%5BCa%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BF%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%282s%29%5E%7B2%7D%3D4.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5C4s%5E%7B3%7D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%5E%7B3%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%20%3D2.2%20%5Ctimes%2010%5E%7B-3%7D%5Ctext%7B%20mol%2FL%7D)
(b) Mass solubility
2. pH
pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52
3. Oxidizing and reducing agents
Zn + Cl₂ ⟶ ZnCl₂
The oxidation number of Cl has decreased from 0 to -1.
Cl has been reduced, so Cl is the oxidizing agent.
4. Oxidation numbers
(a) Al₂O₃
1O = -2; 3O = -6; 2Al = +6; 1Al = +3
(b) XeF₄
1F = -1; 4F = -4; 1 Xe = +4
(c) K₂Cr₂O₇
1K = +1; 2K = +2; 1O = -2; 7O = -14
+2 - 14 = -12
2Cr = + 12; 1 Cr = +6
Answer:
Q = 4.056 J
Explanation:
∴ m = 406.0 mg = 0.406 g
∴ <em>C </em>= 1.85 J/g.K
∴ T1 = 33.5°C ≅ 306.5 K
∴ T2 = 38.9°C = 311.9 K
⇒ ΔT = 311.9 - 306.5 = 5.4 K
⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)
⇒ Q = 4.056 J
Answer:
53.7 grams of HNO3 will be produced
Explanation:
Step 1: Data given
Mass of NO2 = 59.0 grams
Molar mass NO2 = 46.0 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 59.0 grams / 46.0 g/mol
Moles NO2 = 1.28 moles
Step 4: Calculate moles HNO3
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3
Step 7: Calculate mass HNO3
Mass HNO3 = 0.853 moles * 63.01 g/mol
Mass HNO3 = 53.7 grams
53.7 grams of HNO3 will be produced
