Answer:
![\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B1.%20%5Ctext%7B%200.17%20g%2FL%3B%202.%203.52%3B%203.%20Cl%3B%204.%20%28a%29%20%2B3%3B%20%28b%29%20%2B4%3B%20%28c%29%20%2B6%7D%7D)
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
![\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7BSolubility%7D%20%3D%202.2%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctext%7B%20mol%2FL%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B78.07%20g%7D%7D%7B%5Ctext%7B1%20L%20%7D%7D%20%3D%20%5Ctext%7B0.17%20g%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20solubility%20of%20CaF%24_%7B2%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.17%20g%2FL%7D%7D%24%7D)
2. pH
pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52
3. Oxidizing and reducing agents
Zn + Cl₂ ⟶ ZnCl₂
![\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}](https://tex.z-dn.net/?f=%5Crm%20%5Cstackrel%7B%5Chbox%7B0%7D%7D%7B%5Chbox%7BZn%7D%7D%20%2B%20%5Cstackrel%7B%5Chbox%7B0%7D%7D%7B%5Chbox%7B%20Cl%7D_%7B2%7D%20%7D%5Clongrightarrow%20%5Cstackrel%7B%5Chbox%7B%2B2%7D%7D%7B%5Chbox%7BZn%7D%7D%5Cstackrel%7B%5Chbox%7B-1%7D%7D%7B%5Chbox%7BCl%7D_%7B2%7D%7D)
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
![\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B3%7D%24%7D%7D%7B%5Chbox%7BAl%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B-2%7D%7D%7B%5Chbox%7BO%7D_%7B3%7D%7D)
1O = -2; 3O = -6; 2Al = +6; 1Al = +3
(b) XeF₄
![\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B4%7D%24%7D%7D%7B%5Chbox%7BXe%7D%7D%5Cstackrel%7B%5Chbox%7B-1%7D%7D%7B%5Chbox%7BF%7D_%7B4%7D%7D)
1F = -1; 4F = -4; 1 Xe = +4
(c) K₂Cr₂O₇
![\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%7B%2B1%7D%24%7D%7D%7B%5Chbox%7BK%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B6%7D%24%7D%7D%7B%5Chbox%7BCr%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B-2%7D%7D%7B%5Chbox%7BO%7D_%7B7%7D%7D)
1K = +1; 2K = +2; 1O = -2; 7O = -14
+2 - 14 = -12
2Cr = + 12; 1 Cr = +6
Answer:
Celiac disease occurs in the intestines as an allergic response to Gluten in certain cereal grains.
Explanation:
Answer:
ΔS = -0.1076 kJ /kg*K
Explanation:
Step 1: Data given
Initial state = 0.8 m³/kg and 25 °C = 298.15 K
Final state = 0. 3³/kg and 287 °C = 560.15 K
Cv = 0.686 kJ/kg*K
Step 2: Calculate the average temperature
The average temperature = (25°C + 287 °C)/2 =156 °C ( = 429 K)
Step 3: Calculate the ΔS
ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)
ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)
ΔS = -0.1076 kJ /kg*K
Neutralization reaction means the reactants must be one acid and one alkali, and the product will be H2O and metal salt.
The only one satisfying this will be B
Answer:
A
Explanation:
To answer this, we need to use Gay-Lussac's law, which states that:
, where P is pressure and T is temperature
The initial pressure we're given is 4.5 atm (so P1 = 4.5) and the temperature is 45.0°C; however, we need to change Celsius to Kelvins, so add 273 to 45.0: 45.0 + 273 = 318 K (so T1 = 318).
The final pressure is what we want to find, but we do know the final temperature is 3.1°C. Converting this to Kelvins, we get: 3.1 + 273 = 276.1 K, which means T2 = 276.1.
Plug these values in:
![\frac{4.5}{318}= \frac{P_2}{276.1}](https://tex.z-dn.net/?f=%5Cfrac%7B4.5%7D%7B318%7D%3D%20%5Cfrac%7BP_2%7D%7B276.1%7D)
Multiply both sides by 276.1:
≈ 3.9 atm
The answer is thus A.