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
A b A or just the person above me said
<h3>
Answer:</h3>
5.6 L
<h3>
Explanation:</h3>
We are given;
- Initial volume, V1 = 3.5 L
- Initial pressure, P1 = 0.8 atm
- Final pressure, P2 = 0.5 atm
We are required to calculate the final volume;
- According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at a constant temperature.
- That is; P α 1/V
- Mathematically, P=k/V
- At two different pressure and volume;
P1V1 = P2V2
In this case;
Rearranging the formula;
V2 = P1V1 ÷ P2
= (0.8 atm × 3.5 L) ÷ 0.5 atm
= 5.6 L
Therefore, the resulting volume is 5.6 L
Answer:
0
Explanation:
it is 0 I believe nothing
Answer:
0.362 moles
Explanation:
Mass of butane = 21g
Molar mass of carbon = 12g / mol
Molar mass of hydrogen = 1g/mol
Molar mass of butane ? = [(12*4) * (1*10)]
Molar mass of butane = 58g / mole
Number of moles = mass of molecules / molar mass of molecule
Number of moles = 21 / 58
Number of moles of butane = 0.362 moles
The number of moles in 21g of butane is 0.362 moles