Answer:
1. Polar
2 nonpolar
3. Nitrogen or oxygen
4.carbon
5.Repelled by and attracted to
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:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
Saturated fatty acids exhibit a linear structure while unsaturated fatty acids bend, or kink, due to double bonds within the chemical foundation.