Answer:
1.53 L
Explanation:
Step 1: Given data
- Mass of oxygen (m): 11.2 g
- Ideal gas constant (R): 0.0821 atm.L/mol.K
Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
11.2 g × (1 mol/32.00 g) = 0.350 mol
Step 3: Calculate the volume of oxygen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm
V = 1.53 L
1. 5 electrons.

- Therefore, the 3d subshells blanks will be like this:
- ↑ ↑ ↑ ↑ ↑
2. 6 electrons.

- The 3d subshells blanks will be:
- ↑↓ ↑ ↑ ↑ ↑
3. 7 electrons.

- The 3d subshells blanks will be:
- ↑↓ ↑↓ ↑ ↑ ↑
Hope you could understand.
If you have any query, feel free to ask.
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:
272.31× 10²³ atom of strontium
Explanation:
Given data:
Number of moles of strontium = 45.22 mol
Number of atoms = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
45.22 mol × 6.022 × 10²³ atom / 1 mol
272.31× 10²³ atom
D
Is the answer because of the ratio of you fart and the smell in ur nose