<u>Answer:</u> The expression for equilibrium constant is ![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the general chemical equation:

The expression for
is given as:
![K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
For the given chemical reaction:

The expression for
is given as:
![K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%5BHgO.HgCl_2%5D%7D%7B%5BHgO%5D%5E2%5BH_2O%5D%5BCl_2%5D%5E2%7D)
The concentration of solid is taken to be 0.
So, the expression for
is given as:
![K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D%5BCl_2%5D%5E2%7D)
Answers:
Density = 0.8 g/mol.
Given data:
v = 25 ml
m = 20 g
δ = ?
Solution:
Formula for calculating density is given as,
Density = Mass / Volume
putting values
Density = 20.0 g / 25 ml
Density = 0.8 g/mol.
Electromagnetic energy would be your answer :)
your answer is A if i am wrong let me know
Answer:
1.18 × 10²⁴ particles Mg
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
47.7 g Mg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Mg - 24.31 g/mol
<u>Step 3: Convert</u>
<u />
= 1.18161 × 10²⁴ particles Mg
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg