Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of 
at equilibrium = 0.067 mol
Concentration of 
at equilibrium = 0.021 mol
Concentration of 
at equilibrium = 0.040 mol
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 424.3
Answer:
The removal of one chlorine atom and addition of one fluorine atom.
Explanation:
When GeCl4 is transformed into GeCl3F by a substitution process, the main change that is occur in GeCl4 is the removal of one chlorine atom and addition of one fluorine atom. This process is known as substitution process in which two molecules exchange their atoms with each other when they comes in physically contact with each other.
Glucose is the simplest sugar and carbohydrate that provides energy. The simplified model of glucose (C₆H₁₂O₆) shows carbon, hydrogen, and oxygen atoms linked together.
<h3>What is glucose?</h3>
Glucose is an example of a carbohydrate macromolecule that is further classified as a monosaccharide. They are crystalline and fundamental units of carbohydrates.
The molecular formula of glucose is C₆H₁₂O₆ and the mass is 180.156 g/mol. It is an aldohexose that contains an aldehydic functional group. In its structure, there are six oxygen atoms, six carbon atoms, and twelve hydrogen atoms.
Therefore, the glucose molecule is composed of C, H, and O.
Learn more about glucose here:
brainly.com/question/2396657
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Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
