This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
According to the IUPAC convention alkyl substituents on a hydrocarbon chain should be listed in alphabetical without considering prefixes order.
<h3>What is IUPAC convention?</h3>
IUPAC convention of organic chemistry is a method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC).
The name of the compound is written out with the substituents in alphabetical order followed by the base name (derived from the number of carbons in the parent chain).
Types of IUPAC Nomenclature of a few important aliphatic compounds:
- Alkane
- Alkene
- Alkyne
Example :
Ethane, which has 2 carbon atoms and 6 hydrogen atoms, with the molecular formula of = C₂H₆
Formation of alkyl group:
Methane (CH₄) Remove 1 hydrogen (H) convert to methyl (H₃-C-)
Example: Propyl (-CH₃ - CH₂ - CH₂ - )
According to the IUPAC convention alkyl substituents on a hydrocarbon chain should be listed in alphabetical without considering prefixes order.
Learn more about IUPAC convention, Here:
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Answer:
2. Inorganic
Explanation:
All man-made and most carbon-based compounds are inorganic
Answer:
The pH is 4.76 (option C)
Explanation:
pH of a solution = TO BE DETERMNED
[sodium acetate] = 1M ; [acetic acid] = 1M
K = [H+] [A-] / [HA]
K = [H+] [1] / [1]
K = H
pH = pK
⇒ now pH is equal to pK
⇒ The value of pK is given which is 4.76
⇒ This means the value of the pH is also 4.76
Calculate the molar it’s of nitric acid
