take two carbon atoms and join by two springs making a double bond. complete the ethene moleeule by adding two hydrogen atoms to
1 answer:
Ethene (H2C=CH2) is a gaseous by-product of the long-chain hydrocarbons that are cracked to create highly sought-after goods like gasoline and aviation fuels.
<h3>What is ethene used for?</h3>- The uses of ethene include: It is employed in the creation of polythene. It is used to make fruits artificially ripe.
- Ethene is also utilised to create other beneficial molecules such ethanol, oxyethylene flame, and ethylene glycol, which is used as an antifreeze.
- A substance having the chemical formula C2H4 is ethylene, commonly known as ethene.
- It is both the most easily manufactured organic compound in the world and the simplest alkene.
- Unsaturated hydrocarbon ethylene, often known as C2H4, is utilised in industrial facilities and is occasionally found in typical medicine cabinets as a hormone.
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Answer:
7.3
Explanation:
By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that , and pKa = -logKa. Ka is the equilibrium constant of the acid.
Henderson Hasselbalch equation :
Where [HA] is the concentration of the acid, and is the concentration of the anion which forms the acid.
So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X
n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol
When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.
Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of will be
n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol
So, the 0.0075 mol of reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of
, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.
The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.
Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!
So,
6.72 = pKa - log 4
pKa - log4 = 6.72
pKa = 6.72 + log4
pKa = 6.72 + 0.6
pKa = 7.3