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:
The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).
Explanation:
The reaction between an acid and a base is called neutralization, forming a salt and water.
Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.
When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:
V acid *M acid = V base *M base
where V represents the volume of solution and M the molar concentration of said solution.
In this case:
- V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
- M acid= 0.129 M
- V base= ?
- M base= 0.135 M
Replacing:
0.0137 L* 0.129 M= V base* 0.135 M
Solving:
V base=0.0131 L = 13.1 mL
<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>