In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.
Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.
For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.
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PV=nRT<=> P=nRT/V=2,40*R*(273+97)/45 atm.
Calculate it. R is a number that is given, find it and use your math to solve.
Answer:
50 mg.
Explanation:
From the prescription, 240 ml amount is to be prepared containing 30 ml tussin syrup. As the frequency is 5 mL qid, it means 5mL of syrup is to be administered four times daily i.e. 20 mL daily. To calculate the amount of tussin taken daily,
<em> </em><em><u>30 mL tussin</u></em><em> = </em><em><u>X mL of tussin</u></em>
<em> 240 mL solution 20ml solution</em>
⇒ 2.5 mL of tussin
5 ml tussin contains 100mg guafenesin,
So,
⇒ 2.5 ml of tussin contains 100x2.5/5 = 50 mg guaifenesin
Answer: vanadium selenide
Explanation: