The stock solution contains 10.5 moles of HCl per litre. A 5.5 litre solution of 2.5M HCl contains 5.5x2.5 = 13.75moles of HCl. Since every litre of stock solution provides 10.5M HCl, the amount of stock solution needed is 13.75/10.5 = 1.309L. Therefore you would dilute 1.309L of stock solution to 5.5L
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Detailed solution is shown below ask if any doubt :
Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
