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Answer:
B. Four moles of water were produced from this reaction.
Explanation:
I took the test and got it correct
The answer is the letter C
Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.
<h3>What is electrolysis?</h3>
Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.
The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.
Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium
0.1 F will discharge = 0.1/3 × 27 g of aluminium
mass of aluminium deposited = 0.9 g of aluminium.
Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.
Learn more about electrolysis at: brainly.com/question/26050361
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54