Which of the following could be the subshell representation of an element of Group IV?
1 answer:
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Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
8 P₄ + 3 S₈ → 8 P₄S₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- P₄: 8 moles
- S₈: 3 moles
- P₄S₃: 8 moles
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?
<u><em>moles of P₄= 10.667 moles</em></u>
But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.
Learn more about the reaction stoichiometry:
#SPJ1
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
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