Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. ![Na_{2} CO_{3}](https://tex.z-dn.net/?f=Na_%7B2%7D%20CO_%7B3%7D)
Major Species at B:
1. ![Na_{2} CO_{3}](https://tex.z-dn.net/?f=Na_%7B2%7D%20CO_%7B3%7D)
2. ![NaHCO_{3}](https://tex.z-dn.net/?f=NaHCO_%7B3%7D)
Major Species at C:
1. ![NaHCO_{3}](https://tex.z-dn.net/?f=NaHCO_%7B3%7D)
Major Species at D:
1. ![NaHCO_{3}](https://tex.z-dn.net/?f=NaHCO_%7B3%7D)
2. ![H_{2}CO_{3}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D)
Major Species at E:
1. ![H_{2}CO_{3}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D)
Major Species at F:
1. ![H_{2}CO_{3}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D)
b) pH calculation:
At Halfway point B:
pH = pK
+ log[![CO_{3}](https://tex.z-dn.net/?f=CO_%7B3%7D)
]/[H![CO_{3}](https://tex.z-dn.net/?f=CO_%7B3%7D)
]
pH = pK
= 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H![CO_{3}](https://tex.z-dn.net/?f=CO_%7B3%7D)
]/[H2
]
pH = pK
= 10.33
Technically yes but no because you have to fill the 3s orbital before the 5s orbital.
<h3><u>Answer;</u></h3>
A) HNO3 and NO3^-
<h3><u>Explanation;</u></h3>
- <em><u>HNO3 is a strong acid and NO3 is its conjugate base, meaning it will not have any tendency to withdraw H+ from solution.</u></em>
- Buffers are often prepared by mixing a weak acid or base with a salt of that weak acid or base.
- The buffers resist changes in pH since they contain acids to neutralize OH- and a base to neutralize H+. Acid and base can not consume each other in neutralization reaction.
22 grams, 5 x 3 x 1 = 15 / volume. mass (330) ÷ 15 (volume) = 22 grams (density)
Francium oxide Fr₂O
Francium, as well as sodium, is in 1A group. Its valency is equal to unit.