Answer:
CaS, CaBr₂, VBr₅, and V₂S₅.
Explanation:
- The ionic compound should be neutral; the overall charge of it is equal to zero.
- Binary ionic compound is composed of two different ions.
<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.
- CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.
<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.
- VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.
<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>
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Answer:
= 72900 years
Explanation:
- The half-life is the time taken by a radioactive material to decay by half the original amount.
- The half-life of plutonium-239 is 24300 years which means it takes 24300 years to decay by half the original amount.
To calculate the time taken for a mass of 8 kg to decay to 1 kg we use;
New mass = Original mass x (1/2) ^n, where n is the number of half-lives
Therefore;
1 kg = 8 kg × (1/2)^n
1/8 = (1/2)^n
solving for n;
n =3
Therefore;
Time = 3 × 24300 years
= 72900 years
It will, therefore, take 72900 years for 8 kg of plutonium-239 to decay to 1 kg.
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. 
Major Species at B:
1.
2. 
Major Species at C:
1.
Major Species at D:
1.
2. 
Major Species at E:
1.
Major Species at F:
1.
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK = 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H]/[H2]
pH = pK = 10.33
