Answer:
1) There are Four Elements
Li , H , P , O
2) Li have 3 atoms.
3) H have 3 atoms.
4) P have 3 atoms.
5) O have 3 atoms.
Answer:
c. CH₃COO⁻
Explanation:
The task is not very clear but I think this is the original question.
<em>Identify the conjugate base of CH₃COOH in the reaction:
</em>
<em>CH₃COOH + HSO₄⁻ ⇄ H₂SO₄ + CH₃COO⁻
</em>
<em>a. HSO₄⁻
</em>
<em>b. SO₄²⁻
</em>
<em>c. CH₃COO⁻</em>
<em>d. H₂SO₄</em>
<em>e. OH⁻</em>
Let's consider the following reaction
CH₃COOH + HSO₄⁻ ⇄ H₂SO₄ + CH₃COO⁻
According to the Bronsted-Lowry acid-base theory, an acid is a species that donates H⁺ and a base is a species that accepts H⁺.
CH₃COOH is an acid because it donates H⁺ to HSO₄⁻, whereas CH₃COO⁻ is its conjugate base, because it accepts H⁺ from H₂SO₄ to form CH₃COOH.
Another acid-base pair is H₂SO₄/HSO₄⁻
Answer:
m = 0.122 m
Explanation:
Molality(m): A measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
Answer:
Rubidium and cesium
Explanation:
It is noteworthy to say here that larger cations have more stable superoxides. This goes a long way to show that large cations are stabilized by large cations.
Let us consider the main point of the question. We are told in the question that the reason why potassium reacts with oxygen to form a superoxide is because of its low value of first ionization energy.
The implication of this is that, the other two metals that can be examined to prove this point must have lower first ionization energy than potassium. Potassium has a first ionization energy of 419 KJmol-1, rubidium has a first ionization energy of 403 KJ mol-1 and ceasium has a first ionization energy of 376 KJmol-1.
Hence, if we want to validate the hypothesis that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy, we must also consider the elements rubidium and cesium whose first ionization energies are lower than that of potassium.