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⁻
23 would be the answer actually idk I just need to answer some answers
Answer : The molar heat capacity at constant volume and constant pressure is 29.9 J/mol.K and 38.2 J/mol.K respectively.
Explanation :
Formula used for specific heat at constant volume :
![q_v=n\times c_v\times \Delta T](https://tex.z-dn.net/?f=q_v%3Dn%5Ctimes%20c_v%5Ctimes%20%5CDelta%20T)
where,
= heat = 229 J
n = moles of gas = 3.0 mol
= molar heat capacity at constant volume = ?
= change in temperature = 2.55 K
Now put all the given value in the above formula, we get:
![229J=3.0mol\times c_v\times 2.55K](https://tex.z-dn.net/?f=229J%3D3.0mol%5Ctimes%20c_v%5Ctimes%202.55K)
![c_v=29.9J/mol.K](https://tex.z-dn.net/?f=c_v%3D29.9J%2Fmol.K)
The molar heat capacity at constant volume is 29.9 J/mol.K
Now we have to calculate the molar heat capacity at constant pressure.
Formula used :
![c_p-c_v=R](https://tex.z-dn.net/?f=c_p-c_v%3DR)
where,
= molar heat capacity at constant volume = 29.9 J/mol.K
= molar heat capacity at constant pressure = ?
R = gas constant = 8.314 J/mol.K
Now put all the given value in the above formula, we get:
![c_p-29.9J/mol.K=8.314J/mol.K](https://tex.z-dn.net/?f=c_p-29.9J%2Fmol.K%3D8.314J%2Fmol.K)
![c_p=38.2J/mol.K](https://tex.z-dn.net/?f=c_p%3D38.2J%2Fmol.K)
The molar heat capacity at constant pressure is 38.2 J/mol.K