Answer:
0.034 M is the molarity of sodium acetate needed.
Explanation:
The pH of the buffer solution is calculated by the Henderson-Hasselbalch equation:
![pH = pK_a + \log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pK_a%20%2B%20%5Clog%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where:
pK_a= Negative logarithm of the dissociation constant of a weak acid
= Concentration of the conjugate base
[HA] = Concentration of the weak acid
According to the question:

The desired pH of the buffer solution = pH = 5.27
The pKa of acetic acid = 4.74
The molarity of acetic acid solution = [HAc] = 0.01 M
The molarity of acetate ion =![[Ac^-] = ?](https://tex.z-dn.net/?f=%5BAc%5E-%5D%20%3D%20%3F)
Using Henderson-Hasselbalch equation:
![5.27= 4.74 + \log \frac{[Ac^-]}{[0.01 M]}](https://tex.z-dn.net/?f=5.27%3D%204.74%20%2B%20%5Clog%20%5Cfrac%7B%5BAc%5E-%5D%7D%7B%5B0.01%20M%5D%7D)
![[Ac^-]=0.0339 M\approx 0.034M](https://tex.z-dn.net/?f=%5BAc%5E-%5D%3D0.0339%20M%5Capprox%200.034M)
Sodium acetate dissociates into sodium ions and acetate ions when dissolved in water.

![[Ac^-]=[Na^+]=[NaAc]= 0.034M](https://tex.z-dn.net/?f=%5BAc%5E-%5D%3D%5BNa%5E%2B%5D%3D%5BNaAc%5D%3D%200.034M)
0.034 M is the molarity of sodium acetate needed.
Answer:
A) 3.17 g of Zn
Explanation:
Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.
Zn⁺²(aq) + 2e⁻ → Zn(s)
We can establish the following relations:
- 1 min = 60 s
- 1 A = 1 C/s
- The charge of 1 mole of electrons is 96,468 C (Faraday's constant).
- When 2 moles of electrons circulate, 1 mole of Zn is deposited.
- The molar mass of Zn is 65.38 g/mol
The mass of Zn deposited under these conditions is:

Lithium is in the Alkali Metal group or 1A column. The atoms in this group form ions with a 1+ charge. Lithium ion’s charge is 1+.
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT
To find the heat released.
Are u sure this is the right option? Well, antimony can be decomposed. Including octane.