Answer: phosphorus trihyde
Explanation: just took the quiz and got this answer
Answer:

Explanation:
= Initial concentration = 1.28 M
= Final concentration = ![0.17[A]_0](https://tex.z-dn.net/?f=0.17%5BA%5D_0)
k = Rate constant = 0.0632 s
t = Time taken
For first order reaction we have the relation
![kt=\ln\dfrac{[A]_0}{[A]}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{[A]}}{k}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{0.17[A]_0}}{0.0632}\\\Rightarrow t=28.037\ \text{s}](https://tex.z-dn.net/?f=kt%3D%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%7D%7Bk%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B0.17%5BA%5D_0%7D%7D%7B0.0632%7D%5C%5C%5CRightarrow%20t%3D28.037%5C%20%5Ctext%7Bs%7D)
Time taken to reach the required concentration would be
.
Required pH = 4.93
- OH⁻ from NaOH reacts with CH₃COOH giving CH₃COO⁻ and H₂O
- Let the volume of 3.5 M NaOH be x ml
Moles of NaOH = Moles of OH⁻ = Molarity * x ml = 3.5x mmol
- The reaction table for moles is as follows:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Initial 60 3.5x 40
Change -3.5x -3.5x +3.5x
Final (60-3.5x) 0 (40+3.5x)
- Substitute in Henderson equation and solve for x:
pH = pKa + log
![\frac{[CH_{3}COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D%20)
4.93 = 4.76 + log

0.17 = log


x = 5.62 ml NaOH required
A conjugate base is the species that D. Remains after an acid has given up a single proton.
It is the deprotonated form of the acid.