It is actually Group 3A elements. I just took the test.
Answer:
3.6
Explanation:
Step 1: Given data
- Concentration of formic acid: 0.03 M
- Concentration of formate ion: 0.02 M
- Acid dissociation constant (Ka): 1.8 × 10⁻⁴
Step 2: Calculate the pH
We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.
![pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%3D%20-log%201.8%20%5Ctimes%2010%5E%7B-4%7D%20%2B%20log%20%5Cfrac%7B0.02%7D%7B0.03%7D%20%3D%203.6)
Answer:
What is the name of the compound (NH4) 3PO4? (NH)4+ is an ammonium radical and (PO4) 3- is a phospate radical. When we start writing the compound,the valency of phosphate goes to ammonium as subscript,and that of ammonium goes to phosphate. Hence the formula (NH4)3 PO4 and it's name is ammonium phospate.
Explanation:
No, but we can make it conduct energy by adding salt
Answer: 2-Benzyloxyethanol is produced reaction in the fig. below.
Explanation: