<span>c] The golfer sent the golf ball flying toward the cup to score a hole in one.</span>
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
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: a + 2
Explanation: Alkali Earths or Group II has an ionization charge of a + 2. Alkali Metals have a ionization a + 1. Halogens or cold elements have a ionization of a +3.
Answer: The molar mass and molecular weight of 4Fe is 223.38.
