Covalent example ch4 where carbon is covalently bonded with all 4 hydrogens
Sand dunes would be created due to the mixture falling on each other
x
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
The mass is 224 grams of NaOH
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)
