A positive change atom of hydrogen-ion. A normal hydrogen atomic nucleus
1×10^-4 = 0,0001M
pH = -log[H+]
pH = -log0,0001
pH = 4
Answer: 106.905
Explanation: If there are only 2 isotopes, and 1 of them is 48.16%, the second must, by default, be (100 - 48.16%) = 51.84% The final, averaged, atomic mass is 107.868. This is made up of each isotope's atomic mass times the percentage of that isotope in the total sample. The weighted value of the known isotope (109) plus that of the unknown must come to the observed value of 107.868 amu. (107.868 - 52.45 = 55.42). Divide that by the % for that isotope (55.42/0.5184) = 106.90 amu for the second isotope.
<u>Atomic Mass</u> <u>% of Sample</u> <u>Weighted Value</u>
108.905 48.16% 52.45
X 51.84% <u>55.42</u>
107.87
X = (55.42/0.5184) = 106.90 amu
137 g NO) / (30.0061 g NO/mol)x (43kcal / 2 mil NO) = 98kcal
Answer:
In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory. Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space. Although sometimes taught together with the valence shell electron-pair repulsion (VSEPR) theory, valence bond and hybridisation are in fact not related to the VSEPR model.
