The average atomic mass of her sample is 114.54 amu
Let the 1st isotope be A
Let the 2nd isotope be B
From the question given above, the following data were obtained:
- Abundance of isotope A (A%) = 59.34%
- Mass of isotope A = 113.6459 amu
- Mass of isotope B = 115.8488 amu
- Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
- Average atomic mass =?
The average atomic mass of the sample can be obtained as follow:
Thus, the average atomic mass of the sample is 114.54 amu
Learn more about isotope: brainly.com/question/25868336
The rows in the top third - This group consists of elements like Sodium, Magnesium, Potassium and Calcium on the right and Chlorine, Carbon, Nitrogen and Oxygen on the left.
Sodium and Chlorine are components of salt, a very important compound of our blood, essential for transferring electrical signals from the brain to the rest of the body and vice versa. Calcium is the building block of our bones, while Magnesium and potassium ensure proper functioning of our organs.
Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar
