Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-5%7DM)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-3%7DM)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
Answer:
C. 100.7 amu
Explanation:
Isotopes of an element are atoms of an element with the same atomic number but different atomic masses. Each atomic mass of an isotope is known as an isotopic mass. An element that exhibits isotope, that is, that have two or more isotopes has a relative atomic mass that is not a whole number.
Relative atomic mass of X is the sum of the products of the relative abundances of each isotope and its isotopic mass.
For Isotope ¹⁰⁰X: 30% × 100 = 30 amu
For Isotope ¹⁰¹X: 70% × 101 = 70.7 amu
Relative atomic mass of X = (30 + 70.7) amu = 100.7 amu
Therefore, the approximate atomic mass of X is 100.7 amu
Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.
Electron Affinity of Lithium is 59.6 kJ/mol.
Electron Affinity of Caesium is 45.5 kJ/mol.
Electron Affinity of Lithium is 59.6 kJ/mol. Electronegativity of Lithium is 0.98. ... Electron affinities are more difficult to measure than ionization energies. An atom of Lithium in the gas phase, for example, gives off energy when it gains an electron to form an ion of Lithium.
Trends
The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.
The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.
The noble gases possess very high ionisation energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements.
I’m pretty sure it’s none of the above cause I’m googling it a bunch and and it say you use the dideoxy method
