Answer:
K loses one electron to CI
Explanation:
The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.
We have atom of K with one valence electrons
Cl with 7 valence electrons
For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.
This will make both species attain a stable octet;
Hence, K will lose 1 electron and Cl will gain the electrons.
This is a simple chemical change due to what it produces and how it is added together. Hope this helps.
Answer:
- <u>Tellurium (Te) and iodine (I) are two elements </u><em><u>next to each other that have decreasing atomic masses.</u></em>
Explanation:
The <em>atomic mass</em> of tellurium (Te) is 127.60 g/mol and the atomic mass of iodine (I) is 126.904 g/mol; so, in spite of iodine being to the right of tellurium in the periodic table (because the atomic number of iodine is bigger than the atomic number of tellurium), the atomic mass of iodine is less than the atomic mass of tellurium.
The elements are arranged in increasing order of atomic number in the periodic table.
The atomic number is equal to the number of protons and the mass number is the sum of the protons and neutrons.
The mass number, except for the mass defect, represents the atomic mass of a particular isotope. But the atomic mass of an element is the weighted average of the atomic masses of the different natural isotopes of the element.
Normally, as the atomic number increases, you find that the atomic mass increases, so most of the elements in the periodic table, which as said are arranged in icreasing atomic number order, match with increasing atomic masses. But the relative isotope abundaces of the elements can change that.
It is the case that the most common isotopes of tellurium have atomic masses 128 amu and 130 amu, whilst most common isotopes of iodine have an atomic mass 127 amu. As result, tellurium has an average atomic mass of 127.60 g/mol whilst iodine has an average atomic mass of 126.904 g/mol.
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
Answer:
(a) 
(b) Rubidium
Explanation:
Hello,
This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:
Solving for the molarity of base we've got:
Now, we can compute the moles of the base as:
(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:
(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:
So, that atomic mass dovetails to the Rubidium's atomic mass.
Best regards.
