For a and b, you need to divide it by Avogadro’s number to find the answer.
a. (6.022x10^23)/6.022x10^23 = 1 mole of Ne
b. (3.011x10^23)/6.022x10^23 = 0.5 moles of Mg
For c and d, you’ll use the mass provided divided by the molar mass to find the number of moles.
Pb molar mass = 207.2 g/mol
c. (3.25x10^5)/207.2 = 1.57x10^3 moles of Pb
For d, I can’t tell if is Cu, C or something else but you can follow the steps above to solve the problem.
A. Aluminum......................
Answer:
The coordination sphere of a complex consists of <u><em>the central metal ion and the ligands bonded to it.</em></u>
Explanation:
The Coordination Compounds are sets of a central metal ion attached to a group of molecules or ions that surround it. They are also called metal complexes or simply complexes. Then they are compounds that have a central atom surrounded by a group of molecules or ions, the latter called ligands.
The central atom must have empty orbitals capable of accepting pairs of electrons, with the transition metals being the ones with the greatest tendency. Because of this, they can act as Lewis acids (electron pair acceptors). The ligands have unshared electron pairs, then acting as Lewis bases (electron pair donors).
When forming a complex, it is said that the ligands coordinate to the metal and the central metal and the ligands attached to it constitute the coordination sphere of the complex.
Finally, <u><em>the coordination sphere of a complex consists of the central metal ion and the ligands bonded to it.</em></u>
Answer:
D. water undergoing electrolysis
Answer:
Atomic Mass of Element = 59.57 amu
Given Data:
Element X Isotope 1 atomic weight = 59.015 amu
Element X Isotope 2 atomic weight = 62.011 amu
The abundance of lighter isotope = 81.7%
Atomic Mass of Element X = ?
Solution:
81.7% is the abundance of element that is lighter so it is the of the isotope having weight 59.015 amu
Then the remaining abundance = 100 -81.7 = 18.3 %
18.3 % abundance of the isotope weight = 62.011 amu
Formula:
Mass of the element = % of abundance (atomic mass of Isotope 1) + % of abundance (atomic mass of Isotope 2)
Put the values
Atomic mass of Element X = 81.7/100 (59.015) + 18.3 /100 (62.011)
Atomic mass of Element = 0.817 (59.015) + 0.183 (62.011)
Atomic mass of Element = 48.22 + 11.35
Atomic Mass of Element = 59.57 amu