In periodic tables, the relative atomic mass is usually not a whole number, even though the mass of both protons and neutrons is 1. This is due to isotopes, which are atoms of the same element, but they have different amounts of neutrons in their nuclei. So the relative mass of an individual atom is always a whole number, and the relative atomic mass is an average of all of the isotopes of the element.
ANSWER
The number of moles of oxygen formed is 7.06 moles
EXPLANATION
Given that
The number of moles of mercury (II) oxide is 14.12 moles
Follow the steps below to find the number of moles of oxygen
Step 1; Write the balanced equation for the decomposition of the reaction
![\text{ 2HgO }\rightarrow\text{ 2Hg}_{(s)}\text{ + O}_{2(g)}](https://tex.z-dn.net/?f=%5Ctext%7B%20%202HgO%20%7D%5Crightarrow%5Ctext%7B%202Hg%7D_%7B%28s%29%7D%5Ctext%7B%20%2B%20O%7D_%7B2%28g%29%7D)
In the reaction above, 2 moles HgO decompose to produce 2 moles Hg and 1 mole O2
Step 2; Find the number of moles of oxygen using a stoichiometry ratio
Let x represents the number of moles of oxygen
![\begin{gathered} \text{ 2 moles HgO }\rightarrow\text{ 1 mole O}_2 \\ \text{ 14.12 moles HgO}\rightarrow\text{ x mole O}_2 \\ \text{ cross multiply} \\ \text{ 2 moles HgO }\times\text{ x mole O}_2\text{ }=\text{ 1 mole O}_2\times\text{ 14.12 moles HgO} \\ \text{ Isolate x} \\ \text{ x = }\frac{1\text{ mole O}_2\times14.12moles\cancel{HgO}}{2moles\cancel{HgO}} \\ \text{ x = }\frac{14.12}{2} \\ \text{ x = 7.06 moles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7B%202%20moles%20HgO%20%7D%5Crightarrow%5Ctext%7B%201%20mole%20O%7D_2%20%5C%5C%20%5Ctext%7B%20%2014.12%20moles%20HgO%7D%5Crightarrow%5Ctext%7B%20x%20mole%20O%7D_2%20%5C%5C%20%5Ctext%7B%20cross%20multiply%7D%20%5C%5C%20%5Ctext%7B%202%20moles%20HgO%20%7D%5Ctimes%5Ctext%7B%20x%20mole%20O%7D_2%5Ctext%7B%20%7D%3D%5Ctext%7B%201%20mole%20O%7D_2%5Ctimes%5Ctext%7B%2014.12%20moles%20HgO%7D%20%5C%5C%20%5Ctext%7B%20Isolate%20x%7D%20%5C%5C%20%5Ctext%7B%20x%20%3D%20%7D%5Cfrac%7B1%5Ctext%7B%20mole%20O%7D_2%5Ctimes14.12moles%5Ccancel%7BHgO%7D%7D%7B2moles%5Ccancel%7BHgO%7D%7D%20%5C%5C%20%5Ctext%7B%20x%20%3D%20%7D%5Cfrac%7B14.12%7D%7B2%7D%20%5C%5C%20%5Ctext%7B%20x%20%3D%207.06%20moles%7D%20%5Cend%7Bgathered%7D)
Therefore, the number of moles of oxygen formed is 7.06 moles
Answer:
0.00121 atm
Explanation:
Step 1: Given data
- Height of the column of water (h): 12.5 m
- Density of water (ρ): 1.00g/cm³
- Gravitational constant (g): 9.80665 m/s²
Step 2: Calculate the pressure exerted by the column of water in SI units
The pressure exerted by the column of liquid depends on the height of the column, the density of the liquid and the gravitational constant. We can calculate it using the following expression.
P = ρ × g × h
P = 1.00g/cm³ × 9.80665 m/s² × 12.5 m
P = 123 Pa
Step 3: Convert "P" to atm
We will use the conversion factor 101325 Pa = 1 atm.
123 Pa × (1 atm/101325 Pa) = 0.00121 atm
1) Name: copper(II) nitrate
2) molar mass:
Determine the mass of every element by multiplying its subscript by the atomic mass and add the masses of all the elements
Element atomic mass subscript total mass of the element
Cu 63.55 1 63.55
N 14.01 2 2*14.01 = 28.02
O 16.00 3*2=6 6*16.00 = 96.00
Molar mass = 63.55 + 28.02 + 96.00 = 187.57 g/mol
Answer:
Kf
Explanation:
The stability constant Kf of a given complex specie is an equilibrium constant that represents the formation of that particular complex specie in solution. It measures the strength of the interaction between the ligands and metal that form the particular complex specie. The magnitude of Kf shows how easily a complex specie is formed in solution.
Hence if I want to dissolve the bromides or chlorides of silver which are ordinarily insoluble in water by means of complex formation, the magnitude of the stability constant for each particular complex specie is important as it gives information regarding the thermodynamic feasibility of the process.