1. Chemical
2. Physical
1. Exothermic
2. Endothermic
1. Chemical
2. Physical
I need a picture because I don’t know what this is talking about
Not always ammonium salts of weak acids form neutral solutions.
When formic acid reacts with ammonia, ammonium formate is produced:
HCO2H + NH3 ----> NH4HCO2
You already know that the weak conjugate bases of NH3 and HCO2H are NH4+ and HCO2, respectively.
How can the pH of the solution be calculated if the salt's anion causes the pH to rise and the salt's cation causes it to fall? The relative intensities of the basic anion and the acidic cation hold the key to the solution.
As was already established, formate is a weak base and will create hydroxide ions in water, whereas ammonium is a weak acid and will make hydronium ions in water.
NH4⁺ + H2O -----> NH3 + H3O⁺
HCO2⁻ + H2O -----> HCO2H + OH⁻
Since the acid ionization of NH4+ is more favored than the base ionization of HCO2-, the solution will be acidic.
To learn more about ammonium salts:
brainly.com/question/10874844
#SPJ4
In the question, the number of atoms per unit cell is required for:
A) Polonium (Po)
In polonium, the structure is simple cubic, meaning there are 8 corner atoms, which add up to one atom per unit cell.
B) Manganese (Mn)
The structure of the Mn can be considered to be a body centered cubic (BCC) and the number of atoms for this is 8 corner atoms and 1 central atoms, making a total of 2 atoms per unit cell.
C) Silver (Ag)
Silver has a face centered cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
<u>Answer:</u> The percentage abundance of 
and 
isotopes are 77.5% and 22.5% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of isotope be 'x'. So, fractional abundance of isotope will be '1 - x'
- <u>For isotope:</u>
Mass of isotope = 35 amu
Fractional abundance of isotope = x
- <u>For isotope:</u>
Mass of isotope = 37 amu
Fractional abundance of isotope = 1 - x
Average atomic mass of chlorine = 35.45 amu
Putting values in equation 1, we get:
![35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775](https://tex.z-dn.net/?f=35.45%3D%5B%2835%5Ctimes%20x%29%2B%2837%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.775)
Percentage abundance of isotope = 
Percentage abundance of isotope = 
Hence, the percentage abundance of and isotopes are 77.5% and 22.5% respectively.
