To determine the approximate mass of one formula unit of silver nitrate (AgNO₃), follow these steps:
1. Identify the atomic masses of the constituent elements:
- Silver (Ag): 107.8682 g/mol
- Nitrogen (N): 14.0067 g/mol
- Oxygen (O): 15.999 g/mol
2. Calculate the molar mass of AgNO₃:
The formula unit of silver nitrate consists of one atom of silver, one atom of nitrogen, and three atoms of oxygen.
[tex]\[
\text{Molar mass of } AgNO_3 = \text{Atomic mass of Ag} + \text{Atomic mass of N} + 3 \times \text{Atomic mass of O}
\][/tex]
Substituting the atomic masses:
[tex]\[
\text{Molar mass of } AgNO_3 = 107.8682 \, \text{g/mol} + 14.0067 \, \text{g/mol} + 3 \times 15.999 \, \text{g/mol}
\][/tex]
Performing the multiplication and addition:
[tex]\[
\text{Molar mass of } AgNO_3 = 107.8682 + 14.0067 + 47.997
\][/tex]
[tex]\[
\text{Molar mass of } AgNO_3 = 169.8719 \, \text{g/mol}
\][/tex]
3. Express the molar mass in scientific notation:
To express the value [tex]\(169.8719 \, \text{g/mol}\)[/tex] in scientific notation with the correct significant figures:
The value is approximately [tex]\(1.6975 \times 10^2 \, \text{g/mol}\)[/tex].
4. Conclusion:
The mass of one formula unit of silver nitrate ([tex]\(AgNO_3\)[/tex]) is approximately [tex]\(1.6975 \times 10^2 \, \text{grams} \, \text{AgNO_3}\)[/tex].
So, the coefficient is [tex]\(1.6975\)[/tex] and the exponent is [tex]\(2\)[/tex]. Thus, the answer is:
[tex]\[
1 \, \text{f.un.} \, AgNO_3 = 1.6975 \times 10^2 \, \text{grams} \, AgNO_3
\][/tex]
Hence, you should enter 1.6975 in the green blank and 2 in the yellow blank.
