At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To find the mass of [tex]$5.30 \times 10^{22}$[/tex] formula units of barium nitrate [tex]$\left( Ba \left( NO_3\right)_2 \right)$[/tex], we can follow a series of steps systematically. Here's the detailed solution:
1. Given Information:
- Molar mass of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex]: [tex]\( 261.35 \, \text{g/mol} \)[/tex]
- Number of formula units: [tex]\( 5.30 \times 10^{22} \)[/tex]
2. Avogadro's Number:
- Avogadro's number defines the number of formula units in one mole of a substance, which is [tex]\( 6.022 \times 10^{23} \)[/tex] formula units/mol.
3. Calculate the Moles:
- The number of moles of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex] can be found using the formula:
[tex]\[ \text{Moles} = \frac{\text{Number of Formula Units}}{\text{Avogadro's Number}} \][/tex]
- Plugging in the values:
[tex]\[ \text{Moles} = \frac{5.30 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.08801062769843906 \, \text{moles} \][/tex]
4. Calculate the Mass:
- The mass can then be found using the formula:
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \][/tex]
- Using the given molar mass:
[tex]\[ \text{Mass} \approx 0.08801062769843906 \, \text{moles} \times 261.35 \, \text{g/mol} \approx 23.00157754898705 \, \text{g} \][/tex]
5. Select the Closest Answer:
- Among the options given:
\begin{itemize}
\item [tex]$0.0900 \, \text{g}$[/tex]
\item [tex]$12.0 \, \text{g}$[/tex]
\item [tex]$23.0 \, \text{g}$[/tex]
\item [tex]$3,130 \, \text{g}$[/tex]
\end{itemize}
- The calculated mass rounds to [tex]$23.0 \, \text{g}$[/tex], which matches the given option.
Therefore, the mass of [tex]$5.30 \times 10^{22}$[/tex] formula units of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex] is approximately [tex]\(23.0 \, \text{g}\)[/tex].
The correct answer is:
[tex]\[ \boxed{23.0 \, \text{g}} \][/tex]
1. Given Information:
- Molar mass of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex]: [tex]\( 261.35 \, \text{g/mol} \)[/tex]
- Number of formula units: [tex]\( 5.30 \times 10^{22} \)[/tex]
2. Avogadro's Number:
- Avogadro's number defines the number of formula units in one mole of a substance, which is [tex]\( 6.022 \times 10^{23} \)[/tex] formula units/mol.
3. Calculate the Moles:
- The number of moles of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex] can be found using the formula:
[tex]\[ \text{Moles} = \frac{\text{Number of Formula Units}}{\text{Avogadro's Number}} \][/tex]
- Plugging in the values:
[tex]\[ \text{Moles} = \frac{5.30 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.08801062769843906 \, \text{moles} \][/tex]
4. Calculate the Mass:
- The mass can then be found using the formula:
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \][/tex]
- Using the given molar mass:
[tex]\[ \text{Mass} \approx 0.08801062769843906 \, \text{moles} \times 261.35 \, \text{g/mol} \approx 23.00157754898705 \, \text{g} \][/tex]
5. Select the Closest Answer:
- Among the options given:
\begin{itemize}
\item [tex]$0.0900 \, \text{g}$[/tex]
\item [tex]$12.0 \, \text{g}$[/tex]
\item [tex]$23.0 \, \text{g}$[/tex]
\item [tex]$3,130 \, \text{g}$[/tex]
\end{itemize}
- The calculated mass rounds to [tex]$23.0 \, \text{g}$[/tex], which matches the given option.
Therefore, the mass of [tex]$5.30 \times 10^{22}$[/tex] formula units of [tex]$\left( Ba \left( NO_3 \right)_2 \right)$[/tex] is approximately [tex]\(23.0 \, \text{g}\)[/tex].
The correct answer is:
[tex]\[ \boxed{23.0 \, \text{g}} \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
