In the laboratory, a student dilutes 17.8 mL of a 9.80 M hydrobromic acid solution to a total volume of 125.0 mL. What is the concentration of the diluted solution? Concentration = 0.78 M

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**Dilution of HydroBromic Acid in the Laboratory**

In the laboratory, a student dilutes **17.8 mL** of a **9.80 M** hydrobromic acid solution to a total volume of **125.0 mL**. What is the concentration of the diluted solution?

**Concentration** = **0.78 M**

This calculation demonstrates the process of dilution, where a concentrated solution is mixed with a solvent to achieve a lower concentration. The final concentration of the diluted hydrobromic acid solution is **0.78 M**.

To find the concentration of the diluted solution, the following dilution formula is used:

\[ C_1 V_1 = C_2 V_2 \]

Where:
\[ C_1 = 9.80 \, M \, (\text{initial concentration}) \]
\[ V_1 = 17.8 \, \text{mL} \, (\text{initial volume}) \]
\[ V_2 = 125.0 \, \text{mL} \, (\text{final volume}) \]
\[ C_2 = ? \, (\text{final concentration}) \]

Solving for \( C_2 \) (the final concentration),

\[ C_2 = \frac{C_1 \cdot V_1}{V_2} = \frac{9.80 \, M \cdot 17.8 \, mL}{125.0 \, mL} \]

\[ C_2 \approx 0.78 \, M \]

Thus, the concentration of the diluted solution is **0.78 M**.
Transcribed Image Text:**Dilution of HydroBromic Acid in the Laboratory** In the laboratory, a student dilutes **17.8 mL** of a **9.80 M** hydrobromic acid solution to a total volume of **125.0 mL**. What is the concentration of the diluted solution? **Concentration** = **0.78 M** This calculation demonstrates the process of dilution, where a concentrated solution is mixed with a solvent to achieve a lower concentration. The final concentration of the diluted hydrobromic acid solution is **0.78 M**. To find the concentration of the diluted solution, the following dilution formula is used: \[ C_1 V_1 = C_2 V_2 \] Where: \[ C_1 = 9.80 \, M \, (\text{initial concentration}) \] \[ V_1 = 17.8 \, \text{mL} \, (\text{initial volume}) \] \[ V_2 = 125.0 \, \text{mL} \, (\text{final volume}) \] \[ C_2 = ? \, (\text{final concentration}) \] Solving for \( C_2 \) (the final concentration), \[ C_2 = \frac{C_1 \cdot V_1}{V_2} = \frac{9.80 \, M \cdot 17.8 \, mL}{125.0 \, mL} \] \[ C_2 \approx 0.78 \, M \] Thus, the concentration of the diluted solution is **0.78 M**.
**Calculating the Standard Enthalpy Change of a Reaction**

**Problem Statement:**
Using standard heats of formation, calculate the standard enthalpy change for the following reaction:

\[2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)\]

**Calculation Equation:**

\[\Delta H^\circ_{\text{rxn}} = \text{(Box) kJ}\]

**Explanation:**
To find the standard enthalpy change of the reaction (\( \Delta H^\circ_{\text{rxn}} \)), we need to use the standard heats of formation (\( \Delta H_f^\circ \)) of the reactants and products involved in the reaction.

The formula to calculate the standard enthalpy change is:

\[ \Delta H^\circ_{\text{rxn}} = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \]

1. Identify the number of moles and the standard enthalpy of formation, \( \Delta H_f^\circ \), for each compound involved.

2. Multiply the number of moles of each compound by their corresponding \( \Delta H_f^\circ \).

3. Sum the \( \Delta H_f^\circ \) values of the products.

4. Sum the \( \Delta H_f^\circ \) values of the reactants.

5. Subtract the sum of the \( \Delta H_f^\circ \) values of the reactants from the sum of the \( \Delta H_f^\circ \) values of the products to obtain \( \Delta H^\circ_{\text{rxn}} \).

This reaction involves:
- **Reactants:**
  - 2 moles of \( C_2H_6(g) \)
  - 7 moles of \( O_2(g) \)
- **Products:**
  - 4 moles of \( CO_2(g) \)
  - 6 moles of \( H_2O(g) \)

Insert the values of the standard heats of formation for each of these compounds into the equation to find the standard enthalpy change. 

The final answer can then be written in the provided box.
Transcribed Image Text:**Calculating the Standard Enthalpy Change of a Reaction** **Problem Statement:** Using standard heats of formation, calculate the standard enthalpy change for the following reaction: \[2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)\] **Calculation Equation:** \[\Delta H^\circ_{\text{rxn}} = \text{(Box) kJ}\] **Explanation:** To find the standard enthalpy change of the reaction (\( \Delta H^\circ_{\text{rxn}} \)), we need to use the standard heats of formation (\( \Delta H_f^\circ \)) of the reactants and products involved in the reaction. The formula to calculate the standard enthalpy change is: \[ \Delta H^\circ_{\text{rxn}} = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \] 1. Identify the number of moles and the standard enthalpy of formation, \( \Delta H_f^\circ \), for each compound involved. 2. Multiply the number of moles of each compound by their corresponding \( \Delta H_f^\circ \). 3. Sum the \( \Delta H_f^\circ \) values of the products. 4. Sum the \( \Delta H_f^\circ \) values of the reactants. 5. Subtract the sum of the \( \Delta H_f^\circ \) values of the reactants from the sum of the \( \Delta H_f^\circ \) values of the products to obtain \( \Delta H^\circ_{\text{rxn}} \). This reaction involves: - **Reactants:** - 2 moles of \( C_2H_6(g) \) - 7 moles of \( O_2(g) \) - **Products:** - 4 moles of \( CO_2(g) \) - 6 moles of \( H_2O(g) \) Insert the values of the standard heats of formation for each of these compounds into the equation to find the standard enthalpy change. The final answer can then be written in the provided box.
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