15.0 mL of 4.0 M HCI is added to 85.0 mL of water. What is the final concentration of the acid? [?] mol HCI MHCI Enter

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### Problem Statement:

**15.0 mL of 4.0 M HCl is added to 85.0 mL of water. What is the final concentration of the acid?**

**Calculate the final concentration:**

The final concentration is expressed in mol/L, represented by the following:

[ ? ] mol / L \( \text{HCl} \)


Below the problem statement, there is a blank field labeled "M HCl" where the solution can be entered, followed by an "Enter" button to submit the answer.

### Explanation:

To find the final concentration of the HCl solution, you can use the dilution formula:

\[ C_1V_1 = C_2V_2 \]

where:
- \( C_1 \) is the initial concentration (4.0 M)
- \( V_1 \) is the initial volume (15.0 mL)
- \( C_2 \) is the final concentration
- \( V_2 \) is the final volume

Since the final volume \( V_2 \) is the sum of the initial volume of HCl and the volume of water added:

\[ V_2 = 15.0 \, \text{mL} + 85.0 \, \text{mL} = 100.0 \, \text{mL} \]

Rearranging the dilution formula to solve for \( C_2 \):

\[ C_2 = \frac{C_1 V_1}{V_2} = \frac{(4.0 \, \text{M})(15.0 \, \text{mL})}{100.0 \, \text{mL}} \]

\[ C_2 = \frac{60.0 \, \text{M} \cdot \text{mL}}{100.0 \, \text{mL}} \]

\[ C_2 = 0.6 \, \text{M} \]

So, the final concentration of the acid is 0.6 M HCl.

You can enter the calculated final concentration (0.6 M) in the provided field and click the "Enter" button to submit your answer.
Transcribed Image Text:### Problem Statement: **15.0 mL of 4.0 M HCl is added to 85.0 mL of water. What is the final concentration of the acid?** **Calculate the final concentration:** The final concentration is expressed in mol/L, represented by the following: [ ? ] mol / L \( \text{HCl} \) Below the problem statement, there is a blank field labeled "M HCl" where the solution can be entered, followed by an "Enter" button to submit the answer. ### Explanation: To find the final concentration of the HCl solution, you can use the dilution formula: \[ C_1V_1 = C_2V_2 \] where: - \( C_1 \) is the initial concentration (4.0 M) - \( V_1 \) is the initial volume (15.0 mL) - \( C_2 \) is the final concentration - \( V_2 \) is the final volume Since the final volume \( V_2 \) is the sum of the initial volume of HCl and the volume of water added: \[ V_2 = 15.0 \, \text{mL} + 85.0 \, \text{mL} = 100.0 \, \text{mL} \] Rearranging the dilution formula to solve for \( C_2 \): \[ C_2 = \frac{C_1 V_1}{V_2} = \frac{(4.0 \, \text{M})(15.0 \, \text{mL})}{100.0 \, \text{mL}} \] \[ C_2 = \frac{60.0 \, \text{M} \cdot \text{mL}}{100.0 \, \text{mL}} \] \[ C_2 = 0.6 \, \text{M} \] So, the final concentration of the acid is 0.6 M HCl. You can enter the calculated final concentration (0.6 M) in the provided field and click the "Enter" button to submit your answer.
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