### Problem Statement A chemist adds 170.0 mL of a 5.72 M silver nitrate (AgNO₃) solution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Round your answer to 3 significant digits. ### Input Field - **kg**: [Empty text box for the student to input their answer] ### Control Buttons - Clear Input (icon with 'x') - Reset (icon with a circular arrow) - Help (icon with a question mark) ### Additional Information This problem requires students to use the concept of molarity and molar mass to determine the total mass of a solute (silver nitrate) in a specified volume of solution. Here are the necessary steps: 1. **Calculate the number of moles of AgNO₃**: \[ \text{Moles of AgNO}_3 = Molarity \times \text{Volume in Liters} \] 2. **Convert the volume from mL to L**: \[ 170.0 \text{ mL} = 0.1700 \text{ L} \] 3. **Use the given molarity to find the moles**: \[ \text{Moles of AgNO}_3 = 5.72 \text{ M} \times 0.1700 \text{ L} \] 4. **Calculate the mass of AgNO₃ using its molar mass**: \[ \text{Mass} = \text{Moles} \times \text{Molar Mass of AgNO}_3 \] (Molar mass of AgNO₃ ≈ 169.87 g/mol) 5. **Convert the mass from grams to kilograms**: \[ 1 \text{ g} = 0.001 \text{ kg} \] ### Example Calculation 1. Moles of AgNO₃: \[ 5.72 \text{ M} \times 0.1700 \text{ L} = 0.9724 \text{ moles} \] 2. Mass of AgNO₃: \[ 0.9724 \text{ moles} \times 169.87 \text{ g/mol} \approx 165
### Problem Statement A chemist adds 170.0 mL of a 5.72 M silver nitrate (AgNO₃) solution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Round your answer to 3 significant digits. ### Input Field - **kg**: [Empty text box for the student to input their answer] ### Control Buttons - Clear Input (icon with 'x') - Reset (icon with a circular arrow) - Help (icon with a question mark) ### Additional Information This problem requires students to use the concept of molarity and molar mass to determine the total mass of a solute (silver nitrate) in a specified volume of solution. Here are the necessary steps: 1. **Calculate the number of moles of AgNO₃**: \[ \text{Moles of AgNO}_3 = Molarity \times \text{Volume in Liters} \] 2. **Convert the volume from mL to L**: \[ 170.0 \text{ mL} = 0.1700 \text{ L} \] 3. **Use the given molarity to find the moles**: \[ \text{Moles of AgNO}_3 = 5.72 \text{ M} \times 0.1700 \text{ L} \] 4. **Calculate the mass of AgNO₃ using its molar mass**: \[ \text{Mass} = \text{Moles} \times \text{Molar Mass of AgNO}_3 \] (Molar mass of AgNO₃ ≈ 169.87 g/mol) 5. **Convert the mass from grams to kilograms**: \[ 1 \text{ g} = 0.001 \text{ kg} \] ### Example Calculation 1. Moles of AgNO₃: \[ 5.72 \text{ M} \times 0.1700 \text{ L} = 0.9724 \text{ moles} \] 2. Mass of AgNO₃: \[ 0.9724 \text{ moles} \times 169.87 \text{ g/mol} \approx 165
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![### Problem Statement
A chemist adds 170.0 mL of a 5.72 M silver nitrate (AgNO₃) solution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Round your answer to 3 significant digits.
### Input Field
- **kg**: [Empty text box for the student to input their answer]
### Control Buttons
- Clear Input (icon with 'x')
- Reset (icon with a circular arrow)
- Help (icon with a question mark)
### Additional Information
This problem requires students to use the concept of molarity and molar mass to determine the total mass of a solute (silver nitrate) in a specified volume of solution. Here are the necessary steps:
1. **Calculate the number of moles of AgNO₃**:
\[
\text{Moles of AgNO}_3 = Molarity \times \text{Volume in Liters}
\]
2. **Convert the volume from mL to L**:
\[
170.0 \text{ mL} = 0.1700 \text{ L}
\]
3. **Use the given molarity to find the moles**:
\[
\text{Moles of AgNO}_3 = 5.72 \text{ M} \times 0.1700 \text{ L}
\]
4. **Calculate the mass of AgNO₃ using its molar mass**:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass of AgNO}_3
\]
(Molar mass of AgNO₃ ≈ 169.87 g/mol)
5. **Convert the mass from grams to kilograms**:
\[
1 \text{ g} = 0.001 \text{ kg}
\]
### Example Calculation
1. Moles of AgNO₃:
\[
5.72 \text{ M} \times 0.1700 \text{ L} = 0.9724 \text{ moles}
\]
2. Mass of AgNO₃:
\[
0.9724 \text{ moles} \times 169.87 \text{ g/mol} \approx 165](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F41322bcb-03be-44b9-99fe-cebaa9592c9e%2Fc3187e2b-8517-4f9c-aeee-1919b8b7a0bb%2Fpwpwexp_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A chemist adds 170.0 mL of a 5.72 M silver nitrate (AgNO₃) solution to a reaction flask. Calculate the mass in kilograms of silver nitrate the chemist has added to the flask. Round your answer to 3 significant digits.
### Input Field
- **kg**: [Empty text box for the student to input their answer]
### Control Buttons
- Clear Input (icon with 'x')
- Reset (icon with a circular arrow)
- Help (icon with a question mark)
### Additional Information
This problem requires students to use the concept of molarity and molar mass to determine the total mass of a solute (silver nitrate) in a specified volume of solution. Here are the necessary steps:
1. **Calculate the number of moles of AgNO₃**:
\[
\text{Moles of AgNO}_3 = Molarity \times \text{Volume in Liters}
\]
2. **Convert the volume from mL to L**:
\[
170.0 \text{ mL} = 0.1700 \text{ L}
\]
3. **Use the given molarity to find the moles**:
\[
\text{Moles of AgNO}_3 = 5.72 \text{ M} \times 0.1700 \text{ L}
\]
4. **Calculate the mass of AgNO₃ using its molar mass**:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass of AgNO}_3
\]
(Molar mass of AgNO₃ ≈ 169.87 g/mol)
5. **Convert the mass from grams to kilograms**:
\[
1 \text{ g} = 0.001 \text{ kg}
\]
### Example Calculation
1. Moles of AgNO₃:
\[
5.72 \text{ M} \times 0.1700 \text{ L} = 0.9724 \text{ moles}
\]
2. Mass of AgNO₃:
\[
0.9724 \text{ moles} \times 169.87 \text{ g/mol} \approx 165
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