How many moles of Agl will be formed when 75.0 mL of 0.300 M AgNO3 is completely reacted according to the balanced chemical reaction: 2 AgNO3(aq) + Cal₂(aq) → 2 Agl(s) + Ca(NO3)2(aq)
How many moles of Agl will be formed when 75.0 mL of 0.300 M AgNO3 is completely reacted according to the balanced chemical reaction: 2 AgNO3(aq) + Cal₂(aq) → 2 Agl(s) + Ca(NO3)2(aq)
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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can you help me and line it exactly like this because i keep doing it and its says its wrong
![**Titration Practice Problem: Calculating Moles of Precipitate**
**Question 2 of 4**
**Problem Statement:**
Calculate the number of moles of AgI that will be formed when 75.0 mL of 0.300 M AgNO₃ is completely reacted according to the balanced chemical equation:
\[ 2 \text{AgNO}_3(\text{aq}) + \text{CaI}_2(\text{aq}) \rightarrow 2 \text{AgI}(\text{s}) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
**Calculation Steps:**
1. **Starting Amount:**
- Input fields are provided for the starting amounts in terms of volume and molarity.
2. **Reaction Setup:**
- The equation involves a 1:1 mole ratio for AgNO₃ to AgI according to the balanced equation.
3. **Input Factors:**
- Different values are available to use for calculation, such as:
- Volume: 75.0 mL
- Concentration: 0.300 M
- Molar mass values
- Avogadro’s number
4. **Answer:**
- Enter your calculated moles of AgI in the answer box once all necessary calculations are done.
**Interactive Components:**
- **Add Factor:** Enables you to select factors for the calculation.
- **Reset:** Clears the input fields to start over.
**Note:**
Tap for additional resources if extra help is needed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F574bf9f3-609e-4870-b103-4ad8e97b08b1%2Fecb21628-667f-4657-9673-1df6256d2d98%2Foxgr25v_processed.png&w=3840&q=75)
Transcribed Image Text:**Titration Practice Problem: Calculating Moles of Precipitate**
**Question 2 of 4**
**Problem Statement:**
Calculate the number of moles of AgI that will be formed when 75.0 mL of 0.300 M AgNO₃ is completely reacted according to the balanced chemical equation:
\[ 2 \text{AgNO}_3(\text{aq}) + \text{CaI}_2(\text{aq}) \rightarrow 2 \text{AgI}(\text{s}) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
**Calculation Steps:**
1. **Starting Amount:**
- Input fields are provided for the starting amounts in terms of volume and molarity.
2. **Reaction Setup:**
- The equation involves a 1:1 mole ratio for AgNO₃ to AgI according to the balanced equation.
3. **Input Factors:**
- Different values are available to use for calculation, such as:
- Volume: 75.0 mL
- Concentration: 0.300 M
- Molar mass values
- Avogadro’s number
4. **Answer:**
- Enter your calculated moles of AgI in the answer box once all necessary calculations are done.
**Interactive Components:**
- **Add Factor:** Enables you to select factors for the calculation.
- **Reset:** Clears the input fields to start over.
**Note:**
Tap for additional resources if extra help is needed.
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