M Inbox (1,530) - ftantill@udel.edu X ← с app.101edu.co X < STARTING AMOUNT 664 Partly cloudy Mail- Francesca A Tantillo-Out X Homepage CHM150-251 Chen X Question 20 of 24 How many liters of 0.305 M K.PO solution are necessary to completely react with 187 mL 0.0184 M NiCh according to the balanced chemical reaction 2 K.PO.(aq) + 3 NiCl(aq)-Ni-(PO.):(s)+ 6 KCl(aq) 4 ANSWER RESET ADD FACTOR x( ) 2 1000 6.022 x 10 0.0184 0.305 Aktiv Chemistry 7.52 0.0113 0.00752 0.00376 1 0.001 g NiCh mL NiCl L NiCh mol NiCh mL K.PO. HOLD 2 b Answered: Consider the chemical 6 187 3 g K.PO. M NiCl mol K-PO. M K.PO. LK.PO. ✓NNN + 1980 □ X I Submit 10:08 PM 4/11/2022 D
M Inbox (1,530) - ftantill@udel.edu X ← с app.101edu.co X < STARTING AMOUNT 664 Partly cloudy Mail- Francesca A Tantillo-Out X Homepage CHM150-251 Chen X Question 20 of 24 How many liters of 0.305 M K.PO solution are necessary to completely react with 187 mL 0.0184 M NiCh according to the balanced chemical reaction 2 K.PO.(aq) + 3 NiCl(aq)-Ni-(PO.):(s)+ 6 KCl(aq) 4 ANSWER RESET ADD FACTOR x( ) 2 1000 6.022 x 10 0.0184 0.305 Aktiv Chemistry 7.52 0.0113 0.00752 0.00376 1 0.001 g NiCh mL NiCl L NiCh mol NiCh mL K.PO. HOLD 2 b Answered: Consider the chemical 6 187 3 g K.PO. M NiCl mol K-PO. M K.PO. LK.PO. ✓NNN + 1980 □ X I Submit 10:08 PM 4/11/2022 D
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...
Related questions
Question
![### Question 20 of 24
**Problem Statement:**
How many liters of 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction:
\[ 2 \text{K}_3\text{PO}_4(\text{aq}) + 3 \text{NiCl}_2(\text{aq}) \rightarrow \text{Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{KCl}(\text{aq}) \]
**Chemical Equation:**
\[ 2 \text{K}_3\text{PO}_4(\text{aq}) + 3 \text{NiCl}_2(\text{aq}) \rightarrow \text{Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{KCl}(\text{aq}) \]
**Components and Calculation Breakdown:**
- **Starting Amounts:**
- 0.305 M K₃PO₄ solution
- 187 mL (0.187 L) of 0.0184 M NiCl₂ solution
Below the problem statement, a diagram illustrates the process of calculating the necessary amount of K₃PO₄ to react completely with NiCl₂.
**Diagram Details:**
- The left section titled "STARTING AMOUNT" includes two boxes for inputs.
- The next section shows multiplication factors represented in parentheses.
- Below this, a set of conversion factors appears to assist in the calculation. These factors include:
- \( \frac{7.52}{1000} \) and others that appear more detailed in nature.
- Relevant values related to moles and molarity for both K₃PO₄ and NiCl₂, such as 0.00752, 0.0113, 0.00376, etc.
- The conversion options are labeled with units like grams (g NiCl₂), milliliters (mL NiCl₂), liters (L K₃PO₄), mol NiCl₂, and molarity (M K₃PO₄).
The diagram guides the student in applying the conversion factors to find the volume of K₃PO₄ needed.
### Interactive Elements
- **Adapter Selector:** Allows the student to choose](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1027c5e0-f235-434a-9c39-9791c1808f61%2F148e6d0b-0fc2-404d-bd09-d8277d73c44d%2F31stfnf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 20 of 24
**Problem Statement:**
How many liters of 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction:
\[ 2 \text{K}_3\text{PO}_4(\text{aq}) + 3 \text{NiCl}_2(\text{aq}) \rightarrow \text{Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{KCl}(\text{aq}) \]
**Chemical Equation:**
\[ 2 \text{K}_3\text{PO}_4(\text{aq}) + 3 \text{NiCl}_2(\text{aq}) \rightarrow \text{Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{KCl}(\text{aq}) \]
**Components and Calculation Breakdown:**
- **Starting Amounts:**
- 0.305 M K₃PO₄ solution
- 187 mL (0.187 L) of 0.0184 M NiCl₂ solution
Below the problem statement, a diagram illustrates the process of calculating the necessary amount of K₃PO₄ to react completely with NiCl₂.
**Diagram Details:**
- The left section titled "STARTING AMOUNT" includes two boxes for inputs.
- The next section shows multiplication factors represented in parentheses.
- Below this, a set of conversion factors appears to assist in the calculation. These factors include:
- \( \frac{7.52}{1000} \) and others that appear more detailed in nature.
- Relevant values related to moles and molarity for both K₃PO₄ and NiCl₂, such as 0.00752, 0.0113, 0.00376, etc.
- The conversion options are labeled with units like grams (g NiCl₂), milliliters (mL NiCl₂), liters (L K₃PO₄), mol NiCl₂, and molarity (M K₃PO₄).
The diagram guides the student in applying the conversion factors to find the volume of K₃PO₄ needed.
### Interactive Elements
- **Adapter Selector:** Allows the student to choose
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