Suppose a 500. mL flask is filled with 2.0 mol of NO3 and 1.7 mol of NO2. The following reaction becomes possible: NO₂(g) + NO(g) → 2NO₂(g) The equilibrium constant K for this reaction is 0.413 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. M X S ?
Suppose a 500. mL flask is filled with 2.0 mol of NO3 and 1.7 mol of NO2. The following reaction becomes possible: NO₂(g) + NO(g) → 2NO₂(g) The equilibrium constant K for this reaction is 0.413 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. M X S ?
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Chapter1: Chemical Foundations
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Answer with two decimal places please
![**Chemical Equilibrium and Reaction Calculations**
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Suppose a 500. mL flask is filled with 2.0 mol of NO₃ and 1.7 mol of NO₂. The following reaction becomes possible:
\[ \text{NO}_3(g) + \text{NO}(g) \rightleftharpoons 2\text{NO}_2(g) \]
The equilibrium constant \( K \) for this reaction is 0.413 at the temperature of the flask.
Calculate the equilibrium molarity of \( \text{NO}_3 \). Round your answer to two decimal places.
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**Explanation:**
**Chemical Equation:**
The given chemical equilibrium involves the reaction between nitrogen trioxide (NO₃) and nitric oxide (NO), producing nitrogen dioxide (NO₂) in a gaseous state.
\[ \text{NO}_3(g) + \text{NO}(g) \rightleftharpoons 2\text{NO}_2(g) \]
**Condition:**
- The initial amount of NO₃: 2.0 mol
- The initial amount of NO₂: 1.7 mol
- Volume of the flask: 500. mL
- Equilibrium constant (\( K \)): 0.413
**Objective:**
Calculate the equilibrium molarity of \( \text{NO}_3 \).
**Note:** Ensure to round off the final answer to two decimal places as required.
To solve this, you can set up the expression for the equilibrium constant \( K \), considering the changes in molarity of the reactants and products from initial to equilibrium states.
**Steps to solve:**
1. Convert moles to molarity for initial conditions.
2. Set up the expression for the equilibrium constant.
3. Solve for the concentration of \( \text{NO}_3 \) at equilibrium.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20b515c9-5e3a-448e-90a2-ac3b562e1523%2F40c4d44b-96c6-40ce-9d6a-db2db025007b%2Fnzs8o48_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium and Reaction Calculations**
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Suppose a 500. mL flask is filled with 2.0 mol of NO₃ and 1.7 mol of NO₂. The following reaction becomes possible:
\[ \text{NO}_3(g) + \text{NO}(g) \rightleftharpoons 2\text{NO}_2(g) \]
The equilibrium constant \( K \) for this reaction is 0.413 at the temperature of the flask.
Calculate the equilibrium molarity of \( \text{NO}_3 \). Round your answer to two decimal places.
---
**Explanation:**
**Chemical Equation:**
The given chemical equilibrium involves the reaction between nitrogen trioxide (NO₃) and nitric oxide (NO), producing nitrogen dioxide (NO₂) in a gaseous state.
\[ \text{NO}_3(g) + \text{NO}(g) \rightleftharpoons 2\text{NO}_2(g) \]
**Condition:**
- The initial amount of NO₃: 2.0 mol
- The initial amount of NO₂: 1.7 mol
- Volume of the flask: 500. mL
- Equilibrium constant (\( K \)): 0.413
**Objective:**
Calculate the equilibrium molarity of \( \text{NO}_3 \).
**Note:** Ensure to round off the final answer to two decimal places as required.
To solve this, you can set up the expression for the equilibrium constant \( K \), considering the changes in molarity of the reactants and products from initial to equilibrium states.
**Steps to solve:**
1. Convert moles to molarity for initial conditions.
2. Set up the expression for the equilibrium constant.
3. Solve for the concentration of \( \text{NO}_3 \) at equilibrium.
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