Solve for $ x $.

[tex]\[ 3x = 6x - 2 \][/tex]

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$
2 K +2 H _2 O \rightarrow 2 KOH + H _2
$
[tex]$1.00 g$[/tex] of solid [tex]$K$[/tex] is added to [tex]$500 . mL$[/tex] of water initially at [tex]$21.3^{\circ} C$[/tex]. The temperature at the end of the reaction was [tex]$23.5^{\circ} C$[/tex].
$
c _{\text {soln }}=4.18 J / g ^{\circ} C \quad d _{\text {soln }}=1.00 g / mL
$

What is the heat of reaction, [tex]$q_{r \times n}$[/tex] ?
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Response:
Given the reaction:

[tex]\[ 2K + 2H_2O \rightarrow 2KOH + H_2 \][/tex]

[tex]\[ 1.00 \, \text{g} \][/tex] of solid $ K $ is added to $ 500 \, \text{mL} $ of water initially at $ 21.3^{\circ} \text{C} $. The temperature at the end of the reaction was $ 23.5^{\circ} \text{C} $.

[tex]\[ c_{\text{soln}} = 4.18 \, \text{J/g} ^{\circ} \text{C} \quad d_{\text{soln}} = 1.00 \, \text{g/mL} \][/tex]

What is the heat of reaction, [tex]$ q_{\text{rxn}} $[/tex]?

Let's work through this problem step by step.

### Given Data:
1. Mass of solid $ K $ = $ 1.00 $ gram
2. Volume of water = $ 500 $ mL
3. Initial temperature of the water = $ 21.3^\circ C $
4. Final temperature of the solution after reaction = $ 23.5^\circ C $
5. Specific heat capacity of the solution, $ c_{\text{soln}} $ = $ 4.18 \text{ J/g}^\circ \text{C} $
6. Density of the solution, $ d_{\text{soln}} $ = $ 1.00 \text{ g/mL} $

### Steps for Calculation:

#### Step 1: Calculate the mass of the solution
The mass of the solution can be calculated using the volume of water and the density of the solution.

Since the density $ d_{\text{soln}} $ is $ 1.00 \text{ g/mL} $, the mass of the solution is:
[tex]\[ \text{mass}_{\text{solution}} = \text{volume}_{\text{water}} \times \text{density}_{\text{soln}} = 500.0 \text{ mL} \times 1.0 \text{ g/mL} = 500.0 \text{ grams} \][/tex]

#### Step 2: Calculate the change in temperature ($\Delta T$)
The change in temperature ($\Delta T$) is given by the final temperature minus the initial temperature.

[tex]\[ \Delta T = 23.5^\circ C - 21.3^\circ C = 2.2^\circ C \][/tex]

#### Step 3: Calculate the heat absorbed by the solution ($q$)
The heat absorbed by the solution ($q$) can be calculated using the formula:
[tex]\[ q = m \times c_{\text{soln}} \times \Delta T \][/tex]
where:
- $ m $ is the mass of the solution
- $ c_{\text{soln}} $ is the specific heat capacity of the solution
- $ \Delta T $ is the change in temperature

Substitute the known values into the formula:
[tex]\[ q = 500.0 \text{ grams} \times 4.18 \text{ J/g}^\circ \text{C} \times 2.2^\circ C \][/tex]

#### Step 4: Calculate the numerical value of heat ($q$)
[tex]\[ q = 500.0 \times 4.18 \times 2.2 \approx 4598 \text{ J} \][/tex]

### Conclusion
The heat of reaction, [tex]$ q_{r \times n} $[/tex], associated with the dissolution and reaction of [tex]$ 1.00 $[/tex] gram of [tex]$ K $[/tex] in [tex]$ 500 $[/tex] mL of water, resulting in the temperature change from [tex]$ 21.3^\circ C $[/tex] to [tex]$ 23.5^\circ C $[/tex], is approximately [tex]$ 4598 \text{ J} $[/tex].