madey21
Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Solve for \( x \).

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



Format the following question or task so that it is easier to read.
Fix any grammar or spelling errors.
Remove phrases that are not part of the question.
Do not remove or change LaTeX formatting.
Do not change or remove [tex] [/tex] tags.
If the question is nonsense, rewrite it so that it makes sense.
-----
$
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] ?
-----

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]?

Sagot :

GinnyAnswer
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].
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.