Question

6.45 g of C6H12O6 is burned in a bomb calorimeter containing 950. g of water and the temperature goes from 35.0∘C to 42.3∘C. If the bomb has a heat capacity of 933 J∘C, what is the value for q of the reaction? Your answer should have three significant figures. When reporting your answer in scientific notation format, use the multiplication symbol, ×, not the letter x. Do not include a comma in your answer.

Answer 1

The heat produced by the reaction (q of the reaction) when 6.45 g of glucose is burned in a bomb calorimeter containing 950 g of water and with a 933 J/°C heat capacity of the bomb is -3.58 × 10³ kJ.

Step-by-step explanation:

To determine the value for q of the reaction where 6.45 g of C₆H₁₂O₆ is burned in a bomb calorimeter, we must calculate the heat absorbed by the water and the bomb calorimeter. The formula to use is: qrxn = -(qwater + qbomb). Given that the specific heat capacity of water is 4.184 J/g°C, and we have 950 g of water with a temperature increase from 35.0°C to 42.3°C, and the bomb calorimeter has a heat capacity of 933 J/°C, we can calculate the heat absorbed by water (qwater) and the calorimeter (qbomb), and sum them to determine qrxn.

First, calculate the heat absorbed by the water. qwater = (4.184 J/g°C) × (950 g) × (42.3°C - 35.0°C) which equals to 28959.8 J.

Next, calculate the heat absorbed by the bomb calorimeter. qbomb = (933 J/°C) × (42.3°C - 35.0°C) which equals to 6819 J.

Finally, add up the heats to find qrxn: qrxn = -(qwater + qbomb) = -(28959.8 J + 6819 J) = -35778.8 J, or -35.779 kJ, since we are reporting with three significant figures, we round this to -35.8 kJ, written in scientific notation as -3.58 × 10³ kJ.

Answer 2

The value for q of the reaction is approximately 3.54 times
`10^4 \, \text{J} \)` (rounded to three significant figures and in scientific notation).

To calculate the heat (\( q \)) of the reaction, you can use the formula:

`\[ q = mc\Delta T \]`

where:

q is the heat,

m is the mass,

-c is the specific heat capacity, and

`\( \Delta T \)` is the change in temperature.

Given values:

Mass of water
`(\( m \))`: 950 g

Specific heat capacity of water
`(\( c \))`: 4.18 J/g°C

Change in temperature (
`\( \Delta T \)`:
`\( 42.3°C - 35.0°C = 7.3°C \)`

First, calculate the heat
`(\( q_1 \))` absorbed by the water:

`\[ q_1 = m \cdot c \cdot \Delta T \]`

`\[ q_1 = (950 \, \text{g}) \cdot (4.18 \, \text{J/g°C}) \cdot (7.3°C) \]`

Now, calculate the heat
`(\( q_2 \))` released by the bomb calorimeter:

`\[ q_2 = C \cdot \Delta T \]`

`\[ q_2 = (933 \, \text{J/°C}) \cdot (7.3°C) \]`

The total heat (\( q \)) of the reaction is the sum of
`\( q_1 \)` and
`\( q_2 \)`:

`\[ q = q_1 + q_2 \]`

Now, calculate q using the given values:

`\[ q = (950 \, \text{g}) \cdot (4.18 \, \text{J/g°C}) \cdot (7.3°C) + (933 \, \text{J/°C}) \cdot (7.3°C) \]`

Let's calculate the expression:

`\[ q = (950 \, \text{g}) \cdot (4.18 \, \text{J/g°C}) \cdot (7.3°C) + (933 \, \text{J/°C}) \cdot (7.3°C) \]`

`\[ q = (950 \, \text{g}) \cdot (4.18 \, \text{J/g°C}) \cdot (7.3°C) + (933 \, \text{J/°C}) \cdot (7.3°C) \]`

Now, perform the calculations:

`\[ q \approx 28564.82 \, \text{J} + 6815.9 \, \text{J} \]`

`\[ q \approx 35380.72 \, \text{J} \]`

Now, express the result in scientific notation with three significant figures:

`\[ q \approx 3.54 * 10^4 \, \text{J} \]`