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The combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of [tex]3.4^{\circ} C[/tex].

[tex]\Delta H_{\text{comb}} = -210.7 \text{ J/g}[/tex]

What is the heat of reaction?

[tex]q_{\text{comb}} = [?] \text{ J}[/tex]

Enter either a '+' or '-' sign AND the magnitude. Do not round until the end. Use significant figures.


Sagot :

Certainly! Let's solve this problem step-by-step.

1. Understand the given values:
- The mass of milk, [tex]\( \text{mass} = 4.00 \)[/tex] grams.
- The enthalpy of combustion, [tex]\( \Delta H_{\text{comb}} = -210.7 \)[/tex] Joules per gram.

2. Recall the formula for calculating the heat of combustion ([tex]\( q_{\text{comb}} \)[/tex]):
[tex]\[ q_{\text{comb}} = \text{mass} \times \Delta H_{\text{comb}} \][/tex]

3. Substitute the given values into the formula:
[tex]\[ q_{\text{comb}} = 4.00 \, \text{grams} \times (-210.7 \, \text{Joules/gram}) \][/tex]

4. Perform the multiplication to find the heat of reaction:
[tex]\[ q_{\text{comb}} = 4.00 \times -210.7 = -842.8 \, \text{Joules} \][/tex]

So, the heat of reaction, [tex]\( q_{\text{comb}} \)[/tex], is:
[tex]\[ q_{\text{comb}} = -842.8 \, \text{J} \][/tex]

This value indicates that 842.8 Joules of energy are released during the combustion of 4.00 grams of milk, hence the negative sign.