the combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4 °c…

the combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4 °c. δh_comb = -210.7 j/g what is the heat of reaction? q_comb = ? j enter either a + or - sign and the magnitude.

the combustion of 4.00 grams of milk in a bomb calorimeter resulted in a temperature increase of 3.4 °c. δh_comb = -210.7 j/g what is the heat of reaction? q_comb = ? j enter either a + or - sign and the magnitude.

Answer

Explanation:

Step1: Recall the formula for heat of combustion

The heat of combustion ( q_{\text{comb}} ) is given by the product of the mass of the substance (( m )) and the heat of combustion per gram (( \Delta H_{\text{comb}} )). The formula is ( q_{\text{comb}} = m\times\Delta H_{\text{comb}} ).

Step2: Identify the given values

We are given that the mass of milk ( m = 4.00 , \text{g} ) and ( \Delta H_{\text{comb}}=- 210.7 , \text{J/g} ).

Step3: Substitute the values into the formula

Substitute ( m = 4.00 , \text{g} ) and ( \Delta H_{\text{comb}}=-210.7 , \text{J/g} ) into the formula ( q_{\text{comb}}=m\times\Delta H_{\text{comb}} ).

( q_{\text{comb}}=4.00,\text{g}\times(- 210.7,\text{J/g}) )

Step4: Calculate the result

Multiply ( 4.00 ) by ( - 210.7 ):

( 4.00\times(-210.7)=- 842.8 )

Answer:

(-842.8)