The type of element will change and the mass of the atom will change; A and B
Answer : The enthalpy of combustion per mole of
is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28CO_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28C_6H_%7B12%7DO_6%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_6H_%7B12%7DO_6%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%286%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1260%29%2B%286%5Ctimes%200%29%5D%3D-2815.8kJ%2Fmol)
Therefore, the enthalpy of combustion per mole of
is -2815.8 kJ/mol
Answer:
<h3>The answer is 0.34 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

From the question we have

We have the final answer as
<h3>0.34 g/mL</h3>
Hope this helps you
Answer:
X liters of 10% + (30 - x) liters of 22% = 30...