Mass × velocity = momentum
74 x 15 = momentum
The balanced chemical reaction is:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of C3H8 to be used for the reaction. This will be the starting point for the calculations.
73.7 g C3H8 (1 mol C3H8 / 44.11 g C3H8) ( 5 mol O2 / 1 mol C3H8 ) ( 32 g O2 / 1 mol O2 ) = 267.3 g O2
For this question, I think it is the other way around. It is true that chloroacetic acid is stronger in strength than acetic acid. Acid strength is measured as the equilibrium constant of the reaction <span>HA -----> H+ + A-
</span><span> In acetic acid, the anion produced by dissociation is CH3-COO-; in chloroacetic acid it is CH2Cl-COO-. Comparing the two, in the first one the negative charge is taken up mostly by the two oxygen atoms. In the second there is also an electronegative chlorine atom nearby to draw more charge towards itself. Therefore, the charge is less concentrated in the chloroacetate ion than it is in the acetate ion, and, accordingly, chloroacetic acid is stronger than acetic acid. </span>
Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_3H_8%7D%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=-2220.1%3D%5B%283%5Ctimes%20-393.5%29%2B%284%5Ctimes%20-241.8%29%5D-%5B%285%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
Therefore, the heat of formation of propane is 72.4 kJ/mol
