Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:
You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
![Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_%7B3%7D%20%5D%20%2A%5BCl_%7B2%7D%20%5D%20%7D%7B%5BPCl_%7B5%7D%20%5D%7D)
The concentrations are:
- [PCl₃]=
- [Cl₂]=
- [PCl₅]=
Replacing:
Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
weight depends on the force of gravity that is exerted on it.
During Photosynthesis , Plants Transforms Light Energy Into Chemical Energy .
Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
Answer:Chemistry problems can be solved using a variety of techniques.
Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!
