The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
I would say the answer is C) because the thermal energy of the coffee going into the cup has to level out until the cup gets warmer and the coffee gets cooler, and they reach the same temperature.Meaning that the jug which has already been at the same temperature(we can assume based on the verbiage “the coffee was poured from the jug to the cup”) which would give the jug more thermal energy.
Answer: The value of 
for the reaction is, -2512.4 kJ
Explanation:
The chemical equation for the combustion of acetylene follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(4\times (-393.5))+(2\times (-241.8))]-[(2\times (227.4)+(5\times (0))]\\\\\Delta H^o_{rxn}=-2512.4kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%284%5Ctimes%20%28-393.5%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28227.4%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-2512.4kJ)
Therefore, the value of for the reaction is, -2512.4 kJ
Answer:
The dimerization of butadiene to 4-vinylcyclohexene folows second order kinetics and its rate law will be given by :
![R=k[C_4H_6]^2](https://tex.z-dn.net/?f=R%3Dk%5BC_4H_6%5D%5E2)
Explanation:
The rate of the reaction ;
![R=k[C_4H_6]^x](https://tex.z-dn.net/?f=R%3Dk%5BC_4H_6%5D%5Ex)
As given in the question , that graph of time verses ![\frac{1}{[C_4H_6]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BC_4H_6%5D%7D)
was linear but plots of ![[C_4H_6]](https://tex.z-dn.net/?f=%5BC_4H_6%5D)
or ![\ln[C_4H_6]](https://tex.z-dn.net/?f=%5Cln%5BC_4H_6%5D)
was curved.
Generally:
Graph of time verses ![[concentration]](https://tex.z-dn.net/?f=%5Bconcentration%5D)
for zero order reaction is linear with negative slope.
Graph of time verses ![\ln [concentration]](https://tex.z-dn.net/?f=%5Cln%20%5Bconcentration%5D)
for secon order reaction is linear with negative slope.
Graph of time verses ![\frac{1}{[concentration]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Bconcentration%5D%7D)
for secon order reaction is linear with positive slope.
So, the dimerization of butadiene to 4-vinylcyclohexene folows second order kinetics and its rate law will be given by :
