<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
<u>Explanation:</u>
Enthalpy change 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_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_%7B10%7D%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_6%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_%7B%28H_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-2877.6%29%29%5D-%5B%281%5Ctimes%20%28-2540.2%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D234.2J)
Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
<span>0.967 x 10^-6 HZ
This should be correct (:</span>
Answer:
0.675 atm
513 Torr
Explanation:
Given is that, the atmospheric pressure on the surface of Venus is
6.84 X 10⁴ Pa.
1 atm (atmospheric pressure) is equal to 101325 pascal (Pa).
To convert divide the pressure value by 101325.
Pressure in atm = 
= 0.675055 atm
Rounding it off to 3 significant digits: 0.675 atm
Now, one Torr is 133.322 Pa. For conversion, divide the pressure value by 133.322.
Pressure in Torr = 
=513.04219 Torr
Rounding it off to 3 significant digits: 513 Torr
Answer:A.) molecule of atoms forms
Explanation:
Answer:

Explanation:
Hello,
In this case, the described chemical reaction is:

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

Best regards.