<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.
Lithium is a good reducing agent because it is electropositive [it rapidly gains electrons]
fluorine is good oxidizing agent electronegative [it loses electrons fastly]
A carbon iota can bond with four other iotas and is just like the four-hole wheel, whereas an oxygen iota, which can bond only to two, is just like the two-hole wheel.
I believe <span>Na2SO3 is the solution to the problem.</span>
Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>0.7457 g is the mass of the helium gas. </u>
