Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
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_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
Answer:
Ka = 1.39x10⁻⁶
Explanation:
A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
<em>Where Ka is:</em>
Ka = [H⁺] [X⁻] / [HX]
<em>Where [] is the molar concentration in equilibrium of each specie.
</em>
The equilibrium is reached when some HX reacts producing H+ and X-, that is:
[HX] = 1.64M - X
[H⁺] = X
[X⁻] = X
As pH is 2.82 = -log [H⁺]:
[H⁺] = 1.51x10⁻³M:
[HX] = 1.64M - 1.51x10⁻³M = 1.638M
[H⁺] = 1.51x10⁻³M
[X⁻] = 1.51x10⁻³M
And Ka is:
Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]
<h3>Ka = 1.39x10⁻⁶</h3>
Answer:

Explanation:
The formula for efficiency is

Data:
Useful energy = 3 J
Energy input = 30 J
Calculation:

<u>a) Answer: </u>
<em>Number of molecules in 1 mole</em>
<u>Explanation:</u>
a) Whether we take any of the substance among all three of the given substances they will have the same number of molecules in 1 mole of the substance is considered and the value for this will be 
<u>b) Answer: </u>
<em>In the given question </em><em>mass of the substance</em><em> which is </em><em>greatest</em><em> is asked for </em><em>one mole</em><em> and we also know that </em><em>mass of one mole is given by molar mass. </em>
<u>Explanation:</u>
b) It is known that
is the molar mass for oxygen which is greater than that of hydrogen while fluorine has a molar mass of
which on comparison shows that, it is the highest amongst all three.