337.92 moles of Ozone will be produced
1 Oxygen atom is 8 g
1 mole of ozone, O3 = 8 * 3 = 24 g
7.92 * 1024 = 8110.08 g
1 mole = 24 g
? moles = 8110.08 g
? = 337.92 moles
Read more on moles here:
brainly.com/question/15356425
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<u>Answer:</u> For the given equation, only iron has the value of
equal to 0 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^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of
equal to 0 kJ.
Explanation:
SADMEP
-2(bx-5) = 16 distribute
-2bx +10 = 16 subtracte
-10 -10
-2bx = 6
divide by -2x (on both sides)
b = -3x
Answer:
571.81 mL
Explanation:
Assuming constant pressure, we can solve this problem by using <em>Charles' law</em>, which states that at constant pressure:
Where in this case:
We <u>input the data</u>:
- 852 mL * 200 K = V₂ * 298 K
And <u>solve for V₂</u>:
The new volume would be 571.81 mL.