<u>Answer:</u> The enthalpy of the reaction for the production of 
is coming out to be -74.9 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_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%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_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of is coming out to be -74.9 kJ
Answer:
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Explanation:
The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.
The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.
The net ionic equation for this reaction is;
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Answer:
the health codes would not cover the food service complaints
Remember that density refers to the "mass per unit volume" of an object.
So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.
In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is
As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:
density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml
The effect on density would be that it would erroneously appear to be greater
Hope this helps!
Good luck
