Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
#learnwithBrainly
Answer:
Subtract The <u>mean</u> from each Numbers Given
Hope it helpz!
This question is describing the following chemical reaction at equilibrium:

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

Thus, by recalling the Van't Hoff's equation, we can write:

Hence, we solve for the enthalpy change as follows:

Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
Learn more:
The answer for the question is 79
Answer:
![\boxed{\sf D. \ [Ar] \ 3d5 \ 4s2}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20D.%20%5C%20%20%5BAr%5D%20%5C%203d5%20%5C%204s2%7D)
Explanation:
Manganese is a group 7 transition metal with an electronic configuration of [Ar] 3d5 4s2.