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Damm [24]
1 year ago
5

Draw the structure of the product formed when HBr reacts with 2‑methyl‑2‑butene or 2‑methylbut‑2‑ene in the presence of organic

peroxides.
Chemistry
1 answer:
Kruka [31]1 year ago
3 0

the structure of the product formed when HBr reacts with 2‑methyl‑2‑butene in the presence of organic peroxides. Reconstruction debate that had federalism debate that had been an issue since the 1790s.

Reconstruction failed by most other measures: Radical Republican legislation ultimately failed to protect former slaves from white persecution and failed to engender fundamental changes to the social fabric of the South. When President Rutherford B. federalism debate that had been an issue since the 1790s almost mediately . Hayes removed federal troops from the South in 1877, former Confederate officials and slave returned to With the support of a conservative Supreme Court, these newly empowered white southern politicians passed black codes, voter qualifications, and other anti-progressive legislation to reverse the rights that blacks had gained during Radical Reconstruction. The U.S. Supreme Court bolstered this federalism anti-progressive movement federalism  with decisions in the Slaughterhouse Cases, the Civil Rights Cases, and United States v.

Learn more about Reconstruction on:
brainly.com/question/24761999

#SPJ4

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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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