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3 years ago

9

How does the potential energy of the reactants compare to the potential energy of the products in an exothermic reaction, and wh

ere does the energy that is given off come from?

1 answer:

Ksju [112]3 years ago

4 0

In an endothermic reaction the products have less potential energy than the reactants. I'm not sure about the second question though.

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At which part of a river would you have the best luck growing food?

krek1111 [17]

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B. Delta

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Why do solids maintain their shape, whereas fluids do not?

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Write a balanced half-reaction for the reduction of permanganate ion mno−4 to solid manganese dioxide mno2 in acidic aqueous sol

Svetlanka [38]

The half-reaction includes either the reduction or the oxidation reaction of the redox reactions. In acidic solution permanganate ion will react with hydrogen ion to yield manganese ion and water.

<h3>What are Redox reactions?</h3>

Redox or oxidation-reduction reactions are the chemical reactions in which the oxidation and the reduction of the chemical species occur simultaneously.

Permanganate (VII) ion is a strong oxidizing agent and gets easily reduced to manganese ion in presence of the hydrogen ion in an acidic solution.

The balanced half-reaction for reduction is shown as,

$\rm MnO_{4}^{-} \; (aq)+ 8H^{+} \; (aq)+ 5e^{-} \rightarrow Mn^{2+} \; (aq)+ 4H_{2}O \; (l)$

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2 years ago

Giving brainly if correct

mel-nik [20]

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The corect answer would be C.

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The flow rate set at a differentt time would be the correct measurement beecause wats and speed add up to your main answer.

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2 years ago

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Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

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3 years ago