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nikklg [1K]
3 years ago
8

Please help quick the 6th question is “What is a catalyst?” (include all three parts)

Chemistry
1 answer:
allochka39001 [22]3 years ago
4 0
1. The reaction is exothermic.

2. I know this because the delta H is a negative number. (-802.4 kJ)

3. This means that this reaction releases heat.
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ILL PAYPAL YOU 10$ FOR THIS
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Arrange carbon, phosphorus and sulfur in order of reactivity to write their reaction with oxygen.
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Solid phosphorus reacts with gaseous oxygen to produce solid diphosphorus pentaoxide. ... Methanol burns in oxygen to produce carbon dioxide gas and water vapor.

Explanation:

Solid phosphorus reacts with gaseous oxygen to produce solid diphosphorus pentaoxide. ... Methanol burns in oxygen to produce carbon dioxide gas and water vapor.

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Explain why mixing of red paint with white paint does not constitute a chemical reaction even though the product has a different
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Explanation:

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If a horse is moving and then come to a stop is the horse accelerating or decelerating
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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
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