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Setler79 [48]
3 years ago
13

Type the correct answer in the box.

Chemistry
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

<em>a= In scientific notation</em>

6.96×10⁵ Km

<em>b =In  expanded notation</em>

0.00019 mm

Explanation:

Given data:

Radius of sun = 696000 Km

size of bacterial cell = 1.9 ×10⁻⁴ mm

Radius of sun in scientific notation = ?

Size of bacterial cell in expanded notation = ?

Solution:

Radius of sun:

696000 Km

<em>In scientific notation</em>

6.96×10⁵ Km

Size of bacterial cell:

1.9 ×10⁻⁴ mm

<em>In  expanded notation</em>

1.9/ 10000 = 0.00019 mm

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B. Bleach and sea water should be identified as bases

Explanation:

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In the given question, the pH scale measures for bleach is 8 and for sea water it is 13. So, bleach is basic, not neutral and Sea water is basic too instead of acid. So, Bleach and sea water should be identified as bases.

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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
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<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}

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n = number of electrons transferred = 2

F = Faradays constant = 96500 C

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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

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