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3 years ago

15

A __________ ALSO IS MADE UP OF TWO OR MORE ATOMS. BUT IN THIS CASE IT CAN BE THE SAME AS WELL AS DIFFERENT

1 answer:

alina1380 [7]3 years ago

3 0

Answer:

D

Explanation:

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Two new planets with equal masses are discovered. Planet A is twice as far from the sun as Planet B. Which of the following is t

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3 years ago

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Which of the following is not a characteristic of a metal?

jolli1 [7]

Answer:

B. Poor conductor.

Explanation:

It cannot be A, as only 1 metal is not solid at room temp.

It cannot be C, as most metals are ductile.

It cannot be D, as most metals are malleable.

This leaves B, which is not true about metals, as a lot are very good conductors.

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2 years ago

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Answer:

1852 kJ/mol

Explanation:

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3 years ago

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Dmitry_Shevchenko [17]

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3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

<u>For a:</u>

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

<u>Oxidation half reaction:</u> $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

<u>Reduction half reaction:</u> $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u>

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

<u>Oxidation half reaction:</u> $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

<u>Reduction half reaction:</u> $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago