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Aleksandr [31]
3 years ago
13

Which reactants would lead to a spontaneous reaction?

Chemistry
1 answer:
balandron [24]3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

The elements which have excess or deficiency of electrons will react readily.

Atomic number of Mn is 25 and electronic configuration of Mn^{2+} is [Ar]4s^{0}3d^{5}. This configuration is stable.

Atomic number of Cr is 24 and electronic configuration of Cr is [Ar]4s^{1}3d^{5}. This configuration is not stable.

Atomic number of Fe is 26 and electronic configuration of Fe is [Ar]4s^{2}3d^{6}. This configuration is stable.

Atomic number of Cu is 29 and electronic configuration of Cu^{2+} is [Ar]4s^{0}3d^{9}. This configuration is not stable.

Atomic number of Al is 13 and electronic configuration of Al is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}. This configuration is not stable.

Atomic number of Ba is 56 and electronic configuration of Ba^{2+} is [Kr]4d^{10}5s^{2}5p^{6}. This configuration is stable.

Atomic number of Mg is 12 and electronic configuration of Mg^{2+} is 1s^{2}2s^{2}2p^{6}. This configuration is stable.

Atomic number of Sn is 50 and electronic configuration of Sn is [Kr]4d^{10}5s^{2}5p^{2}. This configuration is stable.

Thus, we can conclude that out of the given options, only Fe and Cu^{2+} reactants would lead to a spontaneous reaction as they have incomplete sub-shells. Therefore, in order to gain stability they will readily react.


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covalent bonding

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The molar solubilities of the following compounds (in mol/L) are:
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<u>Answer:</u> The decreasing order of K_{sp} is AgSCN>AgBr>AgCN

<u>Explanation:</u>

  • <u>For AgBr:</u>

The balanced equilibrium reaction for the ionization of silver bromide follows:

AgBr\rightleftharpoons Ag^{+}+Br^-

                 s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][Br^-]

We are given:

s=7.3\times 10^{-7}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}

Solubility product of AgBr = 5.33\times 10^{-13}

  • <u>For AgCN:</u>

The balanced equilibrium reaction for the ionization of silver cyanide follows:

AgCN\rightleftharpoons Ag^{+}+CN^-

                    s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][CN^-]

We are given:

s=7.7\times 10^{-9}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}

Solubility product of AgCN = 5.33\times 10^{-17}

  • <u>For AgSCN:</u>

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

AgSCN\rightleftharpoons Ag^{+}+SCN^-

                     s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][SCN^-]

We are given:

s=1.0\times 10^{-6}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}

Solubility product of AgSCN = 1.0\times 10^{-12}

The decreasing order of K_{sp} follows:

AgSCN>AgBr>AgCN

4 0
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An electrolytic cell transferred 0.10 m o l of electrons when a constant current of 2.0 A was applied. How many hours did this t
marusya05 [52]

Answer:

1.34 hr

Explanation:

From the question given above, the following data were obtained:

Number of mole of electron (e) = 0.1 mole

Current (I) = 2 A

Time (t) =?

Next, we shall determine the quantity of electricity transferred. This can be obtained as follow:

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Therefore,

0.1 mole of electron = 0.1 × 96500

0.1 mole of electron = 9650 C

Thus, 9650 C of electricity was transferred.

Next, we shall determine the time. This can be obtained as follow:

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Quantity of electricity (Q) = 9650 C

Time (t) =?

Q = It

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Divide both side by 2

t = 9650 / 2

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Finally, we shall convert 4825 s to hour. This can be obtained as follow:

3600 s = 1 hr

Therefore,

4825 s = 4825 s× 1 hr / 3600 s

4825 s = 1.34 hr

Thus, the time taken is 1.34 hr

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