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LekaFEV [45]
4 years ago
6

Give the names and charges of the cation and anion in each of the following compounds:________.

Chemistry
1 answer:
BlackZzzverrR [31]4 years ago
7 0

Answer:

(a) Cu^{2+}S^{2-}: copper (II) sulfide or cupric sulfide.

(b)  Ag^+_2(SO_4)^{2-}: silver sulfate.

(c) Al^{3+}(ClO_3)_3^-: aluminum chlorate.

(d) Co^{2+}(OH)^-_2: cobalt (II) hydroxide.

(e) Pb^{2+}(CO_3)^{2-}: lead (II) carbonate.

Explanation:

Hello,

In this case, we proceed by exchanging the subscripts between the cation and anion for each case as follows:

(a) CuS: cation is cupper (II) and anion is sulfide, thus the name is copper (II) sulfide or cupric sulfide.

Cu^{2+}S^{2-}

(b) Ag₂SO₄: cation is silver ion and anion is sulfate, thus the name is silver sulfate.

Ag^+_2(SO_4)^{2-}

(c) Al(ClO₃)₃: cation is aluminum ion and anion is chlorate, thus the name is aluminum chlorate.

Al^{3+}(ClO_3)_3^-

(d) Co(OH)₂: cation is cobalt (II) and anion is hydroxide, thus the name is cobalt (II) hydroxide.

Co^{2+}(OH)^-_2

(e) PbCO₃: cation is lead (II) and anion is carbonate, thus the name is lead (II) carbonate.

Pb^{2+}(CO_3)^{2-}

Regards.

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Answer:  Kc=1.24*10^5^2

Explanation: For the given reaction:

Kc=\frac{[Zn^+^2]}{[Ag^+]^2}

Concentrations of the ions are not given so we need to think about another way to calculate Kc.

We can calculate the free energy change using the standard cell potential as:

\Delta G^0=-nFE^0_c_e_l_l

E^0_c_e_l_l can be calculated using standard reduction potentials.

Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.

E^0_c_e_l_l = E^0_c_a_t_h_o_d_e+E^0_a_n_o_d_e

Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.

E^0_c_e_l_l  = 0.78 V - (-0.76 V)

E^0_c_e_l_l  = 0.78 V + 0.76 V

E^0_c_e_l_l  = 1.54 V

Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .

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\Delta G^0 = -297173.8 J

Now we can calculate Kc using the formula:

\Delta G^0=-RTlnKc

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297173.8 = 2477.572*lnKc

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<u>Given:</u>

Mass of Ag = 1.67 g

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The enthalpy of formation of AgCl(s)

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The reaction is:

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Calculate the moles of Ag and Cl from the given masses

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