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Alina [70]
2 years ago
7

1A: Consider these compounds:

Chemistry
1 answer:
Setler [38]2 years ago
4 0

Solution :

Compound                      Ksp

$PbF_2$                               $3.3 \times 10^{-8}$

$Ni(CN)_2$                        $3 \times 10^{-23}$

FeS                                $8 \times 10^{-19}$

$CaSO_4$                           $4.93 \times 10^{-5}$

$Mg(OH)_2$                      $5.61 \times 10^{-12}$

Ksp of $Ni(CN)_2 and both compounds dissociate the same way. Hence $Mg(OH)_2$ is more soluble than $(B). \  Ni(CN)_2$

$Mg(OH)_2$  is less soluble than $(A). \ \ PbF_2 \ ()Ksp \  PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$

It is not possible to determine CD - $FeS \text{ or} \ CaSO_4$  is more or less soluble than $Mg(OH)_2$  as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

$Ni(OH)_2$  can be directly compared with PbS, $AlPO_4, MnS$

$\text{For } \ Ni(OH)_2$

$AB_2(s) \rightarrow A^{2+} + 2B^{-}$

$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$

100

1-s s 2s

Ksp = [A2+][B-]^2 = s \times (2s)^2 = 4s^3

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

$AB(s) \rightarrow A^{x+}+B^{x-}$

$AgCl(s) \rightarrow Ag^+ + Cl^-$

1 0 0

1 - s s s

Ksp $=[A^{x+}][B^{x-}]=s \times s = s^2$

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

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Is the answer B? Help
Masja [62]

Answer:

A

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

Cl2(g)

Ag(s)

NaCl(aq)

AgNO3(aq)

Pt 2+, 4+, 6+  Though it states Pt is inert

Cl 2-

Ag 1+

Na 1+

NO3-

Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

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