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2 years ago

Which techniques would be best for separating a

Chemistry

1 answer:

pashok25 [27]2 years ago

6 0

Answer: centrifugation, boiling/heating , and long standing

Explanation:

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

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:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

Which compound type is formed when two atoms share two electrons?

Pie

Covalent compounds
All the best

5 0

2 years ago

Which particle makes the atom an unstable isotope?

Galina-37 [17]

Instability of an atoms nucleus can result from an excess of either neutrons or protons . So neutrons and protons .

6 0

2 years ago

Please help me on this!!

Fudgin [204]

Answer:

Explanation:

7 0

3 years ago

Given the reactant side of the total ionic equation for the neutralization reaction of lithium hydroxide (LiOH) with hydrochlori

Sunny_sXe [5.5K]

Answer:


Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)

Explanation:


1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).

LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻

2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).

Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)

3) Finally, you can wirte the total ionic equation:

Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)

8 0

2 years ago