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Alexxandr [17]
3 years ago
10

A battery contains two metals that have different tendencies to attract electrons. If one is lithium with an electron affinity o

f −3.05, and the other is zinc with an electron affinity of −0.76, describe how the electrons will flow. Then, describe how you could make this an even stronger battery.
Chemistry
2 answers:
Vitek1552 [10]3 years ago
4 0

Answer:

I know that the electrons flow would be:  The electrons will flow from lithium to zinc.

Explanation:

But I also dont know what would make this an even stronger battery.

tekilochka [14]3 years ago
4 0

Answer:

Electrons will flow from Zinc to Lithium

Explanation:

Since lithium has an electronegativity value of -3.05, this means that it has a very strong attraction to electrons. So a element with less electronegativity (zinc) will end up having its electrons pulled away, since electrons flow from least electronegative to most electronegative.

As for making the battery stronger, we can change it so that we use different metals with an even greater difference in electronegativity.

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Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.15 mol of PCl3 in the flask. What is the equi
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Answer : The equilibrium concentration of PCl_5 is, 0.50 M

Explanation : Given,

Initial moles of PCl_5 = 0.65 mole

Volume of solution = 1.0 L

Moles of PCl_3 at equilibrium = 0.15 mole

The balanced equilibrium reaction will be,

                          PCl_5\rightleftharpoons PCl_3+Cl_2

Initial moles     0.65        0         0

At eqm.           (0.65-x)     x         x

Moles of PCl_3 at equilibrium = x = 0.15 mole

Moles of Cl_2 at equilibrium = x = 0.15 mole

Moles of PCl_5 at equilibrium = (0.65-x) = (0.65-0.15) = 0.50 mole

Now we have to calculate the concentration of PCl_3,Cl_2\text{ and }PCl_5 at equilibrium.

Formula used : Concentration=\frac{Moles}{Volume}

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M

\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.50mole}{1.0L}=0.50M

Therefore, the equilibrium concentration of PCl_5 is, 0.50 M

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