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Ivenika [448]
3 years ago
13

What is the factor that affects how easily an electron can be removed from an atom?

Chemistry
1 answer:
alexgriva [62]3 years ago
8 0
I think the correct answer among the choices listed above is option D. It is the structure of the atom that is the <span>factor that affects how easily an electron can be removed from an atom. This is because having a stronger structure would mean that electrons have a strong attraction to the atom and it would be harder to remove the electrons.</span>
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Breaking glass is an example of what type of change
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B

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What is the number of protons, neutrons, and electrons for isotope (Ta) with a mass number of 181?
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Answer:

Explanation:

Name: Tantalum

Symbol: Ta

Atomic Number: 73

Atomic Mass: 180.9479 amu

Melting Point: 2996.0 °C (3269.15 K, 5424.8 °F)

Boiling Point: 5425.0 °C (5698.15 K, 9797.0 °F)

Number of Protons/Electrons: 73

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Is 0.2 liters of soda the same measurement as 2000 grams of soda
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G = L x 1000

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Find the potentials of the following electrochemical cell:
melomori [17]

Answer: 0.18 V

Explanation:-

Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0_(Cd^{2+}/Cd)=-0.40V[/tex]

E^0_(Ni^{2+}/Ni)=-0.24V[/tex]

Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}

E^0=-0.24-(-0.40)=0.16V

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}

where,

n = number of electrons in oxidation-reduction reaction = 2

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E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}

E_{cell}=0.18V

Thus the potential of the following electrochemical cell is 0.18 V.

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