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Radda [10]
3 years ago
5

What the symbol for the oxygen isotope with 10 neutrons?

Chemistry
2 answers:
VashaNatasha [74]3 years ago
8 0
The element symbol for oxygen<span> is O and its atomic number is 8. The mass numbers for oxygen must be 8 + 8 = 16; 8 + 9 = 17; 8 + 10 = 18</span>
Likurg_2 [28]3 years ago
4 0
To write the symbol, we need to things, the atomic number and atomic mass.

if you go to the periodic table, you see that Oxygen's atomic number is 8.

for the atomic mass, we need to determine it. remember that atomic mass is equal to neutrons plus protons. Also, remember that the atomic number represent the number of protons, so

atomic number= number of protons= 8
number of neutrons= 10 (given by the question)

atomic mass= 10 + 8= 18

now we can write the symbol, I have drawn it and attach it.

remember than when you draw it, the mass goes on top, atomic number goes bottom.


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8 0
3 years ago
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

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}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

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3 years ago
Which of the following is true: A. Gamma rays are too dangerous to be used in a medicinal way. B. Gamma radiation is used common
ratelena [41]

Answer:

D

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Gamma radiation penetrates the cell wall of prokaryotic organisms such as bacteria and can inhibit their metabolic functions as well as destroy their DNA.

Debunking the other answers:

A is incorrect as Gamma radiation is used in the treatment of cancer via radiotherapy.

B is incorrect as Gamma rays are too small and would just penetrate any smoke particles.

C is incorrect because Gamma rays are used to disinfect food products to prevent food borne illness. Irradiation is safe to use on food and does not make it radioactive.

Thus, D is correct.

6 0
1 year ago
Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.
Citrus2011 [14]

Answer:

Explanation:

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After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu  has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

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Ce⁺²

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The spinal cord is part of the central nervous system

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3 years ago
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