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goldfiish [28.3K]
4 years ago
11

Which of these is not a precaution taken at nuclear power plants to ensure safety? Safety systems are built into them. All parts

are kept in good repair. Enriched uranium is the fuel source. Processes are carefully monitored.
Chemistry
1 answer:
irga5000 [103]4 years ago
8 0

Answer: Enriched uranium is the fuel source.

The nuclear power plants utilizes the radiation energy from the fission of the heavy molecular weight atomic particles. These radiations can cause damage when exposed to the external air, they are capable of causing explosion, mutation in the living beings. Therefore, safety measures should be taken to prevent the damages.

Enriched uranium is the fuel source is not the precaution taken at the nuclear power plant to ensure safety because it is just explaining the fact that the uranium is the source of radiation. All other factors such as safety systems are built into them, all parts are kept in good repair and processes are carefully monitored are relevant precautions.

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The difference in an area with high concentration and an area with low concentration is called.
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The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

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This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

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2 years ago
Convert 281 K to degrees Celsius.<br><br> 281 K = _____<br><br> -8°C<br> 554°C<br> 8°C
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3 0
4 years ago
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]

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

<u>Explanation:</u>

We are given:

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

<u>Reduction half reaction:</u>  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
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