Which is the most valid reason a group of people might oppose the storage of spent fuel rods in their community?

Chemistry

2 answers:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

The most likely reason is that spent fuel bars can infect the community with radiation causing numerous health problems.

Explanation:

A spent fuel is a fuel that has been exposed to irradiation in a nuclear reactor. This fuel expels radiation and this is one of the main reasons why a group of people oppose the storage of spent fuel bars in their community. This is because through this fuel can occur a leak of radiation contaminating the community and generating major health problems.

KATRIN_1 [288]3 years ago

6 0

The fuel rods will remain radioactive for a long time....hope this helps

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Which statement describes the reactions in an electrochemical cell

34kurt

Answer & explanation:

Summary on electrochemical cells and redox reactions:

Electrochemical cells (or batteries) can be defined as devices capable of transforming chemical energy into electrical energy through spontaneous reactions of redox, in which electron transfer occurs.

Redox it is a chemical reaction in which there is the occurrence of oxidation and reduction of atoms of substances (chemical species) present in the process.

Oxidation is the loss of electrons by an atom of a chemical species, while reduction is the gain of electrons by an atom of a chemical species.

Thus, during an oxirreduction reaction, electrons move from the species that loses them towards the species that will receive them. This "movement" results in the formation of an electric current (or electrical energy) as occurs with batteries, for example.

6 0

3 years ago

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

egoroff_w [7]

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

$2Ca+O_2\rightarrow 2CaO$

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

$mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa$

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

$mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa$

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

$m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO$