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2 years ago

Why was Becquerel’s experiment a first step in discovering radioactivity?

2 answers:

6 0

Henri Becquerel in 1896 used naturally fluorescent minerals to examine the characteristics of X-rays. He used potassium uranyl sulfate and placed it on the photographic plates covered in black paper, considering that the uranium captivated energy from the Sun and then emitted it as X-rays.

However, the hypothesis was disapproved due to overcast conditions in Paris. Due to certain reason, Becquerel decided to create his photographic plates, and it was witnessed that the images were clear and robust, claiming that the uranium discharged radiation in the absence of an external source of energy like Sun. Thus, radioactivity got discovered.

Therefore, Becquerel experiment is considered as the initial step in discovering radioactivity because it demonstrated that a substance could continuously discharge radiation even if it was not exposed to light.

galina1969 [7]2 years ago

5 0

This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.

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<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

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