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ozzi
3 years ago
10

An iron nail (Fe) is placed into a solution of copper(II) sulfate (CuSO4). After some time, the blue color of the copper (II) su

lfate disappears and some copper-colored solid (Cu) is found on the surface of the nail. The solution now contains iron (II) sulfate (FeSO4), and the container feels warmer than before the reaction.. 1. Does this reaction absorb or release absorb or release energy? How do you know?. 2. What do the color changes indicate?
Chemistry
2 answers:
Sphinxa [80]3 years ago
8 0
The reaction is a displacement reaction. Iron is more reactive than copper. Hence, it displaces copper,forming iron II sulfate and copper metal. 
This reaction is an exothermic reaction, hence it releases energy. The colour change indicates that iron has displaced copper. 
Virty [35]3 years ago
7 0
This reaction shows that iron is more reactive than copper as it displaces copper from its solution and iron passes into solution as Fe2+ ions and ferrous sulfate solution is formed. It is called a single displacement reaction. There is a release of energy since the container feels warmer. This type of reaction is called exothermic. The color changes indicate that there is a chemical reaction that occurred. The copper from the solution will now be deposited on the nail.
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Answer:

_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

Explanation:

The unbalanced nuclear equation is

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow ?

It is convenient to replace the question by an atomic symbol, _{x}^{y}\text{Z}, where <em>x </em>= the atomic number, <em>y</em> = the mass number, and Z = the symbol of the element.

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

Then your equation becomes

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts and of the subscripts</em> must be the same on each side of the equation.  

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Element 92 is uranium, so the nuclear equation becomes

_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

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