Question

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Answer 1

Members of the alkaline earth elements are $\boxed{{\text{B) metals with two valence electrons}}}$.

After the nuclear fission of U-235, the missing nuclide has $\boxed{{\text{B) 85}}}$ neutrons.

Further Explanation:

The systematic arrangement of various elements in the increasing order of their atomic numbers is termed as the periodic table. It exists in the form of a table, comprising of horizontal rows called periods and vertical columns called groups. There are 18 groups and 7 periods in a periodic table.

Group 2 elements are classified as alkaline earth elements. These are situated at the left part of the periodic table. These elements have the general outermost electronic configuration of  . This group includes beryllium, magnesium, calcium, strontium, and barium. These elements have two electrons in their outermost shells that can be easily removed so these are highly electropositive in nature and are thus called metals. Since these elements can lose electrons, these have chemical reactivity.

Nuclear fission is a nuclear reaction in which the heavier nucleus of an atom breaks down into smaller and lighter nuclei. Here, a large amount of energy is released along with the production of protons and neutrons. During nuclear fission, mass remains conserved.

The nuclear fission of U-235 occurs as follows:

 $_{92}^{235}{\text{U}} \to {\text{X}} + _{36}^{92}{\text{Kr}} + 2_0^1{\text{n}}$

Here, X is the missing nuclide.

Consider A to be the mass of the missing nuclide X. Since mass is conserved during nuclear fission, the total mass of reactant and product side is the same.

Mathematically,

 ${\text{Total mass of reactant side}} = {\text{Total mass of product side}}$

The mass of the reactant side is 235.

The mass of the product side is the sum of A, 92 and 2.

Therefore, the above expression becomes,

$\begin{galined}{\text{235}}&= {\text{A}} + {\text{92}} + {\text{2}} \hfill \\{\text{235}}&= {\text{A}} + 94 \hfill\\\end{aligned}$  

Solving for A,

${\text{A}} = 141$  

So the atomic mass of the missing nuclide (X) comes out to be 141. Therefore the missing nuclide is barium. According to the periodic table, its atomic number is 56. Therefore the number of neutrons can be calculated as follows:

 $\begin{aligned}{\text{Number of neutrons}} &= 141 - 56\\&= 85\\\end{aligned}$

Learn more:

1. Which general statement does not apply to metals? brainly.com/question/2474874
2. Among sodium and chlorine, which has higher ionization energy? brainly.com/question/6324347

Answer details:

Grade: High School

Chapter: Periodic classification of elements

Subject: Chemistry

Keywords: elements, alkaline earth metals, nuclear fission, neutrons, X, A, 85, 141, ns^2, U-235.

Answer 2

First question:
All metals 
All share metallic properties
All solids at room temperature

Second question:
In nuclear fission, heavier nuclei breaks into two or more particles which are relatively stable than the former one.
The reaction can be written as,
U-235 ------> X-k + Kr-92 + 2n
X is the unknown element which is to be determined.
Mass is conserved in Nuclear fission but not in nuclear fusion.

Therefore, mass of X will be 141. Which is an isotope of Barium. Thus the missing element was Barium.
Therefore, X-k = Ba-141, k=141.
Tomic number of Ba is 56. Therefore, number of neutrons of missing nucleus(Barium) = 141 -56 = 85.