After the fission of u-235 takes place, how many neutrons does the missing nucleus have? a) 84 b) 85 c) 141 d) 143 the atomic number is consider a unique atom and element identifier. describe the associated subatomic particle, and how it can be used to identify the atom. a) the number of electrons changes in a unique way as the charge on the atom changes. b) the proton number remains constant through changing the atomic charge, energy, and isotopes. c) as the atom absorbs photons the way the photons interact in the nucleus gives the atomic number. d) all isotopes of the an element have the same number of neutrons allowing it to describe the unique atom.

1. B) 85

2. B) The proton number remains constant through changing atomic charge, energy, and isotopes

Explanation:

Hello,

1. Considering the nuclear fission of uranium-235, it is known that its nucleus absorbs a slow-moving neutron, turning out into a nucleus of Kripton-92, a nucleus of Barium-141 and three extra neutrons.

Next, considering the following statements:

- U's atomic number is 92, therefore it contains 92 protons

- Kr's atomic number is 36, therefore it contains 36 protons

- Ba's atomic number is 56, therefore it contains 56 protons

In such a way, we can compute the number of neutrons in the nucleus of Ba, as we know that the total number of neutrons  are always conserved. Thus:

- On the left side of the reaction: (235-92) neutrons of the uranium + 1 neutron absorbed = 144 neutrons

- On the right side of the reaction: (92-36) neutrons of the krypton + X neutrons of the barium + 3 neutrons = 144

Therefore:

Thus, the nucleus of barium has 85 neutrons.

2. In this case, the correct answer is B) The proton number remains constant through changing atomic charge, energy, and isotopes. We must consider that the atomic number is understood as the number of protons in the nucleus of the considered element. It would change just if the element undergoes a nuclear process.

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2