Question

atomic theory (fill in the blank) give name of scientist and name of model

1. first atomic theory of matter:__ model name:__
2. description of particles:____
3. first periodic table:__ model name:__
4. discovery of electrons:____
5. discovery of the nucleus:____
6. charge of the electron:__ model name:__
7. discovery of orbitals:__ model name:__
8. discover of neutrons:____ (including neutrons, wave theory,

wave theory:____ quantum energy and uncertainty)
quantum of energy:____
uncertainty principle:____
multiple choice

1. chlorine exists as two naturally - occurring isotopes. cl - 35 occurs 75% of the time and cl - 37 occurs 25% of the time. which of the following calculations should be used to calculate the correct average atomic mass of chlorine?

a (35 x.75)+(37 x.25) b (35 x 3)+37÷2 c (35 x 3)+37÷3 d (35 + 37)÷2

1. which of these conclusions can be drawn from rutherfords experiment? each atom...

f contains electrons. g can be split. h contains protons. j is mostly empty space.

1. how does the radioactive isotope c - 14 differ from its stable counterpart c - 12? the isotope c - 14 has...

a two less neutrons. b two more electrons. c two more neutrons. d two less neutrons.

1. chlorine forms a 1 - ion. how many electrons does a chloride ion have?

f 1 g 16 h 17 j 18

1. a neutral atom of aluminum - 27 contains —

a 13 p+ and 27 e - b 14 p+ and 13 n c 13 e -, 13 p+, and 14 n d 13 e -, 14 p+, and 13 n

1. an element has an electron configuration of 1s²2s²2p⁶3s². which will be in the same group as this element?

f 1s²2s²2p⁶ g 1s²2s²2p⁶3s¹ h 1s²2s²2p⁶3s²3p²3d¹⁰4s² j 1s²2s²2p⁶3s²3p⁶

Explanation:

Step1: Calculate average atomic mass

The average atomic mass of an element with isotopes is calculated by multiplying the mass - number of each isotope by its relative abundance and then summing these values. For chlorine, Cl - 35 has a mass - number of 35 and an abundance of 0.75, and Cl - 37 has a mass - number of 37 and an abundance of 0.25. The formula is $(35\times0.75)+(37\times0.25)$.

Step2: Analyze Rutherford's experiment

Rutherford's gold - foil experiment led to the conclusion that an atom is mostly empty space because most of the alpha particles passed straight through the gold foil.

Step3: Compare C - 14 and C - 12

The atomic number of carbon is 6. For C - 12, the number of neutrons is $12 - 6=6$. For C - 14, the number of neutrons is $14 - 6 = 8$. So C - 14 has two more neutrons than C - 12.

Step4: Determine electrons in chloride ion

Chlorine has an atomic number of 17. A chloride ion ($Cl^-$) has gained one electron, so it has $17 + 1=18$ electrons.

Step5: Analyze aluminum - 27 atom

Aluminum has an atomic number of 13. A neutral atom of aluminum - 27 has 13 protons ($p^+$), 13 electrons ($e^-$), and $27-13 = 14$ neutrons ($n$).

Step6: Find elements in the same group

Elements in the same group have the same number of valence electrons. The given electron configuration $1s^22s^22p^63s^2$ has 2 valence electrons in the 3s orbital. The electron configuration $1s^22s^22p^63s^23p^64s^2$ also has 2 valence electrons in the 4s orbital.

Answer:

1. A. $(35\times0.75)+(37\times0.25)$
2. J. is mostly empty space.
3. C. two more neutrons.
4. J. 18
5. C. $13e^-,13p^+$, and $14n$
6. J. $1s^22s^22p^63s^23p^64s^2$