QUESTION IMAGE
Question
- **write the complete symbol for an atom with the following characteristics:
| characteristics | atomic symbol (include atomic #, mass #, and charge) |
|---|
|a. neutral atom that contains 19 electrons and 20 neutrons|
|b. nitrogen atom with 8 neutrons and 10 electrons|
|c. neutral isotope of (b) with 9 neutrons|
|d. contains 20 protons, 19 electrons, and 22 neutrons|
|e. neutral iodine atom with 181 subatomic particles in its nucleus|
- use the data table below to answer the following questions. the two isotopes are of unknown element j.
| isotope | atomic mass (amu) | natural abundance |
|---|---|---|
| 2 | 122.90 | 42.79% |
a. *calculate the atomic mass of this element and determine its identity. make sure to use significant figures!
b. based on your answer above, complete the table below.
| isotope | nuclear symbol | isotope name |
|---|
|1|
|2|
10.
a.
Step1: Determine the atomic number
For a neutral atom, number of electrons = number of protons. Given 19 electrons, atomic number \(Z = 19\), which is potassium (K).
Step2: Calculate the mass number
Mass number \(A=\) number of protons+number of neutrons. Protons = 19, neutrons = 20, so \(A=19 + 20=39\). Charge \(q = 0\) (neutral atom).
The atomic symbol is \(\ce{_{19}^{39}K}\).
b.
Step1: Determine the atomic number of nitrogen
Nitrogen has an atomic number \(Z = 7\).
Step2: Calculate the mass number
Mass number \(A=\) number of protons+number of neutrons. Protons = 7, neutrons = 8, so \(A=7 + 8 = 15\).
Step3: Determine the charge
There are 10 electrons and 7 protons, so the charge \(q=7 - 10=- 3\).
The atomic symbol is \(\ce{_{7}^{15}N^{3 -}}\).
c.
Step1: Recall the atomic number of nitrogen
Atomic number of nitrogen \(Z = 7\) (unchanged as it's still nitrogen).
Step2: Calculate the mass number
Protons = 7, neutrons = 9, so \(A=7+9 = 16\). Charge \(q = 0\) (neutral atom).
The atomic symbol is \(\ce{_{7}^{16}N}\).
d.
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The atomic mass of the element is 121.76 amu and the element is antimony (Sb).
b.
For isotope 1:
Nuclear symbol: \(\ce{_{51}^{121}Sb}\) (since \(Z = 51\) for antimony and \(A = 120.90\approx121\)), isotope name: Antimony - 121.
For isotope 2:
Nuclear symbol: \(\ce{_{51}^{123}Sb}\) (since \(Z = 51\) for antimony and \(A = 122.90\approx123\)), isotope name: Antimony - 123.