Question

1. which is true of the $^{243}am^{3 + }$ ion?

(a) protons 148 electrons 148 neutrons 243
(b) 95 98 243
(c) 95 95 148
(d) 95 92 148
use the mass spectroscopy data below for question 18.

1. the figure above represents the mass spectrometry data for an unknown element. which statement best interprets the data below?

(a) there is only one isotope for the unknown element with a mass of 19 amu. this means the isotope has 10 neutrons.
(b) there is only one isotope for the unknown element with a mass of 19 amu. this means the isotope has 9 neutrons.
(c) there is only one ion for the unknown element with a charge of 19 amu. this means the ion has 10 electrons.
(d) there is only one ion for the unknown element with a charge of 19 amu. this means the ion has 9 electrons.

Explanation:

Step1: Determine the number of protons

The element \(Am\) (Americium) has an atomic number of \(95\). The atomic number is equal to the number of protons. So, the number of protons in \(^{243}Am^{3 +}\) is \(95\).

Step2: Determine the number of electrons

For a neutral atom, the number of electrons is equal to the number of protons. Since the ion has a charge of \(+ 3\), it has lost \(3\) electrons. So, the number of electrons is \(95 - 3=92\).

Step3: Determine the number of neutrons

The mass number is \(243\). The mass number \(A\) is related to the number of protons \(Z\) and neutrons \(N\) by the formula \(A = Z+N\). So, \(N=A - Z\). Substituting \(A = 243\) and \(Z = 95\), we get \(N=243 - 95 = 148\).

for question 18:

Step1: Analyze the mass - spectrometry data

The mass - spectrometry data shows a single peak at \(m/z = 19\). This indicates that there is only one isotope (since mass - spectrometry measures the mass of isotopes, not ions directly in this context of basic interpretation for the given data).

Step2: Calculate the number of neutrons

If the mass number \(A = 19\) and assuming it is a neutral atom (since the data is presented in a way that first considers isotopic mass in a basic analysis), and if we assume it is \(F\) (fluorine, atomic number \(Z = 9\)). Using the formula \(A=Z + N\), we get \(N=A - Z\). Substituting \(A = 19\) and \(Z = 9\), we have \(N=19 - 9=10\). But wait, if we consider the general case for an unknown element with mass \(19\) amu (isotope mass), and if we assume it is \(F\) (a common element with mass close to \(19\) amu). For a neutral atom of \(F\), \(Z = 9\), \(A = 19\), \(N=19 - 9 = 10\). But if we consider the options, we made a wrong assumption. Let's start over.
The mass - spectrometry peak at \(m/z=19\) represents the mass of the isotope. If we assume the element is \(F\) (atomic number \(Z = 9\)). For an isotope with mass \(A = 19\), using \(A=Z + N\), we get \(N=A - Z=19 - 9 = 10\). But let's check the options.
Option A: If \(A = 19\) (isotope mass), \(Z\) (atomic number) for \(F\) is \(9\), \(N=A - Z=19 - 9 = 10\).

Answer:

D. \(95\) Protons, \(92\) Electrons, \(148\) Neutrons