Question

there are 3 known isotopes of a mystery element a. what are acceptable abundances (%s) of them?
50%, 40%, 20%
10%, 10%, 80%
60%, 40%
45%, 35%, 10%
97%, 1%, 1%
the average atomic mass of lithium is 6.941 amu. round this to the nearest whole number.
6
7
a mystery element has isotopes: x - 10 at 30% and x - 11 60%. the average atomic mass is...
between 10 and 11 amu but closer to 10
between 10 and 11 amu but closer to 11
less than 10 amu
more than 11 amu

Explanation:

First Question:

Step1: Check sum of percentages

The sum of abundances of isotopes of an element must be \(100\%\).
For \(50 + 40+20=110
eq100\).
For \(10 + 10 + 80=100\).
For \(60+40 = 100\) (but there are 3 isotopes in the question, so this is invalid).
For \(45+35 + 10=90
eq100\).
For \(97+1+1 = 99
eq100\).

Step1: Round the number

When rounding \(6.941\) to the nearest whole number, we look at the digit in the tenths place. The digit in the tenths place is \(9\). Since \(9\gt5\), we round up the units digit. So \(6.941\approx7\).

Step1: Calculate the weighted - average contribution

The contribution of \(X - 10\) is \(10\times0.3 = 3\).
The contribution of \(X - 11\) is \(11\times0.6=6.6\).
Let the abundance of the third isotope (since sum of abundances of \(X - 10\) and \(X - 11\) is \(30 + 60=90\%\), the abundance of the third isotope is \(10\%\)) be \(x\). Let's assume the third isotope has mass \(m\) (minimum value of \(m = 0\) and maximum value of \(m\) is not given, but if we assume it's non - negative). The average atomic mass \(A=10\times0.3+11\times0.6 + m\times0.1=3 + 6.6+0.1m=9.6 + 0.1m\).
Since \(m\geq0\), \(A\geq9.6\). Also, if we assume the third isotope has a non - negative mass. Another way: using the two given isotopes (ignoring the third for a rough estimate, since \(30\%\) of \(10\) and \(60\%\) of \(11\)): \(\frac{10\times30 + 11\times60}{30 + 60}=\frac{300+660}{90}=\frac{960}{90}\approx10.67\)

Answer:

\(10\%,10\%,80\%\)

Second Question: