Question

1. in which of the following sets are the symbol of the element, the number of protons, and the number of electrons given correctly?

a) in: 40 protons, 48 electrons
b) ca: 56 protons, 132 electrons
c) fe: 26 protons, 30 electrons
d) cl: 17 protons, 17 electrons

1. how many protons, electrons, and neutrons are present in an atom of ar - 40? (6 points)

18p, 22n, 18e

1. how many protons, neutrons and electrons are present in one atom of the following:
2. what is the average atomic mass of chlorine given that:

• the exact mass of ^ { 35 } cl is 34.9689 amu and the relative abundance is 75.53 %
• the exact mass of ^ { 37 } cl is 36.9659 amu and the relative abundance is 24.47 %

1. what is the average atomic mass of ruthenium given the following information:

look it up in the periodic table. is this the answer that you expected?

Explanation:

Step1: Formula for average atomic mass

The formula for average atomic mass \(M\) is \(M=\sum_{i}(m_i\times a_i)\), where \(m_i\) is the exact mass of the isotope and \(a_i\) is the relative - abundance (expressed as a decimal) of the isotope.

Step2: Convert abundances to decimals

For \(^{96}Ru\): \(a_1 = 5.54\%=0.0554\), \(m_1 = 95.907599\)
For \(^{98}Ru\): \(a_2=1.87\% = 0.0187\), \(m_2 = 97.905287\)
For \(^{99}Ru\): \(a_3 = 12.76\%=0.1276\), \(m_3=98.9059389\)
For \(^{100}Ru\): \(a_4 = 12.60\%=0.1260\), \(m_4 = 99.9042192\)
For \(^{101}Ru\): \(a_5=17.06\% = 0.1706\), \(m_5 = 100.9055819\)
For \(^{102}Ru\): \(a_6=31.55\% = 0.3155\), \(m_6 = 101.9043485\)
For \(^{104}Ru\): \(a_7=18.62\% = 0.1862\), \(m_7 = 103.905424\)

Step3: Calculate each term

\(m_1\times a_1=95.907599\times0.0554 = 5.313281\)
\(m_2\times a_2=97.905287\times0.0187 = 1.830829\)
\(m_3\times a_3=98.9059389\times0.1276 = 12.620498\)
\(m_4\times a_4=99.9042192\times0.1260 = 12.587932\)
\(m_5\times a_5=100.9055819\times0.1706 = 17.214492\)
\(m_6\times a_6=101.9043485\times0.3155 = 32.150822\)
\(m_7\times a_7=103.905424\times0.1862 = 19.347190\)

Step4: Sum up the terms

\(M=5.313281 + 1.830829+12.620498+12.587932+17.214492+32.150822+19.347190\)
\(M = 101.064\)

Answer:

The average atomic mass of ruthenium is \(101.064\) amu. Looking at the periodic table, the average atomic mass of ruthenium (\(Ru\)) is approximately \(101.07\) amu. So, this is a close result considering the precision of the calculations.