Question

27.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: 49 protons, 48 electrons
b) cs: 55 protons, 132 electrons
c) fe: 26 protons, 30 electrons
d) cl: 17 protons, 17 electrons
28.how many protons, electrons, and neutrons are present in an atom of ar - 40? (6 points)
18p, 22n, 18e
29.how many protons, neutrons and electrons are present in one atom of the following:
30.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 %

31.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: Calculate average atomic mass formula

The formula for average atomic mass \(M\) is \(M=\sum_{i = 1}^{n}m_i\times a_i\), where \(m_i\) is the mass of the \(i\) - th isotope and \(a_i\) is the relative abundance (in decimal form) of the \(i\) - th isotope.

Step2: Convert abundances to decimals

For \(^{96}Ru\): \(a_1 = 0.0554\), for \(^{98}Ru\): \(a_2=0.0187\), for \(^{99}Ru\): \(a_3 = 0.1276\), for \(^{100}Ru\): \(a_4=0.1260\), for \(^{101}Ru\): \(a_5 = 0.1706\), for \(^{102}Ru\): \(a_6=0.3155\), for \(^{104}Ru\): \(a_7 = 0.1862\)

Step3: Calculate each term

• For \(^{96}Ru\): \(m_1\times a_1=95.907599\times0.0554 = 5.313281\)
• For \(^{98}Ru\): \(m_2\times a_2=97.905287\times0.0187 = 1.830829\)
• For \(^{99}Ru\): \(m_3\times a_3=98.9059389\times0.1276 = 12.620498\)
• For \(^{100}Ru\): \(m_4\times a_4=99.9042192\times0.1260 = 12.587932\)
• For \(^{101}Ru\): \(m_5\times a_5=100.9055819\times0.1706 = 17.224492\)
• For \(^{102}Ru\): \(m_6\times a_6=101.9043485\times0.3155 = 32.150822\)
• For \(^{104}Ru\): \(m_7\times a_7=103.905424\times0.1862 = 19.347190\)

Step4: Sum up the terms

\(M=5.313281 + 1.830829+12.620498+12.587932+17.224492+32.150822+19.347190\)
\(M = 101.074\) amu

Answer:

The average atomic mass of ruthenium is approximately \(101.07\) amu. Looking at the periodic table, the atomic mass of ruthenium (\(Ru\)) is about \(101.07\) amu, so this is the expected answer.