Question

1. circle the correct answer.
2. the element chromium has four naturally occurring isotopes: 4.34% 50cr, 83.79% 52cr, 9.50% 53cr, and the rest is 54cr. calculate the average atomic mass of chromium.

51.996 amu
52.0556 amu
5205.56 amu
5199.6 amu

1. what parts of the atom are in the nucleus?

a. protons and electrons
b. protons and neutrons
c. electrons and neutrons
d. electrons and protons

1. in a regular atom, the atomic number is the same as (circle all that are correct)

a. protons
b. neutrons
c. electrons

1. to find mass of an atom,

a. a = p = e
b. m = a = n
c. a - p = e
d. m - a = n
e. m + a = n

Explanation:

27.

Step1: Calculate the percentage of \(^{54}\text{Cr}\)

The sum of percentages of all isotopes is \(100\%\).
Let \(x\) be the percentage of \(^{54}\text{Cr}\).
\(x = 100-(4.34 + 83.79+9.50)=100 - 97.63=2.37\%\)

Step2: Use the formula for average atomic mass

The formula for average atomic mass \(M=\sum_{i = 1}^{n}m_i\times p_i\), where \(m_i\) is the mass of the \(i\)-th isotope and \(p_i\) is its percentage (in decimal form).
For \(^{50}\text{Cr}\): \(m_1 = 50\), \(p_1=0.0434\)
For \(^{52}\text{Cr}\): \(m_2 = 52\), \(p_2 = 0.8379\)
For \(^{53}\text{Cr}\): \(m_3=53\), \(p_3 = 0.0950\)
For \(^{54}\text{Cr}\): \(m_4 = 54\), \(p_4=0.0237\)

\(M=(50\times0.0434)+(52\times0.8379)+(53\times0.0950)+(54\times0.0237)\)
\(M = 2.17+43.5708+5.035+1.2798\)
\(M=51.9956\approx51.996\) amu

The nucleus of an atom contains protons (positively - charged particles) and neutrons (neutral particles). Electrons are located outside the nucleus in electron shells.

In a neutral atom (a regular atom), the atomic number (\(Z\)) is equal to the number of protons (\(P\)). Also, since the atom is neutral, the number of electrons (\(E\)) is equal to the number of protons. The number of neutrons (\(N\)) is not necessarily equal to the atomic number.

The mass number (\(A\)) of an atom is the sum of the number of protons (\(P\)) and neutrons (\(N\)), i.e., \(A = P+N\). Rearranging for \(N\) gives \(N=A - P\). The mass of an atom (\(M\)) is approximately equal to the mass number (\(A\)) (since the mass of electrons is negligible).

Answer:

\(51.996\) amu

28.