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ch 5 pt 2 hw element atomic # protons electrons k 19 19 5 16 23 boron h…

Question

ch 5 pt 2 hw
element atomic # protons electrons
k 19 19
5
16
23
boron has two isotopes: boron - 10 and boron - 11. which is more abundant, given that the atomic mass of boron is 10.81?
the element copper has naturally occurring isotopes with mass numbers of 63 and 65. the relative abundance and atomic masses are 69.2% for mass = 63 amu, and 30.8% for mass = 65 amu. calculate the average atomic mass of copper.
show work
a(with correct sig figs)
calculate the atomic mass of bromine. the two isotopes of bromine have atomic masses and relative abundance of 78.92 amu (50.69%) and 80.92 amu (49.31%).
show work.
a(with correct sig figs).

Explanation:

Step1: Fill in the element - related table

  • For element K (Potassium), atomic number = 19. In a neutral atom, number of protons = number of electrons = atomic number. So number of protons is 19.
  • If number of electrons is 5, atomic number (number of protons) is 5, and the element is B (Boron).
  • If atomic number is 16, the element is S (Sulfur), and number of protons is 16.
  • If number of protons is 23, atomic number is 23, and the element is V (Vanadium), and number of electrons in neutral atom is 23.

Step2: Determine the more abundant boron isotope

Let the abundance of boron - 10 be $x$, then the abundance of boron - 11 is $1 - x$. The average atomic mass formula is $A = m_1x_1+m_2x_2$. So, $10.81=10x + 11(1 - x)$.
Expand: $10.81=10x+11 - 11x$.
Combine like - terms: $10.81 = 11 - x$.
Solve for $x$: $x=11 - 10.81=0.19$ (abundance of boron - 10), and $1 - x = 0.81$ (abundance of boron - 11). So boron - 11 is more abundant.

Step3: Calculate the average atomic mass of copper

The average atomic mass formula is $A=\sum_{i}m_ix_i$. Here, $m_1 = 63$ amu, $x_1=0.692$, $m_2 = 65$ amu, $x_2 = 0.308$.
$A=(63\times0.692)+(65\times0.308)$
$A = 43.596+20.02$
$A = 63.616\approx63.6$ amu (3 significant figures)

Step4: Calculate the atomic mass of bromine

Using the average atomic mass formula $A=\sum_{i}m_ix_i$. Here, $m_1 = 78.92$ amu, $x_1 = 0.5069$, $m_2 = 80.92$ amu, $x_2=0.4931$.
$A=(78.92\times0.5069)+(80.92\times0.4931)$
$A = 78.92\times0.5069+80.92\times(1 - 0.5069)$
$A = 78.92\times0.5069+80.92-80.92\times0.5069$
$A=(78.92 - 80.92)\times0.5069+80.92$
$A=- 2\times0.5069+80.92$
$A=-1.0138 + 80.92$
$A = 79.9062\approx79.9$ amu (3 significant figures)

Answer:

ElementAtomic #ProtonsElectrons
B555
S161616
V232323

Boron - 11 is more abundant.
Average atomic mass of copper: 63.6 amu
Average atomic mass of bromine: 79.9 amu