Question

1. calculate the average atomic mass for platinum.

isotope	% abundance	mass
pt - 192	0.78	191.961 amu
pt - 194	32.96	193.963 amu
pt - 195	33.83	194.965 amu
pt - 196	25.24	195.966 amu
pt - 198	7.16	197.968 amu

1. fill in the blanks. (1pt each)

symbol	atomic #	mass #	# protons	# electrons	# neutrons
		64	64	64

1. for each of the following elements, write electron configurations only

name	electron configuration
silver	1s 2s 2p 3s 3p 4s 3d 4p 5s 4d
mercury

Explanation:

Step1: Recall average atomic mass formula

The average atomic mass ($A_{avg}$) of an element is calculated using the formula $A_{avg}=\sum_{i = 1}^{n} (m_i\times p_i)$, where $m_i$ is the mass of the isotope and $p_i$ is the percent - abundance (expressed as a decimal) of the isotope.

Step2: Convert percent - abundances to decimals and calculate for platinum

For Pt - 190: $p_1 = 0.01\%=0.0001$, $m_1 = 189.99$ amu; for Pt - 192: $p_2 = 0.78\% = 0.0078$, $m_2=191.961$ amu; for Pt - 194: $p_3 = 32.967\%=0.32967$, $m_3 = 193.963$ amu; for Pt - 195: $p_4 = 33.832\%=0.33832$, $m_4 = 194.965$ amu; for Pt - 196: $p_5 = 25.242\%=0.25242$, $m_5 = 195.964$ amu; for Pt - 198: $p_6 = 7.161\%=0.07161$, $m_6 = 197.967$ amu.
\[

$$\begin{align*} A_{avg}&=(0.0001\times189.99)+(0.0078\times191.961)+(0.32967\times193.963)+(0.33832\times194.965)+(0.25242\times195.964)+(0.07161\times197.967)\\ &= 0.018999+1.497296+63.97377+65.92377+49.47974+14.18549\\ &\approx195.08\text{ amu} \end{align*}$$

\]

For the second table:

• For an atom, the number of protons is equal to the atomic number. If the mass number is $A$ and the atomic number is $Z$, the number of neutrons $N=A - Z$ and for a neutral atom, the number of electrons is equal to the number of protons.
• If the mass number $A = 193$, without knowing the element (assuming it's a neutral atom), if we assume it's iridium (Ir) with atomic number $Z = 77$: number of protons $= 77$, number of electrons $= 77$, number of neutrons $=193 - 77=116$.
• For an atom with 64 protons and 64 electrons (neutral atom), the atomic number $Z = 64$ (gadolinium, Gd), and if the mass number $A = 64$ (this is incorrect as the mass number of Gd is much larger, but if we go by the data given), number of neutrons $=64 - 64 = 0$ (this is an incorrect - data case in a real - world context as mass numbers are usually larger than atomic numbers for non - hydrogen isotopes).

For the electron - configuration part:

• Zirconium (Zr) has atomic number $Z = 40$. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{2}$.
• Silver (Ag) has atomic number $Z = 47$. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{10}$.
• Mercury (Hg) has atomic number $Z = 80$. Its electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}$.

Answer:

• Average atomic mass of platinum: approximately $195.08$ amu
• For mass number 193 (assuming neutral atom): Protons: 77, Electrons: 77, Neutrons: 116
• For atom with 64 protons and 64 electrons (assuming neutral atom): Protons: 64, Electrons: 64, Neutrons: 0 (incorrect data case in real - world context)
• Electron - configuration of Zirconium: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{2}$
• Electron - configuration of Silver: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{10}$
• Electron - configuration of Mercury: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}$