Question

1. calculate the average atomic mass for platinum. isotope pt-190 pt-192 pt-194 pt-195 pt-196 pt-198 % abundance 0.014 0.782 32.967 33.832 25.242 7.163 mass 189.960 amu 191.961 amu 193.963 amu 194.965 amu 195.965 amu 197.968 amu 10. fill in the blanks: (1 pt each) symbol zn+2 i- p3- atomic # 30 37 15 mass # 192 # protons 64 # electrons 64 # neutrons 11. for each of the following elements, write electron configurations only. name zirconium silver mercury electron configuration

Explanation:

Step1: Recall average - atomic - mass formula

The formula for the average atomic mass of an element with multiple isotopes is $\sum_{i}(m_i\times p_i)$, where $m_i$ is the mass of the $i$ - th isotope and $p_i$ is the percent - abundance of the $i$ - th isotope (expressed as a decimal).

Step2: Convert percent - abundances to decimals

For Pt - 190: $p_1 = 0.014/100=0.00014$, for Pt - 192: $p_2 = 0.782/100 = 0.00782$, for Pt - 194: $p_3=32.967/100 = 0.32967$, for Pt - 195: $p_4 = 33.832/100=0.33832$, for Pt - 196: $p_5 = 25.242/100 = 0.25242$, for Pt - 198: $p_6=7.163/100 = 0.07163$.

Step3: Calculate the product for each isotope

For Pt - 190: $m_1\times p_1=189.960\times0.00014 = 0.0265944$.
For Pt - 192: $m_2\times p_2=191.961\times0.00782=1.50113502$.
For Pt - 194: $m_3\times p_3=193.963\times0.32967 = 63.93388221$.
For Pt - 195: $m_4\times p_4=194.965\times0.33832 = 65.9613388$.
For Pt - 196: $m_5\times p_5=195.965\times0.25242 = 49.4645753$.
For Pt - 198: $m_6\times p_6=197.968\times0.07163 = 14.18065784$.

Step4: Sum up the products

$0.0265944 + 1.50113502+63.93388221+65.9613388+49.4645753+14.18065784 = 195.06818357\approx195.08$.

Answer:

$195.08$