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critical thinking questions 5. neon has three different isotopes. 90.51…

Question

critical thinking questions

  1. neon has three different isotopes. 90.51% of neon atoms have a mass of 19.992 amu, 0.27% of neon atoms have a mass of 20.994 amu, 9.22% of neon atoms have a mass of 21.991 amu. what is the average atomic mass of neon?
  2. chlorine - 35 is one isotope of chlorine. (35 is the mass number.) chlorine - 37 is another isotope of chlorine. how many protons and how many neutrons are in each isotope of chlorine?
  3. of all chlorine atoms, 75.771% are chlorine - 35. chlorine - 35 atoms have a mass of 34.96885 amu. all other chlorine atoms are chlorine - 37 and these have a mass of 36.96590. calculate the average atomic mass of chlorine
  4. do your answers for questions 5 and 7 agree with the average atomic masses for neon and chlorine on the periodic table?
  5. a certain element has two isotopes. one isotope, which has an abundance or 72.15% has a mass of 84.9118 amu. the other has a mass of 86.9092 amu. calculate the average mass for this element.
  6. given the following data, calculate the average atomic mass of magnesium.

isotope mass of isotope abundance portion of total
magnesium - 24 23.985 amu 78.70 →.7870

Explanation:

Response
5.

Step1: Convert percentages to decimals

$90.51\% = 0.9051$, $9.22\%=0.0922$, $0.27\% = 0.0027$

Step2: Use the formula for average atomic mass

$Average\ atomic\ mass=(0.9051\times19.992)+(0.0922\times20.994)+(0.0027\times21.991)$
$=18.0883392 + 1.9356468+0.0593757$
$=20.0833617\approx20.18\ amu$

The atomic number of chlorine is 17. Protons = atomic number. Neutrons = mass - number - atomic number.

Step1: Find number of protons

For both $Cl - 35$ and $Cl - 37$, the number of protons is 17 since the atomic number of chlorine is 17.

Step2: Find number of neutrons in $Cl - 35$

Number of neutrons in $Cl - 35=35 - 17 = 18$

Step3: Find number of neutrons in $Cl - 37$

Number of neutrons in $Cl - 37=37 - 17 = 20$

Step1: Convert percentage to decimal

The abundance of $Cl - 35$ is $75.771\%=0.75771$, and the abundance of $Cl - 37$ is $1 - 0.75771 = 0.24229$

Step2: Use the formula for average atomic mass

$Average\ atomic\ mass=(0.75771\times34.96885)+(0.24229\times36.96590)$
$=26.499977+8.959358$
$=35.459335\approx35.46\ amu$

Answer:

$20.18\ amu$

6.