Suppose 2.27g of lead(II) nitrate is dissolved in 300.mL of a 52.0mM aqueous solution of ammonium sulfate. Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved in it.

0.0457M is molarity of nitrate anion in the solution.

Explanation:

Molarity is defined as the ratio between moles of solute (In this case, nitrate ion) and liters of solution (The total volume of the solution is 300.0mL = 0.300L).

Thus, we need to convert mass of Lead(II) nitrate to moles using its molar mass (Molar mass Pb(NO₃)₂: 331.2 g/mol). Moles of Pb(NO₃)₂ = 1/2 Moles of NO₃⁻:

Moles Pb(NO₃)₂ and moles of NO₃⁻:

2.27g * (1mol / 331.2g) = 0.006854 moles Pb(NO₃)₂ * (2moles NO₃⁻ / 1mol Pb(NO₃)₂) = 0.0137moles NO₃⁻

Molarity is:

0.0137 moles NO₃⁻ / 0.300L =

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally