Answer
Neon
Work Step by Step
$6.022 \times 10^{23} particles$ = $1 mol$
If it 25.00% is argon, we have $0.2500$ mol $Ar$
1. Find the mass of argon (Molar mass = 39.95g/mol):
$0.2500 mol \times \frac{39.95 g}{1 mol} = 9.987$ g
2. Calculate the mass of the other element:
25.12 g - 9.987 g = 15.13 g
3. If it is 75.00% the other element, then we have 0.7500 mol $X$. Calculate its molar mass:
$\frac{15.13g}{0.7500mol} = 20.17$ g/mol.
4. Using a periodic table, we can see that the identity of the element with molar mass equal to $20.17$ is "Neon" (Ne).
