Answer:
The correct choice for your inquiry is option A, Argon.
Explanation:
Isotope Atomic mass Percent (%)
1 35.9675 0.337
2 37.9627 0.063
3 39.9624 99.6
To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)
Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)
Average atomic mass = 0.1212 + 0.0239 + 39.8025
Average atomic mass = 39.9476
Theoretical Atomic mass
a) Ar 39.95
b) K 39.10
c) Cl 35.45
d) Ca 40.08
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.
Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
