<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>
8.08 × atoms of hydrogen.
The enthalpy change in this scenario totals 7.205 KJ. The task is to compute the enthalpy variation during the conversion of 10.0 g of ice at -25.0°C into water at 80.0°C, factoring in specific heats and enthalpy for phase transitions.
