Calculate the enthalpy change (in joules) involved in converting 5.00 grams of water at 14.0 °c to steam at 115 °c under a constant pressure of 1 atm. the specific heats of ice, liquid water, and steam are, respectively, 2.03, 4.18, 1.84 j/g-k, and for water δhfusion = 6.01 kj/mole and δhvap = 40.67 kj/mole. hints

Enthalpy change is the term given to the concentration of the heat absorbed or evolved in a reaction carried out at a constant pressure. It is given by the symbol ΔH.

Enthalpy change = m (Cwater × dT + Hvap + Csteam × dT

dHvap = 40.67 kJ/mol = 40.67 × 103/18 = 2259.4 J/g

Enthalpy change = 5.00 [4.18 × (100-14) + 2259.4 + 1.84 × (115-100)]

= 13232.4 J

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

it is a good conductor

explanation: