The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is kj. the specific heats of ice, water, and steam are 2.09 j/g−k, 4.18 j/g−k, and 1.84 j/g−k, respectively. for h2o, δ hfus = 6.01kj/mol, and δhvap = 40.67 kj/mol.

Answer : The enthalpy change for converting 1 mole of ice at to water at is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

Now we have to calculate the enthalpy change.

where,

= enthalpy change

m = mass of water =

= specific heat of ice = 2.09 J/gk

= specific heat of water = 4.18 J/gk

= enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion :

= initial temperature of ice =

= final temperature of ice =

= initial temperature of water =

= final temperature of water =

Now put all the given values in the above expression, we get

    (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at to water at is, 7.712 KJ

where are the choice options?

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72