How much heat is evolved in converting 1.00 mol of steam at 145.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2.01 j/(g⋅∘c) and of ice is 2.09 j/(g⋅∘c).

Heat evolved = 57.9 kJ i.e. -57.9 kJ

Explanation:

The transition from steam at 145 C to ice at -55 C involves the following steps:

Step 1: Heat involved in the change from steam at 145 C to steam at 100 C

Moles of steam (H2O)= 1 mole

Molar mass of H2O = 18 g/mol

Step 2: Phase change from steam at 100 C to water at 100 C

where ΔH(cond) = heat of condensation = -40.7 kJ/mol

n = number of moles

Step 3: Heat involved in the change from water at 100 C to water at 0 C

Step 4: Phase change from water at 0 C to ice at 0 C

where ΔH(fusion) = -6.01 kJ/mol

n = number of moles

Step 5: Heat involved in the change from ice at 0 C to ice at -55 C

Step 6: Total heat evolved

q = -1.628 -40.7 -7.524-6.01-2.069 = -57.9 kJ

answer; i do believe the correct answer is (river); /// good luck

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"