Calculate the quantity of heat gained or lost:

a) 3.50 mol H2O freezes at 0.0 oC
b) 0.44 mol of steam condenses at 100 oC

To calculate the quantity of heat gained or lost, we can use the equation:

q = m * ΔH

Where:
q = quantity of heat gained or lost
m = mass
ΔH = enthalpy change

a) 3.50 mol H2O freezes at 0.0 oC:

First, we need to calculate the mass of H2O.

m = n * M

Where:
n = number of moles
M = molar mass of H2O (18.015 g/mol)

m = 3.50 mol * 18.015 g/mol
m ≈ 63.0525 g

Next, we need to determine the enthalpy change (ΔH) for H2O freezing.

ΔH = -6.01 kJ/mol (enthalpy of fusion for H2O)

Now, let's calculate the quantity of heat gained or lost.

q = m * ΔH
q = 63.0525 g * (-6.01 kJ/mol)
q ≈ -379.029 kJ

Therefore, the quantity of heat lost when 3.50 mol of H2O freezes at 0.0 oC is approximately 379.029 kJ.

b) 0.44 mol of steam condenses at 100 oC:

First, let's determine the mass of steam.

m = n * M

Where:
n = number of moles
M = molar mass of H2O (18.015 g/mol)

m = 0.44 mol * 18.015 g/mol
m ≈ 7.9266 g

Next, we need to determine the enthalpy change (ΔH) for steam condensation.

ΔH = -40.7 kJ/mol (enthalpy of condensation for steam)

Now, let's calculate the quantity of heat gained or lost.

q = m * ΔH
q = 7.9266 g * (-40.7 kJ/mol)
q ≈ -322.440 kJ

Therefore, the quantity of heat lost when 0.44 mol of steam condenses at 100 oC is approximately 322.440 kJ.

To calculate the quantity of heat gained or lost, we can use the formula:

q = m * c * ΔT

where "q" is the heat gained or lost, "m" is the mass, "c" is the specific heat capacity, and "ΔT" is the change in temperature.

a) 3.50 mol of H2O freezes at 0.0 oC:
First, we need to convert moles to mass. The molar mass of water (H2O) is approximately 18.015 g/mol. Therefore, the mass of 3.50 mol of water is:

mass = moles * molar mass
mass = 3.50 mol * 18.015 g/mol

Next, we need to calculate the heat gained or lost during the phase change from liquid to solid. The specific heat of fusion (heat required to change 1 gram of substance from a solid to a liquid or vice versa) for water is 334 J/g. Since this is a change from liquid to solid, it will release heat. Therefore, the change in temperature (ΔT) is 0.0 oC - (-273.15 oC) = 273.15 oC.

Finally, we can substitute these values into the formula:

q = mass * c * ΔT
q = (3.50 mol * 18.015 g/mol) * (334 J/g) * (0.0 oC - (-273.15 oC))

b) 0.44 mol of steam condenses at 100 oC:
First, we need to convert moles to mass. The molar mass of water (H2O) is approximately 18.015 g/mol. Therefore, the mass of 0.44 mol of water is:

mass = moles * molar mass
mass = 0.44 mol * 18.015 g/mol

Next, we need to calculate the heat gained or lost during the phase change from gas to liquid. The specific heat of vaporization (heat required to change 1 gram of substance from a liquid to a gas or vice versa) for water is 40.7 kJ/mol. Since this is a change from gas to liquid, it will release heat. Therefore, the change in temperature (ΔT) is 100 oC - 0.0 oC = 100 oC.

Finally, we can substitute these values into the formula:

q = mass * c * ΔT
q = (0.44 mol * 18.015 g/mol) * (40.7 kJ/mol) * (100 oC - 0.0 oC)

a.

3.50 x molar mass = grams H2O.
q = grams H2O x heat fusion.

b.
0.44 moles H2O x molar mass = ?? grams steam.
q = mass steam x heat vaporization.