Suppose that a small styrofoamtm cooler holds 15.0 l of water heated to 84 ∘
c. calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 33 ∘
c. assume no heat loss to the surroundings.

This can be solve by applying energy balance over the system. The energy gained by the dry ice is equal to the enrgy realesed by the water.

mwCpwdeltaT = mDAHDA

the Cp of water = 4.18 J/gC

enthalpy of sublimation H = 571 J/g

(1000 ml)(1g/ml)(4.18 J/gC)(84 C – 33 C) = mDA(571 j/g)

mDA =373 g dry iceThis can be solve by applying energy balance over the system. The energy gained by the dry ice is equal to the enrgy realesed by the water.

mwCpwdeltaT = mDAHDA

the Cp of water = 4.18 J/gC

enthalpy of sublimation H = 571 J/g

(1000 ml)(1g/ml)(4.18 J/gC)(84 C – 33 C) = mDA(571 j/g)

mDA =373 g dry ice

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: