HELP 1) A 400g sample of alcohol (c = 2.43 J/g°C) at 16°C is mixed with 400g
of water (c = 4.19 J/g°C) at 85°C. What is the final temperature of the
mixture?

Explanation:

Given data:

Mass of alcohol = 400 g

Specific heat capacity of alcohol = 2.43 J/g°C

Initial temperature of alcohol = 16°C

Mass of water = 400 g

Specific heat capacity of water = 4.19 J/g°C

Initial temperature of water = 85°C

Final temperature of mixture = ?

Solution:

Equation:

m₁c₁ (T₂-T₁ ) = m₂c₂(T₂-T₁)

by putting values,

400 × 4.19  × (T₂ -  85°C) = 400 × 2.43  × (T₂ - 16°C)

1676  × (T₂ -  85°C) = 972 × (T₂ - 16°C)

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: