Suppose that 3.33 g of acetone at 25.0 °C condenses on the surface of a 44.0-g block of aluminum that is initially at 25 °C. If the heat released during condensation goes only toward heating the metal, what is the final temperature (in °C) of the metal block?

68.6 °C

Explanation:

From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'

Q = Q'

Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g

Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown

So, mL = m'c(θ₂ - θ₁)

θ₂ - θ₁ = mL/m'c

θ₂ = mL/m'c + θ₁

substituting the values of the variables into the equation, we have

θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C

θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C

θ₂ = 43.56 °C + 25 °C

θ₂ = 68.56 °C

θ₂ ≅ 68.6 °C

So, the final temperature (in °C) of the metal block is 68.6 °C.

the answer would be a

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.