Answer:
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.
Before you begin titration, drops of indicator are added to the Erlenmeyer flask. This happens when preparing to measure pH change.
The substances which will exist in molecular form in solution are-
A. CH₃NH₂
D. CH₃OH
The methyl amine (CH₃NH₂) and methanol (CH₃OH) is organic compounds which are hardly become ionic in nature and prefers to form as molecular state in the solution phase.
On the other hand B. LiOH and C. NH₄OH are inorganic compound and highly ionic in nature and remain as ionic state in solution.
Those are-
LiOH (s) → Li⁺(s) + OH⁻ (s)
NH₄OH→ NH₄⁺ (s) + OH⁻ (s)
Answer: Andre. If it is used as an antifreeze, one would infer that it's freezing point is less than water.
Explanation:
The freezing point of ethylene glycol alone (100%) is -12.9oC. That is somewhat less than the freezing point of water but nowhere near the freezing point of water containing 70% ethylene glycol.
Chloroplasts is the answer
