You mix 125 ml of 0.154 m csoh with 60.0 ml of 0.456 m hf in a coffee cup calorimeter, and the temperature of both solutions rises from 24.4°c before mixing, to 35.9°c after the reaction. csoh (aq) + hf (aq) --> csf (aq) + h2o (l) what is the heat [q] for this reaction per mole of csoh? assume the densities of the solutions are all 1.00 g/ml and the specific heats of the solutions are 4.18 j/(g·k).

the heat of reaction per mole of CsOH is Δh = 461 kJ/mol

Explanation:

Since the number of moles present in the mixture is

n CsOH = 0.154 M *  0.125 L =  0.01925 moles

n HF = 0.456 M *  0.060 L =  0.02736 moles

then since a coffee cup calorimeter is a constant pressure calorimeter , from the first law of thermodynamics applied to enthalpy:

ΔH = Q - ∫VdP , since P=constant →dP=0 → ∫VdP=0, then

ΔH  = Q

where

ΔH = enthalpy change ( heat of reaction)

Q= m*cp*(T final - T initial)

where

m = mass of the solution

cp=  specific heats

T final = final temperature

T initial = initial temperature

then since the density ρ is the same for both components

m = ρ* V total , where V total = volume of the solution

therefore

ΔH  = m*cp*(T final - T initial) = ρ* V total * cp*(T final - T initial)

replacing values

ΔH = ρ* V total * cp*(T final - T initial) = 1.00 g/mL*(125 ml + 60 ml) * 4.18 J/(g·K) * (35.9°C - 24.4°C) = 8892.95 J

the heat of reaction per mole of CsOH is

Δh= ΔH/ n CsOH  = 8892.95 J/ 0.01925 moles = 461971 J/mol = 461 kJ/mol

Δh= 461 kJ/mol

43.452

step-by-step explanation:

284 times .153 = 43.452

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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