Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Hthe heat required to change the temperfature of 100 grams of water from 25 to 55 c is calculated as below
Q(heat) = M(mass) x C(specific heat capacity) x delta T(change in temperature)
M= 100 grams
C= 4.18 j/g/c
delta T= 55-25 =30c
Q=100 g x4.814 j/g/c x 30c = 12552 joules
The freezing point of the sucrose solution is -0.435°C.
<h3>What is the freezing point of the solution?</h3>
The freezing point of the solution is determined from the freezing point depression formula below:
Kf(H₂O) = 1.86 Cm
m is molality of solution = moles of solute/mass of solvent
moles of sucrose = 8.0/342.3 = 0.0233 moles
m = 0.0233/0.1 = 0.233 molal
ΔT = 0.233 m * 1.86°C/m.
ΔT = 0.435 °C.
Freezing point of sucrose solution = 0°C - 0.435°C
Freezing point of sucrose solution = -0.435°C.
In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.
Learn more about freezing point depression at: brainly.com/question/19340523
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mol of Na2CO3 = 2.36 x 10⁻⁴
<h3>Further explanation</h3>
Given
Mass : 0.025 g of Na2CO3
Required
moles
Solution
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass :
mol = mass : molar mass
mass = mol x molar mass
Input the value :
mol = mass : MW Na2CO3
mol = 0.025 g : 106 g/mol
mol = 2.36 x 10⁻⁴
