When 1.010 of sucrose (c12h22o11) undergoes combustions in a bomb calorimeter, the temperature rises from 24.92 ◦ c to 28.33 ◦ c. find derxn for the combustion of sucrose in kj/mol sucrose. the heat capacity of the bomb calorimeter determined in a separate experiment is 4.90 kj/◦ c. ( -5.66 x 103 kj/mole).

Heat of combustion = 5.6 ×10³ kj/mol

Explanation:

Given data:

Mass of sucrose = 1.010 g

Initial temperature = 24.92 °C

Final temperature = 28.33 °C

Heat capacity of calorimeter = 4.90 KJ/°C

Heat of combustion = ?

Solution:

ΔT = 28.33 °C -  24.92 °C = 3.41 °C

Q = - c. ΔT

Q = 4.90 KJ/°C . 3.41 °C

Q = - 16.7 kj

Number of moles of sucrose :

Number of moles of sucrose = mass/ molar mass

Number of moles of sucrose = 1.010 g / 342.3 g/mol

Number of moles of sucrose = 0.003 mol

Heat of combustion:

Heat of combustion = Q/n

Heat of combustion = - 16.7 kj/0.003 mol

Heat of combustion = -5.6 ×10³ kj/mol

answer

the ions in solute supperate from eachouthere

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890