When 0.1523 g of liquid pentane (c 5 h 12 ) combusts in a bomb calorimeter, the temperature rises from 23.7 ∘ c to 29.8 ∘ c. what is δu rxn for the reaction in kj/mol pentane? the heat capacity of the bomb calorimeter is 5.23 kj/ ∘ c?

Q cal = C cal * ΔT

= 5.23 KJ/°C *(29.8-23.7)°C

= 31.903 KJ

This heat is supplied by pentane

Molar mass of C₅H₁₂ = 5*MM(C) + 12*MM(H)

= 5*12.01 + 12*1.008

= 72.146 g/mol

mass of C₅H₁₂ = 0.1523 g

we have below equation to be used:

number of moles of C₅H₁₂,

n = mass of C₅H₁₂/molar mass of C₅H₁₂

=(0.1523 g)/(72.146 g/mol)

= 2.111 *10⁻³ mol

delta U = -Qcal /number of mol

= -31.903 KJ / 2.111*10⁻³ mol

= -15113 KJ/mol

i know insulin is primary but i dont know the others im sorry

explanation: