The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371.2 K. What is the enthalpy of combustion of octane? The specific heat capacity of water is 4.18 J/K g. A. -1226 kJ/mol
B. -5448 kJ/mol
C. 293.25 kJ/mol
D. 1226 kJ/mol

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water =

= initial temperature of water = 293.0 K

= final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

Now we have to calculate the enthalpy of combustion of octane.

where,

= enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

the force on the thing and the distance the force moves the thing

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.