0.07 mol sample of octane, c^6h^18 absorbed 3.5 x 10^3 j of energy. calculate the temp increase of octane if the molar heat capacity of octane is 254.0 j/k* mol

n = number of moles of sample of octane = 0.07 mol

Q = energy absorbed by a sample of octane = 3.5 x 10³ J

c = molar heat capacity of octane = 254.0 J/K* mol

ΔT = increase in temperature of octane = ?

Heat absorbed is given as

Q = n c ΔT

inserting the values

3.5 x 10³ J = (0.07 mol) (254.0 J/K* mol) ΔT

ΔT = (3.5 x 10³ )/((0.07) (254.0))

ΔT = 196.85 K

hence increase in temperature comes out to be 196.85 K

i think so it’s will be b yes i believe its

it's c. elements are placed in rows based on how may valencies electrons have.