Question

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Answer 1

✔✔Here is your answer in the attachment ✔✔

Answer 2

Answer:

$\bold{\huge{\color{red}{\mathfrak {Your\:Solution}}}}$

Explanation:

$Combustion \: of \: octane \: is \: as \: follows:$

C8H18 + 25 /2 O2 →8CO2 + 9H2O

$Mass\:of\:octane, m = 1.25 g\:or\:0.00125g$

$Heat\:capacity, C = 8.93 kJ K ^{ - 1} </p><p>$

$change \: in \: temperature,t = 300.78 - 295.05$

$= 6.73k$

$Heat \: transferred, Q$

= m C ∆T

$= 0.00125 \times 8.93 \times 6.73$

$= 0.075kj$

$Enthalpy \: of \: combustion \: is \: enthalpy \: of \: reaction. \\ In \: reaction \: 1 \: mole \: of \: octane \: is \: used. \\ So, \: enthalpy \: of \: 1 \: mole \: of \: octane \: can \: be \: calculated \: as \: follows:$

$For \: 1.25 \: g \: heat \: transferred \: is \: 0.075 \: kJ \\ For \: 1 \: mole \: or \: 114 \: g \: of \: octane \: heat \: transferred \: will \: be \: \frac{0.075}{1.125} \times 114$

$= 6.45kJ$