Wzorowane na: Chemistry SL Paper 2, specimen 2025, pytanie 4(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Informacja do zadań 4a-4f.
Nonadecane,
The enthalpy of combustion of is .
(i) Calculate the maximum energy released when 1.50 g of is burned completely.
(ii) Determine the maximum temperature increase when of water is heated by burning a 1.50 g sample of
(iii) Outline two assumptions on which the calculation in (b)(ii) relies.
(i) = «1.50 g ÷ [(19 × 12.01) + (40 × 1.01)] g mol = 1.50 g ÷ 268.59 g mol =» 0.005585 / 0.00558 «mol» ✔
«energy = 12 660 kJ mol × 0.005585 mol =» 70.7 «kJ» ✔
(ii) 70 700 J = 400.0 g × 4.18 J g K × Δ ✔
Δ = 42.3 «K» ✔
(iii) Any two of:
all the heat released is transferred to the water / no heat is lost to the surroundings ✔
no water evaporates ✔
the density of water is 1.00 g cm «so 400.0 cm³ is 400.0 g» ✔
combustion is complete ✔
the specific heat capacity used is that of pure water ✔
(i) M1 amount of nonadecane in mol; M2 energy in kJ (accept 70.6 to 70.7 kJ). (ii) M1 substitution into Q = mcΔT with the energy in J (or c in kJ); M2 ΔT = 42.3 K (accept 42.2 to 42.3 K); ECF from (i). (iii) [2 max], one mark per assumption.
Convert the energy to joules (or c to kJ g⁻¹ K⁻¹) before dividing; a temperature change in K is numerically the same in °C.
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