Find the enthalpy change per mole of sodium when sodium reacts with water 8 grams of sodium reacts with 227 cm of water, producing a temperature change from 298 Kto 308.4 K The specific heat capacity of water is 4.18 J/K 9.

The enthalpy change per mole of sodium when sodium reacts with water is 28.36 kJ

Step-by-step explanation:

The given parameters are;

Mass of sodium = 8 grams = 0.008 kg

Volume of water = 227 cm³ = 0.000227 m³

Initial water temperature, T₁ = 298 K

Final water temperature, T₂ = 308.4 K

Specific heat capacity of water = 4.18 J/(K·g)

Density of water = 1000 kg/m³

Mass of water = 1000×0.000227 = 0.227 kg = 227 g

Heat change, ΔH = m×c×ΔT = m×c×(T₂ - T₁)

ΔH = 227*4.18*(308.4 - 298) = 9868.144 J = 9.868 kJ

Molar mass of sodium = 22.989769 g/mol

Therefore, 8 grams contains;

8/22.989769 = 0.348 moles of Na

Which gives the enthalpy change per 0.348 moles of Na when sodium reacts with water = 9.868 kJ

Therefore, when one mole of Na reacts with water we have;

The enthalpy change per 0.348/0.348 = 1 mole of Na when sodium reacts with water = 9.868 kJ/0.348 = 28,358.3 J = 28.36 kJ.

the formula of a slope:

we have the points (-2, 7) and (2, 3). substitute: