55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
ΔHrxn = [(1) -1675.5 ( kJ/mole) + (2) 0 ( kJ/mole)] - [(1) -824.3 ( kJ/mole) + (2) 0 ( kJ/mole)]
Explanation:
ΔHrxn = 2ΔHf (Al₂O₃) - ΔHf (Fe₂O₃)
Remember that for pure elements in their standard state of temperature and pressure by definition their standard heats of formation are zero.
ΔHrxn = 2(-1675.7) - (-824.3) kJ/mol
ΔHrxn = 2527 kJ/mol
Because of the wind and the cold from ur breath
Of course!!! they have as they have mass and occupy volume too...........
