The vapor pressure of water at 50ºC will be greater than that at 10ºC because of the added energy and thus greater movement of the water molecules. If one knows the ∆Hvap at a given temperature, one can calculate the vapor pressure at another temperature. This uses the Clausius-Clapeyron (sp?) equation. It turns out the vapor pressure of water at 10º is 9.2 mm Hg, and that at 50º is 92.5 mm Hg.
To calculate the mass, go from atoms to moles and moles to grams:
Atoms ........ Moles.......... Grams
From atoms to moles, use the Avogadro’s no. which is 6.022 x
10^23 and from moles to grams use atomic mass of Pt that is 195.08g.
So, 4.91 x 10^21 Pt atoms x (1 mole Pt / 6.022x10^23 Pt atoms) x (195.08g Pt
/ 1 mole Pt) = 1.59g Pt
Mass of 4.91 x 10^21 Pt atoms = 1.59g Pt
Answer:
the overall reaction is
P4(s) + 10Cl2(g) => 4PCl5(g)
and its ΔH is
ΔH overall = (-1608 KJ)
Explanation:
given that
P4(s) + 6Cl2(g) => 4PCl3(g) , ΔH1 = –1148 kJ
4PCl3(g) + 4Cl2(g) => 4PCl5(g) , ΔH2 =–460 kJ
summing both equation gives
P4(s) + 6Cl2(g) +4PCl3(g) + 4Cl2(g) => 4PCl3(g) + 4PCl5(g)
but 4PCl3(g) is in both sides , so is cancelled out
P4(s) + 6Cl2(g)+ 4Cl2(g) => 4PCl5(g)
summing the Cl2(g)
P4(s) + 10Cl2(g) => 4PCl5(g) → overall reaction
(can be verified that the reaction is balanced)
since we sum both equations the ΔH of the overall reaction is also the sum of the ΔH of the individual reactions.
ΔH overall = ΔH1 + ΔH2 = (–1148 kJ) + (–460 kJ) = (-1608 KJ)