Answer:
84.30 mm Hg
Explanation:
In 100 cm³ of solution we have: 40 cm³ C6H6 and 60 cm³ CCl4. Given the densities we can calculate their masses and number of moles, and since by Raoult´s law
Ptotal = XAPºA + XBPºB
where XA= mol fraction =na/(na +nb) and PºA vapor pressure pure of pure component A
m C6H6 = 40 cm³ x 0.87865 g/cm³ = 35.146 g
mol C6H6 = 35.146 g/ 78.11 g/mol = 0.45 mol
mass CCl4 = 60 cm³ x 1.5940 g/cm³ = 95.640 g
mol CCl4 = 95.640 g / 153.82 g/mol = 0.62 mol
mol tot = 1.07
XC6H6 = 0.45/ 1.07 = 0.42 XCCl4 = 0.62/1.07 =0.58
Ptot (mmHg) = 0.42 x 74.61 + .58 x 91.32 = 84.30 mmHg
Chemical nomenclature, replete as it is withcompounds with complex names, is arepository for some very peculiar and sometimes startling names. A browse through the Physical Constants of Organic Compounds in the CRC Handbook of Chemistry and Physics (a fundamental resource) will reveal not just the whimsical work of chemists, but the sometimes peculiar compound names that occur as the consequence of simple juxtaposition. Some names derive legitimately from their chemical makeup, from the geographic region where they may be found, the plant or animal species from which they are isolated or the name of the discoverer.
Answer:
ΔHr = -86.73 kJ/mol
Explanation:
Using Hess's law, you can calculate ΔH of any reaction using ΔH°f of products and reactants involed in the reaction.
<em>Hess law: ∑nΔH°f products - ∑nΔH°f reactants = ΔHr</em>
<em>-Where n are moles of reaction-</em>
For the reaction:
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
Hess law is:
ΔHr = ΔH°f Fe(OH)₃ - ΔH°f Fe³⁺ - 3×ΔH°f OH⁻
Where:
ΔH°f Fe(OH)₃: −824.25 kJ/mol
ΔH°f Fe³⁺: −47.7 kJ/mol
ΔH°f OH⁻: −229.94 kJ/mol
Replacing:
ΔHr = −824.25 kJ/mol - (−47.7 kJ/mol) - (3×-229.94 kJ/mol)
<em>ΔHr = -86.73 kJ/mol</em>
Cl, because a higher electronegativity indicates that the atom attracts more electrons.
