Calculate the enthalpy of the reaction: 4 B(s) + 3 O2(g) --> 2 B2O3(s) given the following pertinent information: 1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ 2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ 3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ 4. H2O (l) → H2O (g), ΔHoD = +44 kJ Express your answer numerically, to four significant figures and in terms of kJCalculate the enthalpy of the reaction: 4 B(s) + 3 O2(g) --> 2 B2O3(s) given the following pertinent information: 1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ 2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ 3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ 4. H2O (l) → H2O (g), ΔHoD = +44 kJ Express your answer numerically, to four significant figures and in terms of kJ

ΔH = -2552kJ

Explanation:

Using Hess's law, it is possible to obtain ΔH of a reaction by the sum of similar reactions, thus:

1. B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ

2. 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ

3. H2 (g) + 1/2 O2 (g) → H2O (l), ΔHoC = −285 kJ

4. H2O (l) → H2O (g), ΔHoD = +44 kJ

Subtracting of 2*(2) - 2*(1):

4 B (s) + 6 H2 (g) +  6 O2 (g) → 2 B2O3 (s) + 6 H2O (g)

ΔH' = 2*36kJ - 2*2035kJ = -3998kJ

Now, this reaction - 6*(3):

4 B (s) + 3 O2(g) + 6 H2O(l) → 2 B2O3 (s) + 6 H2O (g)

ΔH'' = -3998kJ - (6*-285kJ)

ΔH'' = -2288kJ

The last reaction - 6*(4):

4 B (s) + 3 O2(g) → 2 B2O3 (s)

ΔH = -2288kJ - 6*44kJ

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.