The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:
They are held together by hydrogen bonds
Explanation:
Hydrogen bonds are special dipole-dipole attractions between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom(oxygen or nitorgen or fluorine).
Such molecules includes water, alkanoic acids, ammonia and amines.
A hydrogen nucleus has a high concentration of positive charge. The bond is actually an electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom(O or N or F).
Hydrogen bonds are very effective in binding molecules into larger units. Most substances that joins with hydrogen bonds have a higher boiling point and lower volatility.
This is why we have a strong intermolecular bond between water molecules.
Answer:
Study the bone structure of limbs.
Explanation:
Study the bone structure of limbs can be used to find out the relatedness and relationship among three species because the pattern of bones structure is similar but the structure is different from one another due to different environmental conditions. This study provides important and useful information whether the three species are closely related or not. If they have similar bone structure than we can say that there are more chances that they are closely related to each other.
Momentum = (mass) x (velocity) = (1,100) x (30) =
33,000
kg-m/sec due east
