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Ber [7]
3 years ago
11

Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose mol

ecules and the water molecules resulting in an increase in the rate of hydrolysis.
True or false?
Chemistry
1 answer:
Sonja [21]3 years ago
8 0

Answer: The given statement is true.

Explanation:

When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.

Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.

Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.

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Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given) (exothermic)

Explanation:

           3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g)  +  17H₂O(g)   +  10CO₂(g)

ΔH⁰(f):  3(-365.6)Kj     1(-301)Kj    14(0)Kj       3(0)Kj    17(-241.8)Kj    10(-393.5)Kj

            = -1096.8Kj     = -301Kj     = 0Kj         = 0Kj      = -4110.6Kj    = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)  

∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

ΔH°(rxn for 14.11g  NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given)

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