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3 years ago

Match the condensed structural diagrams with the correct names

Chemistry

1 answer:

AleksAgata [21]3 years ago

6 0

Answer:

Here's what I get

Explanation:

1. Names

I. CH₃-CH₂-COOH = 49. propanoic acid

II. CH₃-CH₂-OH = 46. ethanol

III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate

IV. H-O-CH₂-CH₂-CH₃ = 48. propan-1-ol

V. H-COO-CH₃ = 51. methyl methanoate

VI. CH₃-COOH = 50. ethanoic acid

2. Precursors

52. methyl propionate ⇒ methanol + propanoic acid

53. ethyl methanoate ⇒ ethanol + methanoic acid

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Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O

trasher [3.6K]

Answer: $\Delta H^{0} = -879.15 kJ/mol$

Explanation: Heats of formation is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

Standard Enthalpy Change is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

$\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}$

For the reaction

2NH₃ + 3N₂O → 4N₂ + 3H₂O

2(-46.2) + 3(82.05) 4(0) + 3(-241.8)

$\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]$

$\Delta H^{0}=-725.4-153.75$

$\Delta H^{0}=-879.15$

The standard enthalpy change for the reaction is $\Delta H^{0}=-879.15$ kJ

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2 years ago

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BabaBlast [244]

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I think this is right hope it helps

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3 years ago

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Answer:

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denis-greek [22]

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