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Natali5045456 [20]
3 years ago
12

What is the approximate bond angle around the central carbon atom in acrolein?

Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

The approximate bond angle around the central carbon atom in acrolein is 120°.

Explanation:

The structure of acrolein is shown in the attachment. From the structure, we can deduce that the central carbon atom is in an sp2 hybridization (Atoms with a double bond hybridize in an sp2 fashion).

Atoms with sp2 hybridization have trigonal planar geometry, in this kind of hybridization, bonds are oriented the farthest away possible from each other, to minimize overlapping and the angle that allows that is 120°. 

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weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

6 0
3 years ago
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Answer:

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Explanation:

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