Mass number = 238
No of protons = 92
No of electrons = 92
No of neutrons = 238-92 = 146
Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
Answer:
The correct option is: b. has a downward projection (on the opposite side from the terminal CH₂OH group).
Explanation:
The Haworth Projection depicts the three-dimensional cyclic structure of the monosaccharides.
The monosaccharide, D-glucose predominantly exists in the pyranose form and is known as D-glucopyranose. The D-glucopyranose has two anomeric forms: α- and the β- form and the carbon-1 (C-1 ) is known as the anomeric carbon.
<u>In the Haworth projection of α-D-glucopyranose</u>, the <u>hydroxyl group of anomeric carbon is projected in the downward direction</u>. Whereas, the <u>terminal CH₂OH group on the carbon-5 (C-5), is projected in the upward direction.</u> Therefore, they are on the <u>opposite sides of the plane of ring</u>.