Molarity of 0.50 mol sugar in 270 mL of solution.
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<h3><u>1. Lewis Structure</u></h3>There are three N-H bonds and one lone pair on the nitrogen atom in the Lewis structure of NH3. On hydrogen atoms, there are no lone pairs that can only hold two electrons.
The lewis structure is as shown in the diagram.
<h3><u>2. Perspective Drawing</u></h3>A molecule is shown from a viewpoint with its atoms' bonds pointing either in your direction (bolded wedge) or away from you (hash wedge).
The perspective drawing of NH3 is given here
<h3><u>3.Number of atoms bonded to central atom </u></h3>Nitrogen is the central atom in the Lewis structure of NH3, which also contains one lone pair and is bound to the three hydrogen atoms.
<h3><u>4.Number of non-bonding electron pairs on the central atom</u></h3>Three bond pairs and two lone pairings exist. This is due to the fact that nitrogen contains five electrons in its outer shell, three of which are bound to hydrogen atoms and two of which are free.
<h3><u>5. Electronic geometry:</u></h3>The electronic geometry of nitrogen is based on a tetrahedral arrangement of electron pairs, and ammonia likewise possesses four electron pairs. There is only one lone pair because there are only three connected groupings. But since the lone pairs are "invisible," the ammonia has a pyramidal form.
<h3><u>6. Molecular geometry with ideal bond angles</u></h3>The molecular geometry of ammonia it has a trigonal pyramidal or distorted tetrahedral structure.
The bond angle in ammonia is less than the standard 109.5⁰. The bond angle is 107⁰
<h3><u>7.Hybridization of central atom</u></h3>The core nitrogen (N) atom in the NH3 molecule has a steric number of (3 + 1) = 4, which causes Sp3 hybridization.
<h3><u>8. Polarity:</u></h3>Because of its asymmetrical form, a trigonal pyramidal structure, and the different electronegativities of N(3.04) and H(2.2), the NH3 (ammonia) molecule is polar in nature .
To know more about ammonia, you can refer to:
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(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
Answer:
<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2><h3>Valence electrons are the electrons in the outermost shell, or energy level, of an atom. </h3><h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>Explanation:
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