The structure of
Alanine is shown below,
Except the carbon atom of carbonyl group which is
Sp² Hybridized, all remaining atoms are
Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Answer:
four (4)
Explanation:
Naphthalein is an organic compound with formula C
10H
8. It is the simplest polycyclic aromatic hydrocarbon, and is a white crystalline solid with a characteristic odor that is detectable at concentrations as low as 0.08 ppm by mass. As an aromatic hydrocarbon, naphthalene's structure consists of a fused pair of benzene rings. It is best known as the main ingredient of traditional mothballs.
The molecule is planar, like benzene. Unlike benzene, the carbon–carbon bonds in naphthalene are not of the same length. The bonds C1−C2, C3−C4, C5−C6 and C7−C8 are about 1.37 Å (137 pm) in length, whereas the other carbon–carbon bonds are about 1.42 Å (142 pm) long. This difference, established by X-ray diffraction is consistent with the valence bond model in naphthalene and in particular, with the theorem of cross-conjugation. This theorem would describe naphthalene as an aromatic benzene unit bonded to a diene but not extensively conjugated to it (at least in the ground state), which is consistent with two of its three resonance structures.
Because of this resonance, the molecule has bilateral symmetry across the plane of the shared carbon pair, as well as across the plane that bisects bonds C2-C3 and C6-C7, and across the plane of the carbon atoms. Thus there are two sets of equivalent hydrogen atoms: the alpha positions, numbered 1, 4, 5, and 8, and the beta positions, 2, 3, 6, and 7. Two isomers are then possible for mono-substituted naphthalenes, corresponding to substitution at an alpha or beta position. Bicyclo[6.2.0]decapentaene is a structural isomer with a fused 4–8 ring system.
Therefore four (4) double bonds will be added to give each carbon atom an octet structure.
Answer:
0.008945 atm
Explanation:
In the reaction:
2H2S(g) ⇌ 2 H2(g) + S2(g)
Kp is defined as:

<em>Where P is the pressure of each compound in equilibrium.</em>
If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:
H2S = 3.00 atm - 2X
H2 = 2X
S2: = X
Replacing:


0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶
Solving for X:
X = 0.008945 atm
As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>
<u>Answer:</u>
Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.
Unit is <em>
</em>
<em>C = Q/m. ∆T</em>
<em>C – Specific heat
</em>
<em>Q- heat energy (J)</em>
<em>M – Mass (Kg)</em>
<em>∆T- change in temperature (K) </em>
<u>Explanation:</u>
<em>Given data:</em>
<em>M= 140 g = 0.14 Kg</em>
<em>Q – 1080 Joules.</em>
<em>∆T – 98.4 – 62.2 = 36.2</em>
Substituting the given data in Equation
<em>Specific heat of Aluminium =
</em>
Answer:
K = 6.5 × 10⁻⁶
Explanation:
C₅H₆O₃ ⇄ C₂H₆ + 3CO
Use PV=nRT to find the initial pressure of C₅H₆O₃
P (2.50) = (0.0493) (0.08206) (473)
P = 0.78atm
C₅H₆O₃ ⇄ C₂H₆ + 3CO
0.78atm 0 0
0.78 - x x 3x
1.63atm = 0.78 - x + x + 3x
P(total) = 0.288atm
C₅H₆O₃ = 0.78 - 0.288
= 0.489atm
C₂H₆ = 0.288atm
CO = 0.846atm

= 0.379


= 6.5 × 10⁻⁶