The answer is 2 mol of H₂O will be produced.
The balanced equation for the chemical reaction is:
<span>c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g)
5 moles of O</span>₂ produces 4 moles of H₂O, and when there is 2.5 mol of O₂, moles of H₂O will be:
2.5 x 4/5 = 2 mol of H₂O
Answer:
i. n = 5
ii. ΔE = 7.61 ×
KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 ×
(
-
)
(1/434 ×
) = -2.178 ×
(
)
⇒ 434 ×
= (1/-2.178 ×
)
But,
= 2
434 ×
= (1/2.178 ×
)
434 ×
× 2.178 ×
= 
⇒
= 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 ×
) ÷ (434 ×
)
= (1.9878 ×
) ÷ (434 ×
)
= 4.58 ×
J
= 4.58 ×
KJ
But 1 mole = 6.02×
, then;
energy in KJ/mole = (4.58 ×
KJ) ÷ (6.02×
)
= 7.61 ×
KJ/mole
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:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)

