Answer:
shorter
longer
Explanation:
The carbon-carbon bond length in ethylene is <u>shorter</u> than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is <u>longer</u> the HCH bond angle in ethane.
The objective of this question is to let us understand the concept of Bond Length and Bond angle among the unsaturated aliphatic hydrocarbons (i.e alkanes, alkenes and alkynes).
The variation in bond angles of unsaturated aliphatic hydrocarbons can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.
The VSEPR model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs consist of the bond pairs and lone pairs. All the electron pairs( lie charge ) will then orient themselves in such a way to minimize the electrostatic repulsion between them.
As the number of the lone pairs increases from zero to 2 ; the bond angles diminish progressively.
However;
Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³ hybrid orbitals etc .
The shape of the geometry of this compound hence determines their bond angle.
The shape of the geometry of ethane is tetrahedral which is 109.5° in bond angle while that of ethylene is trigonal planar which is 120°.
This is why the HCH bond angle in ethylene is longer the HCH bond angle in ethane .
The doubling the amount will change the gibbs free energy as it is an extensive property which depends upon the the amount of the substance
However as asked in question the DeltaG has unit of kcal /mol
So we have already defined the amount of substance to be one mole this means the value per mole will be same irrespective of the amount taken as we are reporting it for a fixed one mole of a substance
Hence answer is
-100 kcal/mol
Answer:
(a) 
(b) 
(c) 
(d) 
(e) 
Explanation:
To calculate de pH of an acid solution the formula is:
![pH = -Log ([H^{+}]) = 1](https://tex.z-dn.net/?f=pH%20%3D%20-Log%20%28%5BH%5E%7B%2B%7D%5D%29%20%3D%201)
were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.
(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.
(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:
![K_{w} =[H^{+} ][OH^{-}]=10^{-14}](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%5D%3D10%5E%7B-14%7D)
clearing the ![[H^{+} ]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D)
![[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
(d) is a weak base so it is necessary to solve the equilibrium first, knowing 
The reaction is 
→ 
so the equilibrium is
clearing the <em>x</em>
![x=[H^{+}]=4.93x10^{-10}](https://tex.z-dn.net/?f=x%3D%5BH%5E%7B%2B%7D%5D%3D4.93x10%5E%7B-10%7D)
Answer:
A. H2O
Explanation:
Let us first define the three types of bonds:
1. Nonpolar Covalent: electronegativity difference < 0.4
2. Polar Covalent: electronegativity difference between 0.4 and 1.8
3. Ionic: electronegativity difference > 1.8
This will help us eliminate choices C and D:
-NaCl has a electronegativity difference of 3.0 - 0.9 = 2.1 (ionic bond)
-Cl2 has a electronegativity difference of 3.0 - 3.0 = 0 (nonpolar covalent bond)
However, we still have two more options, A and B, but they are not diatomic for us to use the electronegativity differences with.
We must now consult their geometries. Because CO2 has a linear geometry (O=C=O), the two sides will cancel each other out, resulting in a nonpolar covalent bond. At this point, by process of elimination, we can already determine the answer to be A. H2O. We can verify this by looking at the geometry of H2O, which is bent (H-O-H; imagine the O is above the H's, I cannot draw it in this response). H2O's bent geometry classifies it as polar covalent; the electrons are slightly more attracted towards the O, the more electronegative element. Side note: this makes the O slightly more negative in charge, whilst the H's are slightly more positive in charge.
P.S. I apologize for not being able to draw and demonstrate that last paragraph, but I hope you get a general idea. You can search up the "H2O geometry" and "CO2 geometry" to get a better idea! :)
