Answer:
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Answer:
1) C2H4(OH)2
Explanation:
A 1,2-ethanediol has an ethane structure consisting of two Carbon atoms with a hydrogen from each carbon substituted by a hydroxyl group. This makes it a 1,2-diol.
Explanation:
The O atom is sp3 in a water molecule, with two sigma bonds and two lone pairs of electrons like that in water. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp hybridised into two identical bonds and two identical bonds in acetylene.
The steric integer is therefore 2 because only sigma bonds are engaged in deciding hybridization, and its structure is linear.
The C atom is sp2 hybridised in ethene with single pi bond and three sigma identical bonds.
Thus the steric integer is 3, and its structure is planar trigonal.
The C atom is sp2 hybridized in ethene, with one pi bond and three sigma identical bonds.
The steric integer would therefore be 3 and its structure is planar trigonal.
The O atom is sp3 in a water molecule with two bond pairs and two lone pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp3 in a methane ring, with 4 bond pairs and no solitary pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
It is non-polar molecule.
CF₄ - dipole moment = 0D
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361
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