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KIM [24]
1 year ago
7

the double bond in ethene is made up of a pi bond and a sigma bond formed by lateral overlap of two p orbitals

Chemistry
1 answer:
Deffense [45]1 year ago
3 0

The unhybridized pz orbitals on each carbon overlap to a π bond (pi).The sigma bond framework of the ethylene molecule is produced by the overlap of hybrid orbitals or by the interaction of a hybrid orbital and a 1s hydrogen orbital.

Each carbon still has its unhybridized pz orbital, though. Sigma bond are typically the only types of single bonding between atoms. One sigma bond and two pi bonds make up triple bonds. One sigma () bond makes up a single bond, one and one pi () bond makes up a double bond, and one and two bonds make up a triple bond.

To learn more about bond, click here.

brainly.com/question/10777799

#SPJ4

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NaNO3     and  H2O

The  equation  for  reaction  is      as folows
NaOH  + HNO3  =  NaNO3  +  H2O 
that  is  1  mole  of  NaOH  reacted  with  1  mole  of  HNO3  to  form  1  mole  of  NaNO3  and  1  mole  of  H2O
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The atomic radius of main group elements generally increases down a group because
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3 years ago
A student mixes a 10.0 ml sample of 1.0 m naoh(aq) with a 10.0 ml sample of 1.0 m hcl(aq) in a polystyrene container. the temper
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5 0
2 years ago
A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

7 0
2 years ago
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