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zimovet [89]
3 years ago
8

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

xsafari

Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

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What is the pH of a 0.640 M solution of C₅H₅NHBr (Kb of C₅H₅N is 1.7 × 10⁻⁹)?
Elden [556K]

The pH of a 0.64 M solution of pyridine (C₅H₅N) is 9.52.  

<h3>What is pH ?</h3>

A figure expressing the acidity or alkalinity of a solution on a logarithmic scale on which 7 is neutral, lower values are more acid and higher values more alkaline.

The equation for the protonation of the base pyridine is the following:

C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻   (1)

Kb = 1.7 × 10⁻⁹ (Given)

To calculate the pH of the solution we need to use the following equation:

pH + pOH = 14

<em>pH = 14 - pOH</em>

     =14 - [-log[OH⁻]]

    = 14 + log[OH⁻]

Now, we need to find the concentration of the OH⁻ ions. Since pyridine is a weak base, at the equilibrium we have (eq 1):

C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

1.7 × 10⁻⁹  =[C₅H₅NH⁺]  [OH⁻]  /  [C₅H₅N]

                = x . x / 0.64-x

1.7 × 10⁻⁹ (0.64-x) - x² = 0

Solving the above quadratic equation for x, we have :

  • x₁ = -3.32 x 10⁻⁵
  • x₂ = 3.32 x 10⁻⁵

Now, We can calculate the pH, after taking the positive value, x₂, (concentrations cannot be negative) and entering into above equation :

<em />

<em>pH = </em>14 + log[OH⁻]

     = 14 + log (3.32 x 10⁻⁵)

 

     = 9.52

Therefore, the pH of the solution of pyridine is 9.52.

Find more about pH here:

brainly.com/question/8834103?referrer=searchResults

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1 year ago
Assuming 1 mol of Fe3+ and 2 mol of SCN- were allowed to react and reach equilibrium. 0.5 mol of product was formed. The total v
zaharov [31]

Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

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a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

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