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erica [24]
3 years ago
11

What is the molecular geometry of the left carbon atom (circled) in acetic acid? The left carbon atom is attached to three hydro

gen atoms and one carbon atom through single bonds. The right carbon atom is attached to a carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond.

Chemistry
1 answer:
levacccp [35]3 years ago
4 0

Answer:  Tetrahedral geometry

Explanation: The structure of the acetic acid is shown in the image below.

The left carbon atom is the one which is attached to three hydrogen atoms through single bonds and to one carbon atom through single bond as well.

Thus the orbitals which are used in the process of the formation of the chemical bond between these 4 are sp3 orbitals. And these orbitals results in the formation of the tetrahedral geometry.

The right carbon atom that is attached to a  carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond has trigonal planar geometry which involves the sp2 orbitals for the formation of the bond.

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The products of a neutralization reaction are ________ and ________ a salt.
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Explanation:

Neutralization reaction is a type of chemical reaction in which an acid and a base react together to form a salt <u><em>and water</em> </u>as products.

7 0
2 years ago
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Calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) where ΔH∘ = -687 kJ and ΔS∘ = -169 J/
aivan3 [116]

Answer:

-0.129V

Explanation:

The change in free energy is obtained from the given parameters after which the value is now applied to obtain the cell potential in volts from the formukar shown in the solution below.

6 0
3 years ago
The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.
Scrat [10]
We can set up an ICE table for the reaction:                      
                      HClO          H+     ClO-
Initial              0.0375       0        0
Change         -x               +x      +x
Equilibrium    0.0375-x     x        x

We calculate [H+] from Ka:     
     Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to         
     3.0x10^-8 = (x)(x)/0.0375     
     3.0x10^-8 = x2/0.0375     
     x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9     
     x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:     
     pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
6 0
3 years ago
93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP
Dahasolnce [82]
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
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n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

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what about  3.72  x10^-3 moles

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3 0
3 years ago
Dilute 0.407M &amp; 2.56L solution to 7.005L. CF?
Alenkinab [10]

Answer:

C₂ = 0.149 M

Explanation:

Given data:

Initial concentration = 0.407 M

Initial volume = 2.56 L

Final volume = 7.005 L

Final concentration = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Initial concentration

V₁ = Initial volume

C₂ = Final concentration

V₂ =Final volume

Now we will put the values.

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C₂ = 1.042 M.L / 7.005 L

C₂ = 0.149 M

5 0
3 years ago
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