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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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1, 2 and 3 ........
Orlov [11]
2.) Average atomic mass =Σ (abundance x molar mass) /100 = (12.64 x 302.04 + 18.23 x 304.12 + 69.13 x 305.03) /100 =304.486 u(Dalton) . This is avg atomic mass.

8 0
3 years ago
Balance the following reaction in KOH (under basic conditions). What are the coefficients in for C3H8O2 and KMnO4 in the balance
GrogVix [38]

Answer:

Coefficient of C_3H_8O_2=3

Coefficient of KMnO_4=8

Explanation:

We are given that  a reaction in which C_3H_8O_2 reacts with KMnO_4

We have to find the coefficient of each reactants in balanced reaction

3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O

Coefficient is defined the constant  value multiplied with a reactant in a reaction.

Coefficient of C_3H_8O_2=3

Coefficient of KMnO_4=8

Coefficient of C_3H_2O_4K_2=3

Coefficient of MnO_2=8

Coefficient of H_2O=8

Coefficient of KOH=2

Hence, Coefficient of C_3H_8O_2=3 and coefficient of KMnO_4=8

7 0
3 years ago
HELP 10pts!
masha68 [24]

IM IN THAT CLASS TOO

Im abt to do it and ill send u all the answers when im

done, add me on sxxap or instaxxxgram so i can send it to u

my userneamen both: Bellaberrygood

5 0
3 years ago
What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?
MArishka [77]
Combined gas law is 
       PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
     P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm                                     P₂ = standard pressure = 1 atm
V₁ = 720.0 mL                                    T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
  
By substituting,
 3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
                                          V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL
5 0
3 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
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