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liubo4ka [24]
3 years ago
11

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? label each atom with the appropriate hybridization?

Chemistry
1 answer:
Alchen [17]3 years ago
8 0
In the structure of propane, CH3-CH2-CH3, there are 10 total bonds. Two carbon-carbon single bonds, and eight carbon-hydrogen single bonds. Each of these single bonds is also known as a sigma bond. Every hydrogen atom in this molecule uses an s-orbital to form bonds.

Each carbon atom in this molecule forms four sigma bonds. The electrons of each carbon atom are found in one s-orbital and three p-orbitals. However, when forming sigma bonds, the carbon atoms combine the four atomic orbitals into four molecular orbitals. This results in each carbon now having four hydridized sp3 orbitals. Therefore, each carbon atom is sp3 hybridized.

Now, when forming a sigma bond, the hybrid orbitals are used in the bonding. So for each C-H bond in this structure, the sigma bond is composed of an sp3 orbital from the carbon, and an s-orbital from the hydrogen. In the case of the C-C bonds, each sigma bond is simply a combination of two sp3 orbitals, one from each carbon.
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What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
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Answer:

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Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

Molarity\ of\ NaCl= 0.3374 M

<u>Thus, the concentration of NaCl = 0.3374 M</u>

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