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tresset_1 [31]
2 years ago
7

What atomic or hybrid orbitals make up the sigma bond between c1 and c2 in ethane, ch3ch3 ?

Chemistry
1 answer:
lutik1710 [3]2 years ago
8 0

Atomic or hybrid orbitals make up the sigma bond between c1 and c2 in ethane, ch3ch3 are one s orbital and 2 p orbital

Sigma bonds are the most powerful type of covalent chemical link. They are generated by atomic orbitals overlapping head-on. For diatomic molecules, sigma bonding is most readily explained using the language and tools of symmetry groups.

Chemical covalent bonds are sigma bonds and pi bonds. The overlap of atomic orbitals creates sigma and pi bonds. Sigma bonds are produced by overlapping from end to end, whereas Pi bonds are created when the lobe of one atomic orbital overlaps another.

Sigma bonding can be seen in the link between two hydrogen atoms. Sigma bonds are also formed between the sp3 orbitals of hybridized carbon and the s orbitals of hydrogen in methane.

To learn more about sigma bonds please visit -
brainly.com/question/14018074
#SPJ4

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Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

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What's the <em>theoretical</em> yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of CaCO₃ available:

\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}.

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}.

Molar mass of CO₂:

M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}.

Mass of the 0.400655 moles of \text{CO}_2 expected for the 40.1 grams of CaCO₃:

m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}.

What's the <em>percentage</em> yield of this reaction?

\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%.

7 0
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6 0
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solmaris [256]

Answer:

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Explanation:

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