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2 years ago

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

1 answer:

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.

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The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

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Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

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3 years ago

The specific spatial arrangement of amino acid residues that are close to each other in the polypeptide chain is called______the

natulia [17]

Answer:

c. tertiary.

Explanation:

In this case, we can review the definition of each level of structuration in the proteins:

Primary structure

In the primary structure, the amino acids are linked by peptide bonds. That is, the order of the amino acids is the criterion that defines this type of structure.

Secondary structure

In the secondary structure, we have to look at the way in which the protein is folded. The options are:

-) Beta-laminar: A structure in which the protein has a planar shape.

-) Alpha-helix: A structure in which the protein has a cross-strand form.

Tertiary structure

In the tertiary structure, the R groups that the amino acids have in the primary structure can generate interactions with each other. Interactions such as hydrogen bridges, dipole-dipole, hydrophobic interactions. This makes the protein have a very specific three-dimensional structure, on which its function depends.

Quaternary structure

In the quaternary structure, several subunits may be attached, or there may be prostatic groups (metals that can help to attach various protein units).

With all these in mind, the deffinition that fits with the description in the question is the tertiary structure.

I hope it helps!

4 0

3 years ago