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gizmo_the_mogwai [7]
2 years ago
6

What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?

Chemistry
1 answer:
zhuklara [117]2 years ago
5 0

This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time,  define the molecular geometry.

<h3>Molecular geometry:</h3>

In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.

Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.

In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)

Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

Learn more about molecular geometry: brainly.com/question/7558603

Learn more about the VSEPR theory:  brainly.com/question/14225705

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If the pressure of 50.0 mL of oxygen gas at 100°C increases from 735 mm Hg to 925 mm Hg, what is
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β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa
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The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

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

\pi=icRT

where,

\pi = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

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

c = concentration of solute = ?

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273.15 +25]=298.15 K

Putting values in above equation, we get:

0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L

The concentration of solute is 2.4278\times 10^{-5} mol/L

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

C=\frac{n}{V(L)}

n=2.4278\times 10^{-5} mol/L\times 0.137 L

n=3.3261\times 10^{-6} mol

To calculate the molecular mass of solute, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of β‑Galactosidase = 3.3261\times 10^{-6} mol

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

3 0
3 years ago
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