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Artist 52 [7]
3 years ago
9

1. For each of the molecules below, determine the electron geometry, molecule geometry, and bond

Chemistry
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

CCl4- tetrahedral bond angle 109°

PF3 - trigonal pyramidal bond angles less than 109°

OF2- Bent with bond angle much less than 109°

I3 - linear with bond angles = 180°

A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°

Explanation:

Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.

Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.

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HC]A laboratory experiment requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M
andreyandreev [35.5K]

The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.

3 0
3 years ago
Vitamin D is produced in the skin when 7-dehydrocholesterol reacts with UVB rays (ultraviolet B) having wavelengths between
goblinko [34]

The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J

The energy of the photon is given by;

E = hc/λ

E = energy of the photon

h = Plank's constant

c = speed of light

λ = wavelength of light

For the upper boundary range;

E = ?

h = 6.6 × 10^-34 Js

c = 3  × 10^8 m/s

λ = 270 × 10^-9

E =  6.6 × 10^-34 Js × 3  × 10^8 m/s / 270 × 10^-9

E = 7.33 × 10^-19 J

For the lower range;

E = ?

h = 6.6 × 10^-34 Js

c = 3  × 10^8 m/s

λ =300 × 10^-9

E =  6.6 × 10^-34 Js × 3  × 10^8 m/s / 300 × 10^-9

E = 6.6 × 10^-19 J

Hence, the energy range 6.6 × 10^-19 J < E <  7.33 × 10^-19 J

Learn more: brainly.com/question/24857760

8 0
2 years ago
How many grams of KCN are in 10.0 ml of a 0.10 M solution?
attashe74 [19]

Explanation:

As it is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = \frac{no. of moles}{Volume in liter}

As it is given that molarity is 0.10 M and volume is 10.0 ml. As 1 ml equals 0.001 L. Therefore, 10.0 ml will also be equal to 0.01 L.

Hence, putting these values into the above formula as follows.

                  Molarity = \frac{no. of moles}{Volume in liter}

                  0.10 M = \frac{no. of moles}{0.01 L}

                        no. of moles = 0.001 mol

As molar mass of KCN is equal to 65.12 g/mol. Therefore, calculate the mass of KCN as follows.

                 No. of moles = \frac{mass}{molar mass}

                                 0.001 mol = \frac{mass}{65.12 g/mol}

                                 mass = 0.06152 g

Thus, we can conclude that 0.06152 grams of KCN are in 10.0 ml of a 0.10 M solution.

3 0
3 years ago
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