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neonofarm [45]
3 years ago
8

Determine the electron geometry, molecular geometry, and idealized bond angles for each molecule. In which cases do you expect d

eviations from the idealized bond angle? b. NF a. CF4 d. H2S c. OF2 Which species has the smaller bond angle, H3O or H2O? Explain. Explain why CO2 and CC14 are both nonpolar even though they contain polar bonds. CH&F is a polar molecule, even though the tetrahedral geometry often leads to nonpolar molecules. Explain.
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
8 0

Answer:

The electron geometry, molecular geometry and idealized bond angles for these molecules respectively are:

a. CF4: tetrahedral, tetrahedral and 109.5 degrees

b. NF3 tetrahedral, trigonal pyramidal and 102.5 degrees

c. OF2 tetrahedral, angular and 103 degrees

d. H2S tetrahedral, angular and 92.1 degrees

Explanation:

The electron geometry considers the bound atoms and unbound electron pairs to determine the geometry. The four molecules have four bound atoms and/or unbound electrons pairs, thus they have a tetrahedral geometry. On the other hand, the molecular geometry only considers the position of bound atoms to determine the geometry.

Between H3O and H2O, H2O has a smaller bond angle due to the two unbound electron pairs. The bond angle decrease as the number of unbound electron pairs increases in every molecule.

CO2 and CCl4 are both nonpolar because of the 3D geometry of the molecule. Each individual bond is polar but both molecules have symmetrical geometry so the dipole bonds are canceled.

CH3F  is a polar molecule because the dipole between the C-H and C-F bonds are differents thus, besides the symmetrical geometry the dipole bonds are not canceled.

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4 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
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Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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Electronegativity of carbon = 2.5

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Thus the difference in electronegativity of carbon and oxygen is=(3.5-2.5)= 1.0

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