All categories

3 years ago

How would you describe the strength of the bonds that result from resonance in SO3?

Chemistry

1 answer:

Sladkaya [172]3 years ago

7 0

Answer:

All bonds are equivalent in length and strength within the molecule.

Gaseous SO3 is a trigonal planar molecule that exhibit a D3h symmetry group.

Sulfur has sp2 hybridization and it has 6 outer electrons which make the bonds with the oxygen.

Its constituent sulfur atom has an oxidation state of +6 and a formal charge of 0.

The Lewis structure is made up of one S=O double bond and two S–O dative bonds that doesn't not engage the d-orbitals. ( Thus, SO3 molecule has three double bonded oxygen to the central sulfur atom). This explains the strength.

It gaseous form had a zero electrical dipole moment because of the 120° angle between the S-O bonds.

Explanation:

You might be interested in

What function does residual CO2 serve

Luba_88 [7]

The function does residual Co2 plays big part and in maintaining the body’s homeostasis. The addition of respiratory to reserve volume or residual volume. The lung is the one that who protect the organs, so in exchange of oxygen and carbon dioxide.

6 0

3 years ago

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

Zinc put in concertrated sodium hydroxide

Readme [11.4K]

What are you asking for

8 0

3 years ago

Which statement describes a reason to consider air a mixture? A) The major constituents of air are gaseous elements. B) The comp

umka21 [38]

The correct answer is - A) The major constituents of air are gaseous elements.

With the statement ''the major constituents of air are gaseous elements'' we can easily conclude that the air is a mixture. The reason for that is that we have a plural usage of the word element, elements, which mean that there are multiple elements that make up the air.

The air is indeed predominantly a mixture of gaseous elements. The most abundant gas in the air being the nitrogen with 78.9%, oxygen with 20.95%, argon 0.93%, and carbon dioxide 0.04%, with lesser amounts of other gases also be present in it. The water vapor is also present in the air, though it is variable, being around 1% at sea level, but only 0.4% over the entire atmosphere.

3 0

3 years ago

A chunk of sulfur has a volume of 8.33 cm3. What is the mass of this sulfur? (Density of sulfur = 2.07 g/cm3.)

nata0808 [166]

Explanation:

equations to note:

density= mass/volume

mass= volume *density

volume= mass/density

you have a volume- 8.33cm3

you have a density- 2.07 g/cm3

Answer:

8.33cm3 * 2.07g/cm3= 17.24g

mass= 17.24g

5 0

2 years ago