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Darya
1 month ago
12

If the balloon described in Question 10.24 is released into the air and rises to an altitude of 10,000 ft where the atmospheric

pressure is 523 torr and the temperature is 7.50ºC, then what is its new volume, in L?
Chemistry
1 answer:
castortr0y [3K]1 month ago
3 0

The new volume will be 33.5 L.

Explanation:

The kinetic theory of gases indicates that the space occupied by gas molecules is directly proportional to their temperature while being inversely proportional to the pressure. Assuming the number of moles is n = 1, the equation for gases can be written as:

PV = nRT

In this equation, P stands for pressure, V represents volume, R is the gas constant, and T denotes temperature.

Given that P = 523 Torr, T = 7.50 °C = 7.50 + 273.15 = 280.65 K, and the gas constant R = 62.363 Torr L mol⁻¹K⁻¹.

523*V = 1 * 62.363*280.65 \\\\V = 33.5 L

Consequently, the new volume will be 33.5 L.

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The thermal energy from the soup is transferred to Greg's hands. 
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1 month ago
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Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which
Anarel [2989]

Response:

a) Representation - (in attachment)

b) The geometry is tetrahedral

c) sp^{3} hybrid orbitals involve sigma bonding.

\pi orbitals arise from the overlap of sulfur's d-orbitals with the p-orbitals of oxygen.

Clarification:

a)

Representation in attachment.

The total charge in both structures is -2. Structure (b) is preferred according to the resonance forms since the formal charges within the species are preserved.

Thus, structure (b) represents the sulfate ion more accurately.

b) In the sulfate ion, the sulfur atom forms connections with four distinct oxygen atoms. Using VSEPR theory, the sulfate ion is tetrahedral in shape.

There are four sigma bonds and no lone pairs surrounding the central atom.

Consequently, the sulfur atom's hybridization is sp^{3}

c)

The s and p orbitals of sulfur undergo hybridization to form sigma bonds. The orbitals involved in \pi bonding are

Therefore, \pi bonds result from the overlapping of sulfur's d-orbitals with the p-orbitals of oxygen.

5 0
1 month ago
ASAP PLEASE HELP:
Anarel [2989]
Thalia needs to factor in the sphere's size, its mass, the necessary permits, and the overall expenses for its fabrication.
5 0
26 days ago
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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Tems11 [2777]

Response:

ΔH = -793.6 kJ

Reasoning:

The ΔH for this particular reaction can be calculated through Hess's law, which allows one to combine the ΔH values of the half-reactions to find the overall ΔH:

The half-reactions are as follows:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The calculation involves summing (4) + 4×(3) - (2) - 2×(1) - 2×(5):

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369.2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356.9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483.6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

<pThus, ΔH is:

ΔH = -639.1 kJ -369.2 kJ -356.9 kJ +483.6 kJ +88.0 kJ

ΔH = -793.6 kJ

I trust this clarifies things!

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1 month ago
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When examining its electronic configuration, the last digit should be five, suggesting configurations like (2,5), (2,8,5), or (2,8,8,5), among others.

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