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cestrela7 [59]
3 years ago
14

Which of these will change if the air in a

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

The air pressure in the bottle will change.

Explanation:

According to Gay-Lussac's Law, one of the main gas laws, the pressure of gas is directly proportional to the temperature of the gas. This means that as the temperature of the gas increases, the pressure of the gas will increase as well. Therefore, if you heat a closed bottle, the air pressure inside the bottle will change (increase).

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What volume would 8.01×1022 molecules of an ideal gas occupy at STP?
Nataly_w [17]

Explanation:

8.01 \times  {10}^{22}  \times  \frac{1}{6.02 \times  {10}^{23} }  \times  \frac{22.4}{1}  = 2.9804

5 0
2 years ago
Consider the Gibbs energies at 25 ∘C.
zysi [14]

Answer:

A)Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

B)Ksp=1.75×10^−10

C)Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=​Δ​G​f​,​p​r​o​d​u​c​t​s​∘​−​Δ​G​f​,​r​e​a​c​t​a​n​t​s​∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​C​l​(​a​q​)​−​∘​]​−​Δ​G​f​,

A​g​C​l​(​s​)​⇌​A​g​(​a​q​)​+​+​C​l​(​a​q​)​−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​B​r​(​a​q​)​−​∘​]​−​Δ​G​f​,

Δ​G​r​x​n​∘​=​[​77.1​k​J​/​m​o​l​+​(​−​104.0​k​J​/​m​o​l​)​]​−​(​−96.90kj/mol

Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K

Ksp=5.45×10^−13

8 0
2 years ago
An aluminum can weighing 10 g absorbs 106.8 J of heat and warms by 12 degrees C. What is the specific heat of the aluminum can?
Sladkaya [172]

Answer:

c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}

Explanation:

The heating process is modelled after the First Law of Thermodynamics:

Q = m \cdot c_{Al}\cdot \Delta T

The specific heat of the aluminium can is:

c_{Al} = \frac{Q}{m\cdot \Delta T}

c_{Al} = \frac{106.8\,J}{(10\,g)\cdot (12^{\circ}C)}

c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}

8 0
3 years ago
You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,20}, thus,

Thus, you need to add:

[H⁺] = 10^{-7,2} -10^{-8,0} = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × \frac{1H_{2}SO_4 mol}{2H^{+} mol} = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × \frac{98,1 g}{1 mol} × \frac{100 gSolution}{12 gH_{2}SO_4 } × \frac{1 mL}{1,080 g}  =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0
3 years ago
In part A, when you added water to the sodium acetate in the flask, some of the chemical dissolved into the water. Would you cal
viktelen [127]

Answer: The solution is a SATURATED solution.

Explanation:

Although most substances are soluble in water, some are more soluble than others,that is , their solubilities differ. SOLUBILITY is a means of comparing the extent to which different solutes can dissolve in a particular solvent at a definite temperature.

From the question above, when water was added to the sodium acetate in the flask, SOME of the chemical dissolved into the water, meaning that some remained undissolved. This is because a given volume of water can only dissolve a certain amount of chemical in it at room temperature. If more chemical is added to such a solution, the chemical will remain undissolved. Such a chemical solution is said to be a SATURATED SOLUTION.

A saturated solution of a solute at a particular temperature is on which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles.

Unsaturated solution is a type of solution that dissolves all its solutes with no presence of undissolved solute.

Supersaturated solution is one which contains more of the solute than it can normally hold at that temperature. It is an unstable solution which crystallizes out when disturbed.

4 0
3 years ago
Read 2 more answers
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