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DanielleElmas [232]
3 years ago
5

At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba

lloons has a volume of 6.25 L. What would be the volume of the balloon if it were submerged in a swimming pool to a depth where the pressure is 922.3 mmHg and the temperature is 26.35°C?
Chemistry
2 answers:
Iteru [2.4K]3 years ago
7 0

Answer:

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg ____ (755.4 / 760) = 0.993 atm

922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

vagabundo [1.1K]3 years ago
6 0

Answer: The volume of the balloon if it were submerged in a swimming pool to a depth where the pressure is 922.3 mmHg and the temperature is 26.35°C is 5.08 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 755.4 mmHg

P_2 = final pressure of gas = 922.3 mmHg

V_1 = initial volume of gas = 6.25 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 28.50^oC=273+28.50=301.50K

T_2 = final temperature of gas = 26.35^oC=273+26.35=299.35K

Now put all the given values in the above equation, we get:

\frac{755.4 \times 6.25}{301.50K}=\frac{922.3\times V_2}{299.35K}

V_2=5.08L

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Amiraneli [1.4K]

Answer:

I believe your answer would be grams.

4 0
2 years ago
At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the
Elis [28]

Answer:

P₂  = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂ 

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂  =  25.83 atm .L.  K  / 293 K . L

P₂  = 0.09 atm

7 0
3 years ago
If 200.4g of water is mixed with 101.42g of salt the mass of the final solution would be reported as
d1i1m1o1n [39]

Answer:

301.8 g

Explanation:

We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.

m(solution) = m(solute) + m(solvent)

m(solution) = 200.4 g + 101.42 g

m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)

8 0
3 years ago
Please explain how to do it as well!
kogti [31]

Answer:

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△H=−72 kcal

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b.

\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}

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7 0
3 years ago
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Naya [18.7K]
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
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molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
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therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
8 0
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