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Rina8888 [55]
4 years ago
11

A gas-filled balloon having a volume of 3.50 L at 1.20 atm and 18°C is allowed to rise to the stratosphere (about 30 km above th

e surface of the Earth), where the temperature and pressure are −45°C and 5.70 × 10−3 atm, respectively. Calculate the final volume of the balloon.
Chemistry
1 answer:
kolezko [41]4 years ago
5 0

Answer:

620.71 L the final volume of the balloon.

Explanation:

Initial volume of the gas in the balloon= V_1=3.50 L

Initial pressure of the gas in the balloon= P_1=1.20 atm

Initial temperature of the gas in the balloon= T_1=18^oC =291.15 K

Moles of gases = n

n=\frac{P_1V_1}{RT_1}...[1]

Final volume of the gas in the balloon = V_2=3.50 L

Final pressure of the gas in the balloon = P_2=5.70\times 10^{-3} atm

Final temperature of the gas in the balloon = T_2=-45^oC =228.15 K

Moles of gases = n

n=\frac{P_2V_2}{RT_2}...[2]

[1]  =  [2]

\frac{P_2V_2}{RT_2}=\frac{P_1V_1}{RT_1}

V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}

V_2=\frac{1.29 atm \times 3.50 L\times 228.15 K}{5.70\times 10^{-3} atm\times 291.15 K}=620.71 L

620.71 L the final volume of the balloon.

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How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at s
nevsk [136]

Answer:

40.5 g of P₄O₁₀ are produced

Explanation:

We state the reaction:

P₄ + 5O₂ → P₄O₁₀

We do not have data from P₄ so we assume, it's the excess reactant.

We need to determine mass of oxygen and we only have volumne so we need to apply density.

Density = mass / volume, so Mass = density . volume

Denstiy of oxygen at STP is: 1.429 g/L

1.429 g/L . 16.2L = 23.15 g

We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles

5 moles of O₂ can produce 1 mol of P₄O₁₀

Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles

We determine the mass of product:

0.138 mol . 292.88 g/mol = 40.5 g

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3 years ago
Which is a higher temperature 20°c or 25?
Lunna [17]
25 °c is the higher temperature.

Hope This Helps You!
Good Luck Studying :)
3 0
4 years ago
Read 2 more answers
A feather falls through the air more slowly than a brick because of ____.
NeTakaya

Answer:

Gravity

Explanation:

Gravity is what moves things

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4 years ago
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En condiciones normales 1g de aire ocupa un volumen de 773 mL ¿ qué volumen ocupará la misma masa de aire a 0 ºC y la presión a
emmainna [20.7K]

Answer:

El volumen que ocupará la misma masa de aire es 839.49 mL.

Explanation:

Las condiciones normales de presión y temperatura (abreviado CNPT) o presión y temperatura normales (abreviado PTN o TPN), son términos que implican que la temperatura referenciada es de 0ºC (273,15 K) y la presión de 1 atm (definida como 101.325 Pa).

La ley de Boyle dice que “El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”  y se matemáticamente como

Presión*Volumen=constante

o P*V=k

La ley de Charles es una ley que dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

\frac{V}{T} =k

La ley de Gay-Lussac​ establece que la presión de un volumen fijo de un gas, es directamente proporcional a su temperatura. Se expresa matemáticamente como:

\frac{P}{T} =k

Combinando estas tres leyes se obtiene:

\frac{P*V}{T} =k

Siendo un estado inicial 1 y un estado final 2, la expresión anterior queda determinada como:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

En este caso:

  • P1=  101325 Pa
  • V1= 773 mL
  • T1= 273.15 K
  • P2= 93,3 kPa= 93300 Pa
  • V2= ?
  • T2= 0°C= 273.15 K

Reemplazando:

\frac{101325 Pa*773 mL}{273,15 K} =\frac{93300 Pa*V2}{273.15 K}

y resolviendo obtenes:

V2=\frac{273.15 K}{93300 Pa} *\frac{101325 Pa*773 mL}{273,15 K}

V2= 839,49 mL

<u><em>El volumen que ocupará la misma masa de aire es 839.49 mL.</em></u>

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3 years ago
Question 41 (1 point)
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Answer:

Characteristics that can be observed or measured without changing the chemical nature of the substance. but if you go to quizzlet it could help you too it has flash cards that could help

Explanation:

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