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Tanzania [10]
3 years ago
14

At 570. mmHg and 25 °C, a gas sample has a volume of 2270 mL. What is the final pressure (in mmHg) at a volume of 1250 mL and a

temperature of 175 °C? 1560 mmHg 210 mmHg 7000 mmHg 690 mmHg 470 mmHg
Chemistry
1 answer:
marusya05 [52]3 years ago
3 0

Answer:

1560 mmHg is the final pressure

Explanation:

To solve this problem we need to apply the Ideal Gases Law Equation:

P . V = n .  R . T

First of all, we need to convert the volume to L, T° to Absolute T° (T°C + 273) and pressure from mmHg to atm

570 mmHg . 1atm/760 mmHg = 0.75 atm

25°C + 273 = 298K

175°C + 273 = 448K

2270 mL . 1L/1000mL = 2.27L

1250 mL . 1L/1000mL = 1.25L

P.V = n . R. T is the main equation but in both cases n and R are constant. We can ignore them, so we make this formula for each situation:

P . V / T = n . R  →  P₁ . V₁ / T₁ = P₂ . V₂ / T₂

We replace data: 0.75 atm . 2.27L / 298K = P₂ . 1.25L / 448K

(0.75 atm . 2.27L / 298K). 448K = P₂ . 1.25L

2.56 atm.L = P₂ . 1.25L → P₂ = 2.56atm.L / 1.25L = 2.05 atm

We finally convert the atm to mmHg to reach the answer

2.05 atm . 760 mmHg / 1atm = 1556 mmHg ≅ 1560 mmHg

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If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?
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Read 2 more answers
4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-m
jeka94

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

<h3>What is Combined gas law?</h3>

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

  • Initial volume V₁ = 14.5L
  • Initial pressure P₁ = 0.980atm
  • Initial temperature T₁ = 20.0°C = 293.15K
  • Final volume V₂ = 14.3L
  • Final pressure P₂ = 740.mmHg = 0.973684atm
  • Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

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