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arlik [135]
3 years ago
14

A gas has a pressure of 450 mmHg at 100 degrees Celsius. What will its new pressure be when the temperature rises 200 degrees Ce

lsius
Chemistry
1 answer:
Ahat [919]3 years ago
3 0

Answer:

P2 = 900 mmHg.

Explanation:

Given the following data;

Initial pressure = 450 mmHg

Initial temperature = 100°C

Final temperature = 200°C

To find the final pressure, we would use Gay Lussac's law;

Gay Lussac states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

PT = K

\frac{P1}{T1} = \frac{P2}{T2}

Making P2 as the subject formula, we have;

P_{2}= \frac{P1}{T1} * T_{2}

P_{2}= \frac{450}{100} * 200

P_{2}= 4.5 * 200

Final pressure, P2 = 900 mmHg.

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Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after
Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is 2.83\times 10^{22}

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the air = 37^oC=[37+273]K=310K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of air = ?

Putting values in above equation, we get:

1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol

Hence, the number of moles of air present in the RV is 0.047 moles

  • <u>For b:</u>

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.047 moles of air will contain (0.047\times 6.022\times 10^{23})=2.83\times 10^{22} number of gas molecules.

Hence, the number of molecules of gas is 2.83\times 10^{22}

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3 years ago
A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction
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The half-life of the reaction is 50 minutes

Data;

  • Time = 43 minutes
  • Type of reaction = first order
  • Amount of Completion = 45%

<h3>Reaction Constant</h3>

Let the initial concentration of the reaction be X_0

The reactant left = (1 - 0.45) X_0 = 0.55 X_0 = X

For a first order reaction

\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1

<h3>Half Life </h3>

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\

Substitute the values

t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min

The half-life of the reaction is 50 minutes

Learn more on half-life of a first order reaction here;

brainly.com/question/14936355

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Answer:

I think this is probably the answer you are looking for.

Explanation:

https://youtu.be/PY431ZC5uDc

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ANSWER

36.12 degrees fahrenheit
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