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A 32.4 L gas sample at STP is compressed to a volume of 28.4 L, and the temperature is increased to 352 K. What is the new press

Sindrei [870]

Answer:

1.47 atm

Explanation:

Step 1: Given data

Initial volume (V₁): 32.4 L

Initial pressure (P₁): 1 atm (standard pressure)

Initial temperature (T₁): 273 K (standard temperature)

Final volume (V₂): 28.4 L

Final pressure (P₂): ?

Final temperature (T₂): 352 K

Step 2: Calculate the final pressure of the gas

We can calculate the final pressure of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

P₂ = P₁ × V₁ × T₂ / T₁ × V₂

P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm

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3 years ago

The first method of determining the chemical composition of substances in space was to

ehidna [41]

<span>The first method to determine the chemical composition of a substance in space was using light. By determining red shift in the observed spectrum of light they could determine the elements they were observing. Different elements change the way light behaves and from this scientists can determine the makeup of things such as stars and nebulas.</span>

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3 years ago

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Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

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3 years ago

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How are the conditions at which phases are in equilibrium represented on a phase diagram?

insens350 [35]

How are the conditions at which phases are in equilibrium represented on a phase diagram?

Image result for How are the conditions at which phases are in equilibrium represented on a phase diagram?

Along the line between liquid and solid, the melting temperatures for different pressures can be found. The junction of the three curves, called the triple point, represents the unique conditions under which all three phases exist in equilibrium together. Phase diagrams are specific for each substance and mixture.

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3 years ago

think about putting a glass of water on the counter, in the freezer, and on a hot stove how does the temperature of the water af

Shtirlitz [24]

The kinetic energy and the physical state of water depend strongly on the temperature;

Firstly, The kinetic energy of water on a hot stove is higher than that on the counter in the freezer; that the kinetic energy is directly proportional to the temperature according to the relation: $K = \frac{3RT}{2NA}$ ; where R is the universal gas constant, T is the temperature and NA is Avogadro number.

As the temperature increases, the speed of colliding molecules increases and the kinetic energy increases.

Secondly, The physical state of water depends on the temperature; water has three states (gas, liquid and solid) depends on the temperature.
If a glass of water is putt on the counter in the freezer, it will be converted to the solid state (ice).
And, as if it is putt on a hot stove, it will be vapor (gaseous state).

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