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How many molecules are in 2.00 moles of H20????????

Elis [28]

2x 6.022x10^23= 1.204x10^24

7 0

3 years ago

Scientist name: ______________

nikitadnepr [17]

This is easy… like you can’t take 20 min to search this up in Googl

5 0

2 years ago

Read 2 more answers

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

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

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

6 0

2 years ago

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What type of energy uses turbines

labwork [276]

Answer:

The answer is : D. Wind energy

3 0

2 years ago

Suppose you work at a theme park. Your supervisor wants you to make a sign displaying the maximum weight that a roller coaster t

grin007 [14]

Answer:

amusement parks. Each day, we flock by the millions to the nearest park, paying a sizable hunk of money to wait in long lines for a short 60-second ride on our favorite roller coaster. The thought prompts one to consider what is it about a roller coaster ride that provides such widespread excitement among so many of us and such dreadful fear in the rest? Is our excitement about coasters due to their high speeds? Absolutely not! In fact, it would be foolish to spend so much time and money to ride a selection of roller coasters if it were for reasons of speed. It is more than likely that most of us sustain higher speeds on our ride along the interstate highway on the way to the amusement park than we do once we enter the park. The thrill of roller coasters is not due to their speed, but rather due to their accelerations and to the feelings of weightlessness and weightiness that they produce. Roller coasters thrill us because of their ability to accelerate us downward one moment and upwards the next; leftwards one moment and rightwards the next. Roller coasters are about acceleration; that's what makes them thrilling. And in this part of Lesson 2, we will focus on the centripetal acceleration experienced by riders within the circular-shaped sections of a roller coaster track. These sections include the clothoid loops (that we will approximate as a circle), the sharp 180-degree banked turns, and the small dips and hills found along otherwise straight sections of the track.

3 0

1 year ago