# A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

Assuming that both helium and argon act like ideal gases, the total pressure after mixing would be approximately .

Explanation:

By the ideal gas equation, , where

• is the pressure of the sample.
• is the volume of the container.
• is the number of moles of gas particles in the sample.
• is the ideal gas constant.
• is the temperature of the sample.

Rewrite to obtain:

• , and
• .

Assume that the two samples have the same temperature, . Also, assume that mixing the two gases did not affect the temperature.

Apply the equation to find the number of moles of gas particles in each container:

• In the helium container, and . Hence, .
• In the argon container, and . Hence, .

After mixing, . Assuming that temperature stays the same.

.

Apply the equation to find the pressure after mixing.

.

The total pressure is 229 atm

Explanation:

Step 1: Data given

Volume of helium flask = 3.0 L

Pressure helium flask = 145 mm Hg

Volume of argon flask = 2.0 L

Pressure argon flask = 355 mmHg

total volume = 5.0 L

Step 2: Partial pressure helium

pHe = 145 *(3/5) = 87.0 atm

Step 3: Calculate pressure argon

pAr = 355*(2/5) = 142.0 atm

Step 4: Calculate total pressure

Total pressure = 87.0 + 142.0 atm

Total pressure = 229 atm

The total pressure is 229 atm

## Related Questions

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0  * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

A scientist performs an experiment involving the collision of two masses on a flat surface. She believes she has made a startling discovery: the total kinetic energy is not conserved. What might she have overlooked?A.
Some gravitational potential energy may have been gained.
B.
Some gravitational potential energy may have been lost.
C.
Some energy may have been gained due to friction and/or sound.
D.
Some energy may have been lost due to friction and/or sound.

c

Explanation:

The reaction of ethyl acetate with sodium hydroxide, CH3COOC2H5(aq)+NaOH(aq)⇌CH3COONa(aq)+C2H5OH(aq) is first order in CH3COOC2H5 and first order in NaOH. If the concentration of CH3COOC2H5 was increased by half and the concentration of NaOH was quadrupled, by what factor would the reaction rate increase?

so the reaction rate increases by a factor 6.

Explanation:

For the given equation the reaction is first order with respect to both ester and sodium hydroxide

So we can say that the rate law is

now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration

The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.

The new rate law is

the final rate = 6 X initial rate

so the reaction rate increases by a factor 6.

If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost if the final weight of the crucible and anhydrous salt is 5.022?

For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.

The weight of the water lost is 0.693 g.

### Explanation:

To calculate the weight of the water that is lost, we need to find the weight of the anhydrous salt. The anhydrous salt is the crucible with the added hydrate, minus the weight of the crucible. So, the weight of the anhydrous salt is 5.022 g - 3.715 g = 1.307 g. Since the weight of the hydrate is 2.000 g, the weight of the water that is lost is equal to the difference between the weight of the hydrate and the weight of the anhydrous salt, which is 2.000 g - 1.307 g = 0.693 g.

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