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coldgirl [10]
3 years ago
12

If a certain gas occupies a volume of 20. L when the applied pressure is 10. atm , find the pressure when the gas occupies a vol

ume of 5.0 L . Express your answer to two significant figures and include the appropriate units. View Available Hint(s)
Chemistry
1 answer:
labwork [276]3 years ago
6 0

Answer:

The new pressure is 40 atm

Explanation:

Step 1: Data given

Volume of the gas = 20.0 L

Pressure of the gas = 10.0 atm

The volume decreases to 5.0 L

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure of the gas = 10.0 atm

⇒with V1 = the initial volume of the gas = 20.0 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the decreased volume = 5.0 L

10.0 atm * 20.0 L = P2 * 5.0 L

P2 = 200 / 5.0

P2 = 40 atm

Since the volume is 4x smaller, the pressure is 4x more

The new pressure is 40 atm

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Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptur
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Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

Hank's Garage has an air compressor with a holding tank that contains a volume of 200L (V₁) of compressed air at a pressure of 5200 torr (P₁).

One day a hose ruptured and all the compressed air was released. The final pressure was the atmospheric pressure (1 atm = 760 torr) (P₂).

We can calculate the new volume (V₂) in these conditions using Boyle's law, which states there is an inverse relationship between the volume and the pressure of an ideal gas.

P_1 \times V_1 = P_2 \times V_2\\\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{5200 torr \times 200L}{760torr} = 1370 L

Hank's Garage has an air compressor with a holding tank that contains 200L of compressed air at 5200 torr. One day a hose ruptured and all the compressed air was released to a volume of 1370 L at atmospheric pressure.

Learn more: brainly.com/question/1437490

4 0
3 years ago
The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the
Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and NH_3.

\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles

\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl

From the balanced reaction we conclude that

As, 1 mole of NaCl react with 1 mole of NH_3

So, 0.243 mole of NaCl react with 0.243 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and NaCl is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaHCO_3

From the reaction, we conclude that

As, 1 mole of NaCl react to give 1 mole of NaHCO_3

So, 0.243 moles of NaCl react to give 0.243 moles of NaHCO_3

Now we have to calculate the mass of NaHCO_3

\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3

Molar mass of sodium bicarbonate = 84 g/mol

\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

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