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labwork [276]
2 years ago
15

A 12-liter tank contains helium gas pressurized to 160 \rm atm.How many 3-liter balloons could the 12-L helium tank fill? Keep i

n mind that an "exhausted" helium tank is not empty. In other words, once the gas inside the tank reaches atmospheric pressure, it will no longer be able to fill balloons.
Chemistry
1 answer:
Vikentia [17]2 years ago
8 0

Answer:

636 balloons

Explanation:

If we assume that helium gas follows an ideal gas behaviour, we can use the ideal gas law to solve this problem as follows:

  1. We consider two different states, the initial given by the conditions of the problem statement and the final, when the tank reaches atmospheric pressure and it's no longer able to fill balloons:P_{1}=160 atm\\V_{1}=12 L\\P_{2}=1 atm\\V_{2}= ?
  2. To find out what would be this volume 2, we use the Boyle's Law: P_{1}V_{1}=P_{2}V_{2}\\V_{2}=\frac{P_{1}V_{1}}{P_{2}} \\V_{2}=\frac{160 atm \times 12L}{1 atm}\\V_{2}=1920 L
  3. Now we find the available volume to fill the balloons by substracting both, volume 2 and volume 1: V_{b}=V_{2}-V_{1}=1920L-12L=1908 L
  4. Finally, we determine the quantity of ballons by dividing that available volume between the volume of each ballon:B=\frac{1908L}{3L} =636 balloons
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3 years ago
A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben
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<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

The chemical equation for the reaction of benzoic acid and methanol is:

\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = \frac{1}{1}\times 0.0295=0.108 moles of methyl benzoate

  • Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

  • To calculate the percentage yield of methyl benzoate, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

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3 years ago
How many moles in 30.0 grams of h3po4
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