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melomori [17]
2 years ago
8

Pop is made up of three major parts (water, carbon dioxide, and sugar). Assume that all other ingredients are present in insigni

ficant amounts. Explain how could you determine the moles and particles of water, carbon dioxide, and sugar in your pop. Be sure to show any necessary formulas in your answer.
Chemistry
1 answer:
Lady_Fox [76]2 years ago
7 0

If you were given the the total # of grams of all three compounds with the % of each molecular compound - H2O, CO2 and sugar [you would need to know the type of sugar in the Pop such as C12H22O11 as to find its MM (molar mass)]

To begin take the % of each compound x the total grams of all the compounds. For illustrative purposes let's say it works out to be 50 grams of H2O in the POP. The same would be. done for CO2 and for the sugar.

Step 1) With the mass of each you could determine the # of moles of each:

Example if the number grams of water in the sample is 50g, to determine the # of moles of water you would do the following - 50g H20 x 1 mol/18g H20 = 2.8 mol H2O The same technique would be used for the other compounds to find the # of moles.

Step 2) To find the representative particles of each(molecules, atoms) you would do the following:

as the example given of above for H20 - 50 grams you calculated as shown above to be the number of mol of H20 = 2.8mol

From the number of mol of H20, to determine the # of molecules of water you would set up the following:

2.8 mol H20 x 6.02 x 10^23/1 mol H2O = 1.69 X 10^24 molecules of H20.

The same would be done for CO2 and the sugar.

Step 3) Now to find the number of atoms of element of the compound taking for example the H2O example above:

Take the # of molecules of H2O found above and set it up in the following manner:

1.69 X 10^24 molecules H2O x 2 atoms H/1 molecule H2O = 3.38 x 10^24 atoms H

1.69 x 10^24 molecules H2O x 1 atom O)/1 molecule H2O = 1.69 X 10^24 atoms O

The same would be done for CO2 and for the sugar compound.

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
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If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

\frac{7.59 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.367 g

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

\frac{2.11 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.102 g

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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