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melomori [17]
1 year 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]1 year 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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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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