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lisov135 [29]
3 years ago
13

A 2.026g sample of a hydrate of sodium carbonate (Na2CO3) is heated to remove water. After heating, the mass of the sample is 0.

750 g. Calculate the percentage by mass of water in the sample.
Chemistry
2 answers:
OLEGan [10]3 years ago
7 0

Answer:

62.98 % of the sample of hydrate is water

Explanation:

Step 1: Data given

Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams

After heating, the mass of the sample is 0.750 g

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mass of water

Mass water = mass of hydrate - mass of sample after heating

Mass water = 2.026 grams - 0.750 grams

Mass water = 1.276 grams

Step 3: Calculate mass % percent of water

Mass % of water = (mass of water / total mass hydrate) * 100 %

Mass % of water = (1.276 grams / 2.026 grams) *100 %

Mass % of water = 62.98 %

62.98 % of the sample of hydrate is water

s2008m [1.1K]3 years ago
3 0

Answer:

percentage mass of water = 62.98%

Explanation:

The reaction is a decomposition reaction whereby  hydrate of sodium carbonate was heated to remove water . The actual weight of the hydrated compound is 2.026 grams . After heating, the mass of the sample was now 0.750 grams.  The percentage of the water sample can be calculated as follows:

total mass of hydrated sample  = 2.026 grams

mass of sample after heating = 0.750 grams

mass of water removed = 2.026 - 0.750 = 1.276 grams

percentage mass of water = mass of water/mass of hydrated compound × 100

percentage mass of water = 1.276/2.026 × 100

percentage mass of water = 127.6/2.026

percentage mass of water = 62.9812

percentage mass of water = 62.98%

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TiliK225 [7]

Taking into account the definition of dilution:

  • you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
  • If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

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<h3>Volume of stock solution</h3>

In this case, you know:

  • Ci= 7 M
  • Vi= ?
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Replacing in the definition of dilution:

7 M× Vi= 0.480 M× 0.120 L

Solving:

Vi= (0.480 M× 0.120 L)÷ 7 M

<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)

Finally, you will need 8.23 mL of the stock solution.

<h3>Concentration of the diluted solution</h3>

In this case, you know:

  • Ci= 7 M assuming the stock solution is 7.00 M HNO₃
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  • Cf= ?
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Replacing in the definition of dilution:

7 M× 0.02 L= Cf× 0.270 L

Solving:

(7 M× 0.02 L)÷ 0.270 L= Cf

<u><em>0.518 M= Cf</em></u>

Finally, the concentration is 0.518 M.

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