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Doss [256]
3 years ago
14

A chemical reaction was used to separate a sample into its components. What conclusion about the sample can be formed based on t

he technique used?
A: It was a pure substance
B: It was a compound
C: It was a heterogeneous mixture
D: It was a homogeneous mixture
Chemistry
2 answers:
masya89 [10]3 years ago
6 0

Answer:

B: It was a compound

Explanation:

If a chemical reaction was used to separate a sample into its components,  then it was a compound. For example, water (H2O, a compound) can be separated by electrolysis into oxygen and hydrogen (its components). the chemical reaction is:

2 H2O -> O2 + 2 H2

Heterogeneous and homogeneous mixtures can be separated by physical ways. Pure substance can be either a compound or an element, so this option is not specific enough to be the correct answer.

Dvinal [7]3 years ago
3 0
C, a compound because it means it was made up of more than one element but, still able to be separated from the other compound(s)
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For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou
Alexxx [7]
<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

                        = 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol ×  82.08 g/mol

                                             = 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

                   = (81.4 % × 31.6) ÷ 100

                  = 25.72 g

The percent yield of H₂SO₃ is 25.72 g

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