A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.

5 0

2 years ago

in the early studies of chemistry, scientists used properties and changed to help identify compounds. this is still done today.

Kaylis [27]

The salt water is a mixture because the salt and the water can be separated from each other. The dissolution of salt inside water is a physical change. At the same time, the salt water is an electrolyte, because it conducts electricity. The electricity that was pass through the water led to the decomposition of the water; those bubbles that form inside the water represent oxygen gas which is a product of decomposition of water, therefore the electricity that was passed through the salt water has caused chemical change to occur inside the salt water.<span />

6 0

2 years ago

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Rama09 [41]

<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

= 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

= 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>

We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

= 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0

2 years ago