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Fofino [41]
3 years ago
13

How is a pure substance different from a mixture?

Chemistry
2 answers:
Troyanec [42]3 years ago
8 0

Answer:

A pure substance is different from a mixture in that <u><em>mixtures are made up of more than on component</em></u>

Explanation:

A pure substance is a substance that is homogeneous and invariable (fixed), and it is not possible to separate or divide it into more substances. For example, water has the formula H2O and is always the same. This indicates that it is formed by molecules in which there are 2 hydrogen atoms and 1 oxygen atom. If I changed that formula, it would be another different substance.

A pure substance cannot be broken down into other simpler substances using physical methods.

A pure substance has its own or defined characteristic properties.

A pure substance can be both a chemical element and a compound, which are formed by atoms or molecules of the same nature.

On the other hand, a mixture is the union of two pure substances or compounds (which have their individual properties), and have a variable chemical composition. They can be separated by chemical or physical means and obtain the components that form the mixtures. They are formed by atoms and molecules of different nature.

The mixtures can be homogeneous (whose composition and properties are uniform in any part of a given sample, such as sugar dissolved in water) or heterogeneous (those in which at a glance it is easy to identify the different components that make them up; for example water and oil or water and sand)

Given the above, it is possible to say that <u><em>a pure substance differs from a mixture in that mixtures are made up of more than on component</em></u>

svlad2 [7]3 years ago
3 0

b.) mixtures are made up of more than on component. A pure substance is like a diatomic element, like O2, while a mixture is more like a salad, and can be separated by physical means.

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Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and
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Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

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Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56

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6 0
2 years ago
Where do you find an elements energy levels
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8 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0
3 years ago
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