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spin [16.1K]
3 years ago
10

Examples of pure and impure substances

Chemistry
2 answers:
erastova [34]3 years ago
8 0
Pure- table salt
Impure- vegetable oil
Sladkaya [172]3 years ago
8 0

Examples of pure substances are water, diamond, table salt

Examples of impure substances are muddy water, air, sugar in water

<u>Explanation</u>:

  • A pure substance is one that is made up of only one element or one chemical compound. It has no impurities. They can be elements or compounds.
  • It has definite properties and composition.
  • Examples are water, diamond, table salt, etc.
  • An impure substance is one that is made up of two or more elements or compounds that are not chemically bonded. It is a mixture.
  • It has no definite composition or properties.
  • Examples are muddy water, sugar in water, air, etc.
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I believe that there are 12 oxygen atoms in aluminum sulfate, And 1 Sulfur and 2 aluminum. If you need anything else let me know.
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3 years ago
Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su
Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

<em>1</em> C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • C₅H₁₂ is the only <em>reactant</em> that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
  • CO₂ is the only <em>product</em> that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

<em>1 </em>C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + ? H₂O

  • Similarly, C₅H₁₂ is the only <em>reactant</em> that contains H atoms. H₂O is the only <em>product</em> that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

<em>1</em> C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

  • Both CO₂ and H₂O are <em>products</em> that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the <em>product</em> side corresponds to 8 O₂ molecules on the reactant side.

<em>1</em> C₅H₁₂ + <em>8</em> O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

  • All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", <em>GCSE Chemistry (Single Science)</em>, BBC Bitesize.

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A permanent magnet can affect:
ruslelena [56]

A. both permanent magnets and electromagnets.

Explanation:

A permanent magnet can affect and attract any other permanent magnet and even electromagnet.

They also affect any magnetic materials especially metals that can be magnetized.

In the vicinity of such substances, an attractive or repulsive force sets in and they both interact in the presence of the force field in place.

Permanent magnets cannot magnetize non-magnets.

An electromagnet is a magnet produced by the passage of electric current through a wire wound round a metallic core.

learn more:

Electromagnet brainly.com/question/2191993

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Calculate the radius of a tantalum (ta) atom, given that ta has a bcc crystal structure, a d ensity of 16.6 g/cm3, and an atomic
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solution:

Ta is BCC, a=\frac{4}{\sqrt{3}r},n=2atoms/cell\\p=16.6gram/cm^3,A=180.9gram/mol\\P=\frac{A\times n}{N_{a}\times v}=\frac{180.9\frac{gram}{mol}\times 2 \frac{atom}{cell}}{6.023\times10^23\frac{atom}{mol}\times(\frac{4}{\sqrt{3}\times r})^3\frac{cm^3}{cell}}\\=16.6\frac{gram}{cm^3}\\r=(\frac{180.9\times2}{16.6\times6.023\times10^23\times(\frac{4}{\sqrt{3}})^3})^\frac{1}{3}\times10^8


4 0
3 years ago
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