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3 years ago

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

Explanation:

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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<h2>Answer:-</h2>

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2 years ago

What amount of energy is required to completely ionize 27.8 grams of carbon atoms in the gas phase (c(g)) if the ionization ener

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I think it would be 2,510 kJ, or 2,510,000 J.

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2 years ago

In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After s

mart [117]

Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

$0.3g Mg$ * $\frac{16gO}{24.3gMg}$ = 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

$0.2g O$ * $\frac{1 mol O}{16g O}$ = 0.01mol O

$0.3g Mg$ * $\frac{1mol Mg}{24.3g Mg}$ = 0.01mol Mg

The moles of MgO can be obtained from:

$0.5g MgO$ * $\frac{1mol MgO}{40.3g MgO}$ = 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.

4 0

2 years ago

30g of naoh is dissolved in 1.5 liter solution the active mass of naoh is?

Pepsi [2]

Answer:

0.5mol/L

Explanation:

First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the question = 30g

Number of mole = Mass /Molar Mass

Number of mole = 30/40 = 0.75mol

Volume = 1.5L

Active mass = mole/Volume

Active mass = 0.75mol/1.5L

Active mass = 0.5mol/L

7 0

3 years ago

I need help ASAP!

MaRussiya [10]

Calcium reacts gently with water to give hydrogen and calcium hydroxide, which is only slightly soluble, thus slows down the reaction.

It will be assumed that hydrochloric acid used is a dilute aqueous solution.

However, calcium reacts with hydrochloric acid to give calcium chloride which is readily soluble in water, and hydrogen, being a typical reaction of relatively active metals with acids.
Ca(s) + 2HCl(aq) -> CaCl2(aq) +H2(g) ↑ + heat

The clues that it is a chemical reaction could be:
- formation of a new substance, gaseous hydrogen
- disappearance of a metallic solid in the solution
- heat formed during the vigorous reaction.

As silver is below hydrogen in the electrochemical series, it will not be expected to react with dilute hydrocloric acid. (however, it dissolves in oxidizing acid such as nitric acid, but not displacing hydrogen as a product).

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3 years ago