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Natasha2012 [34]
3 years ago
10

Describe the difference between pure substance and mixtures ( give an example each)

Chemistry
2 answers:
kondaur [170]3 years ago
4 0

Answer:

a pure substance consists only of one element or one compound

a mixture consists of two or more different substances, not chemically joined together.

examples:

pure substance : Hydrogen gas - Diamond - Gold metal.

mixture : water and oil - mixtures of sand and water - trail mix

Explanation:

forsale [732]3 years ago
4 0
Few examples of pure substances include steel, iron, gold, diamond, water, copper, and many more. Air is also often considered as a pure substance.

Some examples of mixtures are:

Sand and water.
Salt and water.
Sugar and salt.
Ethanol in water.
Air.
Soda.
Salt and pepper.
Solutions, colloids, suspensions.
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Consider the electronic configuration for the following elements:
zavuch27 [327]

Answer:

Explanation:

P. Group 6 , block p

Q. Group 1 , block s

R. Group 3, block s

S . Group 5.block p

Q2p

Q3S

5 0
3 years ago
1.6x10^23 lead atoms. Find the weight in grams
Ber [7]
Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.

Weight of lead = moles x atomic weight of lead
                         =  0.265x207.2
                         =  54.908 grams.

Hope this helps!
5 0
3 years ago
How many moles of sodium chloride (NaCl) solute are in 155 grams of an 88.5 percent by mass solution?
Reil [10]
Its going to be 2.27 mol NaCl
5 0
4 years ago
The water of crystallization is responsible for what?
Ber [7]

Answer:

The shapes of the crystals

Explanation:

6 0
4 years ago
Rank these transition metal ions in order of decreasing number of unpaired electrons.
lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

  • Atomic number of Fe is 26.

Fe^{3+} - [Ar] 3d^{5}

So, there is only 1 unpaired electron present in Fe^{3+}.

  • Atomic number of Mn is 25.

Mn^{4+} - [Ar]3d^{3}

So, there are only 3 unpaired electrons present in Mn^{4+}.

  • Atomic number of V is 23.

V^{3+} - [Ar] 3d^{2}

So, there are only 2 unpaired electrons present in V^{3+}.

  • Atomic number of Ni is 28.

Ni^{2+} - [Ar] 3d^{8}

So, there will be 2 unpaired electrons present in Ni^{2+}.

  • Atomic number of Cu is 29.

Cu^{+} - [Ar] 3d^{10}

So, there is no unpaired electron present in Cu^{+}.

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

7 0
3 years ago
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