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Alchen [17]
3 years ago
7

What are the two main properties of matter?

Chemistry
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

physical and chemical.

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At 298 K, oxygen (O2) and ozone (O3) have different properties because their
Paha777 [63]

Answer: (3) molecules have different molecular structures.


Explanation:


1) Oxygen (O₂) and ozone (O₃) are allotropes of each other.


2) Allotropes are different structural forms of a same element with different structures and properties, when they are in the same state: solid, liquid, gas.


3) The bonds is what define the structure and properties of the substances, so since O₂ has only two bonds and O₃ has three bonds, the properties and behaviors of the element are different.


4) Other example of allotropes are graphite and diamond: two different forms of carbon. Both, graphite and diamond are formed only by carbon atoms, but they are bonded differently so, as you know, diamond and graphite have different properties: graphite is very soft while diamond is one of the hardest known substances.

5 0
3 years ago
The following data was collected when a reaction was performed experimentally in the laboratory.
Lera25 [3.4K]

Answer:

Mass = 279.23 g

Explanation:

Given data:

Number of moles of Fe₂O₃ = 3 mol

Number of moles of Al = 5 mol

Maximum amount of iron produced by reaction = ?

Solution:

Chemical equation:

Fe₂O₃ + 2Al    →     Al₂O₃  +  2Fe

Now we will compare the moles of iron with Al and iron oxide.

                          Fe₂O₃     :       Fe

                             1           :        2

                             3          :       2×3 = 6 mol

                            Al          :          Fe

                              2         :           2

                               5        :           5 mol

The number of moles of iron produced by Al are less so Al is limiting reacting and it will limit the amount of iron so maximum number of iron produced are 5 moles.

Mass of iron:

Mass = number of moles × molar mass

Mass =   5 mol  ×  55.845 g/mol

Mass = 279.23 g

 

3 0
3 years ago
The half-life of 131I is 80 days. How much of a 500.0 mg sample remains after 3 half lives?
I am Lyosha [343]

Answer:

62.5 mg

Explanation:

Just multiply the original amount by  1/2   three times:

500 mg  x  1/2  x  1/2  x  1/2 = 62.5 mg

6 0
2 years ago
Answer pls ASAP will mark the brainliest
harina [27]
First one……………………………….
6 0
3 years ago
Read 2 more answers
Can somebody help? Do not answer if you don't know.
blondinia [14]

Answer:

10.8amu

Explanation:

Given parameters:

Abundance of B - 10  = 20% = 0.2

Abundance of B - 11  = 80% = 0.8

Unknown:

Atomic mass of Boron = ?

Solution:

The atomic mass of Boron can be can be calculated using the expression below;

 Atomic mass  = (abundance of B - 10 x mass of isotope B - 10 ) +( abundance of  B - 11 x mass of isotope B- 11)

 Atomic mass  = (0.2 x 10) + (0.8 x 11) = 2 + 8.8  = 10.8amu

5 0
3 years ago
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