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Ira Lisetskai [31]
3 years ago
5

Substance X has a fixed volume, and the attraction between its particles is strong. Substance Y has widely spread out particles

and can be compressed. What can most likely be concluded about these substances?
Chemistry
1 answer:
DedPeter [7]3 years ago
4 0

Answer: X is a Solid; Y is a Gas

Explanation:

There are three (3) states of matter. They are: Solid, Liquid and Gases.

Substance X and Y, belong to the states of matter.

A Solid is a substance that retains its SIZE and SHAPE without need of a container (as opposed to a liquid or gas).

Thus, it will most likely be concluded that: substance X is a Solid; while Y is a Gas

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Decomposition when an organism dies, ocean release and respiration
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A) How are physical and social systems alike?<br><br>b) How are they different?
tresset_1 [31]

Answer:

no

Explanation:

physical is human touch for example. while social is taking and hanging out or just talking on the phone.

4 0
3 years ago
(20 Points) Which definition of "coarse" matches its use in the phrase "coarse focus?"
vlada-n [284]

Answer by YourHope:


Hi! :)


Which definition of "coarse" matches its use in the phrase "coarse focus?"


A) "rough in texture"


Have a BEAUTIFUL day~

7 0
3 years ago
Read 2 more answers
Example of oxidizing agents include
Ivahew [28]

Explanation:

Substances generally tends to specialize as either oxidizing or reducing agents.

An oxidizing agent is an electron acceptor which causes a co-reactant to be oxidized in a reaction.

Examples are:

   Non-metals especially oxygen and the halogens.

Other examples are H₂SO₄ , HNO₃, KMnO₄, K₂Cr₂O₇

learn more:

Oxidizing and reducing agents brainly.com/question/5558762''

#learnwithBrainly

4 0
4 years ago
When 69.9 g heptane is burned it releases __ mol water.
professor190 [17]

Answer:

1) When 69.9 g heptane is burned it releases 5.6 mol water.  

2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>

This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.

  • We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>

n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.

<em><u>Using cross multiplication:</u></em>

1.0 mol of heptane releases → 8 moles of water.

0.7 mol of heptane releases → ??? moles of water.

<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>

<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>

<em />

5 0
3 years ago
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