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mylen [45]
3 years ago
5

How did Democritus contribute to the development of the atom?

Chemistry
1 answer:
matrenka [14]3 years ago
6 0

Democritus developed the atomic model. He theorized that atoms were specific to the material which they composed. He also believed that the atoms different in size and shape, were in constant motion in a void, collided with each other; and during these collisions, could stick together.

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What best explains why sodium is more likely to react with another element than an element such as neon
goblinko [34]

Answer: Neon is not reactive (full valence shell)

Explanation:

Neon is a noble gas and has a stable structure (8 valence electrons) -therefore, is not very reactive.

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Fied the emicicol formula of a compoun containing<br> 2.44% hydrogen, 39.02% sulphur, 58.54% oxygen
Viefleur [7K]
The answer is in the attachment below:

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2 years ago
Which of the following is the best thermal conductor?<br><br> metal<br> wood<br> paper<br> air
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Answer:

metal

Explanation:

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Read 2 more answers
(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the
djyliett [7]

Answer:

Explanation:

1. The amount of CaCO3 must be so small that  

P

CO

2

 is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full  

P

CO

2

 required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)  

K

c

=

[

CH

3

OH

]

[

H

2

]

2

[

CO

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)  

K

c

=

[

CO

]

[

H

2

]

[

H

2

O

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

3 0
3 years ago
What two categories are used in classifying particulate matter?
Allushta [10]
<span>The two categories for classifying particulate matter are through analysis of the intensive and extensive properties. Intensive properties are independent properties that can be measured independent of the amount of matter while extensive properties are measured dependent on the amount.</span>
3 0
3 years ago
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