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liq [111]
3 years ago
14

100 POINTS!!! ANSWER ASAP

Chemistry
2 answers:
andreyandreev [35.5K]3 years ago
7 0

Answer:

1eV

Explanation:

diamong [38]3 years ago
3 0

Answer:

1ev

Explanation:

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In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il
Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.        

7 0
3 years ago
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Kobe is testing whether 100 g of substance A, 100 g of substance B, or 100 g of substance C produces more oxygen in a chemical r
skad [1K]
C, the amount of oxygen produced is the dependent variable since that is what is being measured and it is dependent on which substance is being tested
4 0
3 years ago
Cual es la relación de la temperatura y la solubilidad?
DanielleElmas [232]

Answer:

A medida que aumenta la temperatura de un líquido, la solubilidad de los gases en ese líquido disminuye. Podemos usar la Segunda Ley de la Termodinámica para explicar por qué. Calentar una solución de un gas permite que las partículas de gas se muevan más libremente entre la solución y la fase gaseosa. La Segunda Ley predice que cambiarán al estado más desordenado, más altamente disperso y, por lo tanto, más probablemente gaseoso.

Explanation:

8 0
3 years ago
What volume would 2.25 moles of Ne gas occupy at STP?
Dafna1 [17]
As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

Solution for problem:

When 1 mole Neon (Ne) occupies 22.4 dm³ at STP then the volume occupied by 2.25 moles of Neon is calculated as,

                                             = ( 22.4 dm³ × 2.25 moles ) ÷ 1 mole
                
                                             = 50.4 dm³                   1dm³ = 1 L

Result:

So
, 50.4 dm³ (Liter) volume will be occupied by 2.25 moles of Neon gas if it acts ideally at STP.
8 0
3 years ago
Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point.
givi [52]

Answer:

The amount of heat required to vaporize 2.58 kg of water at its boiling point is 5,830.8 kJ.

Explanation:

A substance undergoes a change in temperature when it absorbs or gives up heat to the environment around it. However, when a substance changes phase it absorbs or gives up heat without causing a change in temperature. The heat Q that is necessary for a mass m of a certain substance to change phase is equal to:

Q = m*L

where L is called the latent heat of the substance.

In this case:

  • m=2.58 kg
  • The heat of vaporization of water is L=2260*10³ J/kg

Replacing:

Q= 2.58 kg* 2260*10³ J/kg

Q= 5,830,800 J = 5,830.8 kJ (Being 1,000 J= 1 kJ)

<u><em>The amount of heat required to vaporize 2.58 kg of water at its boiling point is 5,830.8 kJ.</em></u>

3 0
3 years ago
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