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sineoko [7]
3 years ago
10

Two liquids are shaken together in a test tube to produce a uniform liquid that does not separate into layers. Which of the foll

owing best describes the behavior of the above pair of substances?
soluble
insoluble
miscible
immiscible
Chemistry
2 answers:
Sever21 [200]3 years ago
6 0

Answer:

The answer is:

Miscible

Explanation:

When we talk about miscible we are referring to the liquids property to mix in any proportion in a homogeneous phase, forming a solution. This term can be used to other phases as gases or solids too but it is commonly referred to liquids

When the mix of two liquids doesn't produce a uniform phase we can say that  substances are immiscible and they are not able to form a homogeneous phase.

Alexxandr [17]3 years ago
5 0
It most likely would be "miscible" because "soluble" refers to something being able to dissolve in water, "insoluble" is something not being able to dissolve or mix such as oil and water and immiscible is something not being able to mix without forming layers.
Answer is: Miscible.
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A chemist measures the energy change ΔH during the following reaction: C3H8 (g) +5O2 (g) →3CO2 (g) +4H2O (l) =ΔH−2220.kJ Use the
statuscvo [17]

Answer:

The reaction is exothermic.

Yes, released.

The heat released is 4,08x10³ kJ.

Explanation:

For the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The ΔH is -2220 kJ, As ΔH is <0, <em>The reaction is exothermic.</em>

As the reaction is exothermic, the heat of the reaction will be <em>released.</em>

The heat released in 81,0g is:

81,0g C₃H₈×\frac{1mol}{44,1g}×\frac{2220kJ}{1mol}= <em>4,08x10³ kJ</em>

<em>-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-</em>

I hope it helps!

3 0
3 years ago
What can be said about an exothermic reaction with a negative entropy change?.
pashok25 [27]
Spontaneous at low temperatures.
3 0
3 years ago
Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25
Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

  • We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,
  • where, Cgas is the solubility if gas,
  • K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),
  • P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).
  • Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

  • Decreasing the temperature will increase the solubilty of O₂ gas in water.
  • When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.
  • Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.


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