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kari74 [83]
3 years ago
7

The solubility of a solute (a dissolved substance) in a solvent (the dissolving medium) is the most important chemical principle

.
a. True
b. False
Chemistry
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

False

Explanation:

The most important chemical principle is not just one, there are several, such as the constitution of an atamo among others.

The important thing about this principle is that solutes can be dissolved in solvents where as a product they give a solution.

According to the amount of solute, three types of solutions can be observed:

Supersaturated solutions, where the solute is in excess and part cannot be integrated into the solvent, precipitating in the lower part of the container.

Saturated solutions, where the solute is in the right concentration so that it does not precipitate and constantly interacts in a cyclical way with the solvent.

Subsaturated solutions, where the amount of the solute is small.

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5<br> What is a physical property of matter?
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Characteristic of matter that is not associated with its change in chemical composition.

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Suppose a friend says that we don't need to worry about the rising temperatures associated with global climate change. Her claim
ludmilkaskok [199]

Answer:

Carbonic acid could be formed.

Explanation:

Hello,

Based on her claim, it would be a really useful strategy to prevent global warming, nevertheless, there would be a problem if a increasing amount of carbon dioxide is not buried at the bottom of the ocean yet it flows freely along the sea and probably reacting with the water, causing carbonic acid to be formed  and subsequently cutting back the sea's pH (increasing its acidity).

It would be useful, but a constant monitoring of the sea's pH must be needed because this could cause some species to be affected not only by the temperature but for the acid pH as well.

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3 years ago
Need help setting the problem up
Sveta_85 [38]

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

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3 years ago
Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
liubo4ka [24]

The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

  • ²³²Th emits α particles, it is the most abundant 99.9%
  • ²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

  • On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
  • The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0
3 years ago
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