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PilotLPTM [1.2K]
3 years ago
13

3. The dry air in the Alps helped to prevent damage to the cells in the Iceman's body.

Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

False

Explanation:

because they do not prevent damage they make damage

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Elements 3 to 10 (Li to Ne) show a more or less steady increase in IE. What does this tell you about the energy level that each
Karo-lina-s [1.5K]

Elements 3 to 10 (Li to Ne) show a more or less steady increase in ionization energy.

<h3>What is ionisation energy?</h3>

The amount of energy required to remove an electron from an isolated atom or molecule.

The major difference is the increasing number of protons in the nucleus as you go from lithium to neon. That causes greater attraction between the nucleus and the electrons and so increases the ionization energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus.

Learn more about the ionisation energy here:

brainly.com/question/20658080

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6 0
2 years ago
Find density of nitrogen dioxide at 75*C and 0.805 atm.
Eva8 [605]

Answer:

1 (348) (D2) = 273 (2.05) (0.805) D2= 1.29 g/L

Explanation:

5 0
3 years ago
Which state of matter would be described as having highly energized, charged particles that move extremely fast
uranmaximum [27]
There are four states of matter, solid, liquid, gas and plasma. Their formation is as when solid is heated it converts into liquid, liquid on heating converts into gases and gases on heating converts into plasma.

Plasma:
           Plasma is the fourth state of matter. It is the highest energy state of matter.

Composition:
           Plasma is made up of negatively charged and positively charged particles.

Result:
           The answer to your question is Plasma.

8 0
3 years ago
When do replicated chromosomes are aligned in the center
Art [367]

Answer:

<u>During Metaphase</u>

<u></u>

Explanation:

During Metaphase the cell chromosome align themselves in the middle of the cell.This occur due to a cellular process called "Tug of War".

The chromosome which have been replicated and joined at the central point  called centromere are called sister chromatids

Prior to metaphase , Kinetochore type of protein are formed around the centromere. Long protein filament called kinetochore are extended from poles to other end of the cell attached to kinetochore.

Therse is important checkpoint in the middle of mitosis called<u> metaphase.</u>

At this point cell ensure that , it is ready to divide or not.

Once the cell ensure that everything is ready to divide. Only after then , yhe cell enters the fourth phase called <u>anaphase.</u>

7 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
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