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Sloan [31]
4 years ago
14

How do energy levels change when atoms collide?

Chemistry
1 answer:
Savatey [412]4 years ago
5 0
Depends.
<u>If you're in normal chem</u>, I think your answer will be that energy levels don't change.
<u>If you're in AP chem</u>, see below:
     If the collisions are completely elastic collisions, like in ideal gases, energy        level does not change.
     However, if the atoms have IMF's(intermolecular forces) then kinetic energy      is lost.

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Which of the following statements regarding physical and chemical changes is true?
skad [1K]
There’s no choices to pick from
8 0
3 years ago
A car driving at 105 km/h. what is the speed in mi/min?<br> (1 mi = 1.61 km)
DedPeter [7]

Answer:

1.087 mi/min

Explanation:

Given data:

Speed of car in Km/h = 105

Speed of car in miles/min = ?

Solution:

It is given that one miles = 1.61 Km

while it is known that one hour = 60 minutes

Thus in order to convert the km/h into mi/min we will divide the given value by 96.561.

105 / 96.561

1.087 mi/min

3 0
4 years ago
A 7.0 L sample of gas begins at 2.5 atm and 320. K. What is the new pressure if the temperature is changed to 273 K and the volu
enyata [817]

Answer:

2.1 atm

Explanation:

We are given the following variables to work with:

Initial pressure (P1): 2.5 atm

Initial temperature (T1): 320 K

Final temperature (T2): 273 K

Constant volume: 7.0 L

We are asked to find the final pressure (P2). Since volume is constant, we want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

\frac{P_{1}}{T_1} =\frac{P_{2}}{T_2} \\

We can rearrange the law algebraically to solve for P_{2}.

{P_{2}} =\frac{(T_2)(P_{1} )}{T_1} \\

Substitute your known variables and solve:

P_2 = \frac{273K(2.5atm)}{320K}  = 2.1 atm

4 0
3 years ago
What is ATP<br> plz be helpfull
Basile [38]
<span>Association of Tennis Professionals 

hope this was help helpful <33</span>
5 0
3 years ago
The pKa of H2O is 15.7 and the pKa of CH3COOH is 4.8. What label and reason correctly describe a reaction between H2O and CH3COO
Damm [24]

Answer:

CH3COOH will be the acid, because it is the stronger acid.

Explanation:

When an acid deprotonates in a solution, an equilibrium reaction will occur between the protonated form and the deprotonated form (conjugate base) of it. The equilibrium is characterized by the value of Ka, the equilibrium constant, which is the multiplication of the concentration of the products divided by the concentration of the acid.

As higher is the value of Ka, more acid is deprotonated, and stronger will be the acid. The value of pKa = -logKa, and so, as higher is the Ka, as low is the pKa. Thus, strong acids have low pKa values.

In the reaction of CH3COOH and H2O, the first one has a low value of pKa, so it is a strong acid. The water is a substance that can work as an acid or as a base, and, because of the other substance is a strong acid, it works as a base.

5 0
3 years ago
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