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tatiyna
3 years ago
13

Suppose a system of two particles, represented by circles, have the possibility of occupying energy states with 0, 10, or 20 J.

Collectively, the particles must have 20 J of total energy. Are there any other energetically equivalent configurations?
Chemistry
1 answer:
MrMuchimi3 years ago
4 0

Answer:

Explanation:

The possible energy states for the particles are 0, 10 and 20 J.  

The constraint in the system is that the total energy of the particles must be 20 J.

One given configuration where the total energy is 20 J is if both the particles occupy the 10 J state.

Hence, (10;10) is the given configuration.

Another possibility is if one of the particle is in 0 J state and another is in 20 J state. Hence, the system has a total energy of 0+20 = 20 J.

Hence, the possible configuration can be written as (0;20) or (20;0) which are energetically equivalent to the given configuration. Note that if the circles are indistinguishable, then the configuration (0,20) and (20,0) is the same thing.

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7. Indicate if the following are correct or incorrect.
solong [7]

Answer:

1s2: correct.

6s3: incorrect because the subshell s can hold up to 2 electrons.

5f10: correct.

4d3: correct.

2d4: incorrect because the second energy level just has the s and p subshells.

3p10: incorrect because the p subshell can hold up to 6 electrons.

Explanation:

Hello there!

In this case, according to the attached file, which shows the correct orbitals and electrons, we can proceed as follows:

1s2: correct.

6s3: incorrect because the subshell s can hold up to 2 electrons.

5f10: correct.

4d3: correct.

2d4: incorrect because the second energy level just has the s and p subshells.

3p10: incorrect because the p subshell can hold up to 6 electrons.

Best regards!

5 0
3 years ago
A student mixed two chemicals to allow them to
Westkost [7]

Answer:

B .it is an exothermic reaction

Explanation:

i need brainleyest

4 0
3 years ago
A 5.00 cm^3 sample of gold has a mass of 96.5 g. Calculate the density of gold.
Sholpan [36]

Answer:

\boxed {\boxed {\sf 19.3 \ g/cm^3}}

Explanation:

We are asked to calculate the density of gold. The density of a substance is its mass per unit volume. It is calculated using the following formula.

\rho= \frac{m}{v}

The mass of the sample is 96.5 grams. The volume of the sample is 5.00 cubic centimeters.

  • m= 96.5 g
  • v= 5.00 cm³

Substitute the values into the formula.

\rho= \frac{ 96.5 \ g}{5.00 \ cm^3}

Divide.

\rho= 19.3 \ g/cm^3

The density of gold is <u>19.3 grams per cubic centimeter.</u>

8 0
3 years ago
Is o-toluic acid soluble in ether, NaOH?, Is Napthalene soluble in NaOH?
natima [27]
Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.

Naphthalene is insoluble in NaOH.
8 0
4 years ago
Mannose (C6H12O6) is a simple sugar found in many fruits and vegetables. How many oxygen atoms are in 7. 15 x 10^23 molecules of
maxonik [38]

Number of oxygen atoms present in 7.15×10²³ molecules of mannose are 1.92≈2 a m u .

<h3>What is mannose sugar?</h3>

D-mannose is a simple sugar found in many fruits. It is related to glucose. In some cells it occurs naturally in the body.

Mannose is a six-carbon sugar found in a variety of fruits and vegetables. This sugar is not found free in foods. It is a part of polysaccharide chains attached to a variety of proteins.

To calculate number of oxygen atoms present in 7.15×10²³  molecules of mannose-

Since,

1 molecule of oxygen=2.303×10²³ no. of atoms

Here,  total number of molecules of oxygen is 6.

Therefore, 6 molecule of oxygen =6×2.303×10²³÷7.15×10²³

=1.92≈2 a m u .

Hence, the total number of oxygen atoms in 7.15×10²³ molecules of mannose is 1.92 ≈2 a m u.

Learn more about mannose sugar, here:

brainly.com/question/7008130

#SPJ4

6 0
2 years ago
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