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Margaret [11]
3 years ago
13

Two nitro (NO_2) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but the

y can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO_2 group can have is required to be a multiple of epsilon, where epsilon = 1.0 times^-24 J. In other words, each NO_2 group could have epsilon of rotational kinetic energy, or 2 epsilon, or 3 epsilon, and so forth - but it cannot have just any old amount of rotational kinetic energy. Suppose the total rotational kinetic energy in this system is initially known to be 39 epsilon. Then, some heat is added to the system, and the total rotational kinetic energy rises to 59 epsilon. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol.

Chemistry
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

Explanation:

The detailed and step by step analysis is as shown in the atached file with appropriate substitution.

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Calculate the activity coefficients for the following conditions:
uysha [10]

<u>Answer:</u>

<u>For a:</u> The activity coefficient of copper ions is 0.676

<u>For b:</u> The activity coefficient of potassium ions is 0.851

<u>For c:</u> The activity coefficient of potassium ions is 0.794

<u>Explanation:</u>

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

where,

\gamma_i = activity coefficient of ion

Z_i = charge of the ion

\mu = ionic strength of solution

\alpha _i = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)        ......(2)

where,

C_i = concentration of i-th ions

Z_i = charge of i-th ions

  • <u>For a:</u>

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

Now, calculating the activity coefficient of Cu^{2+} ion in the solution by using equation 1:

Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

Hence, the activity coefficient of copper ions is 0.676

  • <u>For b:</u>

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

Hence, the activity coefficient of potassium ions is 0.851

  • <u>For c:</u>

We are given:

0.02 M K_2SO_4 solution:

Calculating the ionic strength by using equation 2:

C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794

Hence, the activity coefficient of potassium ions is 0.794

6 0
3 years ago
Which element is least likely to undergo a
aleksley [76]
The element which is least likely to undergo a chemical reaction is (4) neon.
The reason for this is because neon is a noble gas, which means that it doesn't react with other elements as well.
5 0
3 years ago
Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic add. Ka = 6.5 x 10^-5.
slava [35]

Answer:

correct option is (a)

The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over

(C6H5COOH)

Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)

Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350

X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5

x = V2.275 x 10^-5

X = 1.5083 x 10^-5 moles per liter H+

pH = -log(H+) = -log 1.5083 x 10^-5 which

= 4.6215

7 0
3 years ago
To determine the concentration of an EDTA solution, 4.11 g of Zn metal was used. The volume of EDTA solution needed to reach the
USPshnik [31]

Answer:

2.23M

Explanation:

Molarity of a solution is calculated thus

Molarity = number of moles (n) ÷ volume (V)

According to this question, 4.11g of Zn metal was used in order to reach a volume of EDTA solution of 28.26 mL.

28.26mL = 28.26/1000

= 0.02826L

Using mole = mass/molar mass to calculate no. of moles of Zn

Mole = 4.11/65.4

mole = 0.0628mol

Molarity = 0.0628 ÷ 0.02826

Molarity = 2.23M

The concentration of the EDTA solution used is 2.23M

7 0
3 years ago
AHHHHH HELP ME WITH THESE QUESTIONS FOR MY QUIZ
mash [69]

Question 6: True

Question 7: Earth

Hope this helps :)

4 0
3 years ago
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