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tester [92]
3 years ago
13

Given the reaction that occurs in an electrochemical cell:

Chemistry
2 answers:
KatRina [158]3 years ago
8 0

Answer:

c) +2 to 0

Explanation:

SO4 has a charge of -2, so the Cu attached to that has to be a +2 since the polyatomic molecule has no overall charge

Cu(s) is a solid metal and they have no charge, therefore it is zero

Copper undergoes Oxidation (gain of electrons)

jolli1 [7]3 years ago
7 0
Answer:
c) +2 to 0

explanation
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A gas sample has volume 3.00 dm3 at 101
gavmur [86]

Answer:

V₂ = 21.3 dm³

Explanation:

Given data:

Initial volume of gas = 3.00 dm³

Initial pressure = 101 Kpa

Final pressure = 14.2 Kpa

Final volume = ?

Solution;

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 Kpa × 3.00 dm³ = 14.2 Kpa × V₂

V₂ = 303 Kpa. dm³/  14.2 Kpa

V₂ = 21.3 dm³

Explanation:

3 0
2 years ago
jelani and Mikah are watching an approaching storm from their living room window. All of a sudden they see a bright flash of lig
Shtirlitz [24]

Answer: Heat causes surrounding air to rapidly expand and vibrate, which creates the pealing thunder we hear a short time after seeing a lightning flash.

4 0
3 years ago
What variable represents specific heat in the equation Q = mcAT?
Elan Coil [88]

Answer: A. The variable c

============================================

Explanation:

Q = heat transferred

m = mass

c = specific heat

\Delta T = delta T = change in temperature

4 0
3 years ago
What is one way to separate a mixture?
Dmitry_Shevchenko [17]
1. <span>Substances in a </span>mixture<span> can be separated using different </span>methods, for example <span>distillation (separation of two liquids).
2. </span>Mass<span> is the amount of matter an object contains.
</span>Volume is t<span>he amount of space occupied by a three-dimensional object.
3. d(wallet) = 15g</span>÷5cm³ = 3g/cm³.
4. Weight of an astronaut also increases. W=m·g.
7 0
3 years ago
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
2 years ago
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