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lukranit [14]
2 years ago
12

A student constructs an electrochemical cell. A diagram of the operating cell and the unbalanced ionic equation representing the

reaction occurring in the cell are shown below. The blue color of the solution in the copper half-cell indicates the presence of Cu2+ ions. The student observes that the blue color becomes less intense as the cell operates.
Identify the type of electrochemical cell represented by the diagram.
Chemistry
1 answer:
vivado [14]2 years ago
4 0

Answer:

A voltaic cell

Explanation:

A voltaic cell is a device which converts chemical energy to electrical energy. The chemical reactions that take place inside the cell causes electrons to flow from anode to cathode hence, electricity is produced. A simple voltaic cell is made by placing two different metals in contact with an electrolyte separated by a salt bridge. The cathode is the negative electrode while the anode is the positive electrode. It is also called a galvanic cell.

In a voltaic cell having a copper/copper solution half cell, reduction occurs at the cathode. Hence, at the cathode copper II ions accept two electrons and become reduced to ordinary metallic copper. This causes the blue colour of the solution to become discharged (fade) as the cell continues to function.

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The molar mass of an imaginary molecule is is 93.89 g/mol. Determine its density at STP.
Igoryamba

Answer:

Density = 4.191 gm/L

Explanation:

Given:

Molar mass = 93.89 g/mol

Volume(Missing) = 22.4 L (Approx)

Find:

Density at STP

Computation:

Density = Mass/Volume  

Density = 93.89 / 22.4  

Density = 4.191 gm/L

3 0
1 year ago
A 50.0 mL Erlenmeyer flask filled with 1-hexanol has a mass of
Iteru [2.4K]
Yes it is correct for that answer
4 0
2 years ago
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Compare Solution X to Solution A, Solution B, and Solution C. Beaker with a zoom view showing eight shperes, four groups of part
Levart [38]

Answer:

SOLUTION A

Explanation:

7 0
2 years ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

7 0
2 years ago
One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
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