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olga2289 [7]
3 years ago
7

A standard galvanic cell is constructed in which a Cu^2+ | Cu^+ half cell acts as the cathode Which of the following statements

are correct?
(Choose all that apply)
A. Cr^3+|Cr could be the other standard half cell.
B. As the cell runs, anions will migrate from the other compartment to the Cu^2+|Cu^+ compartment.
C. I_2|I^- could be the other standard half cell.
D. Cu^+ is oxidized at the cathode.
E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment
Chemistry
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment

Explanation:

In a galvanic cell, electrons flow from anode to cathode. We must remember that oxidation occurs at the anode and reduction occurs at the cathode. Hence the process; M(s) -------> M^+(aq) + e occurs at the anode.

The electrons lost at the anode are conveyed to the cathode where they are accepted by other chemical species and are reduced to the corresponding reduced species according to the reaction; M^+(aq) + e -----> M(s)

Hence the Cu^2+ | Cu^+ half cell, being the cathode accepts electrons from the other half cell for this reduction reaction to take place, hence the answer

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The equations are 2NaHCO3 -> Na2CO3 + CO2 + H20, NaHCO3 -> NaOH + CO2, 2NaHCO3 -> Na2O + 2CO2 + H20
cricket20 [7]

Answer:

The product made is Na2CO3.

The % yield is 91.8 %

Explanation:

Step 1: Data given

Mass of baking soda (NaHCO3) = 5.0 grams

Molar mass of NaHCO3 = 84.0 g/mol

Each of the equations calculated is left with 2.4g of NaOH, 1.86 g Na2O, and 3.18g Na2Co3.

Step 2: The balanced equations

2NaHCO3 → Na2CO3 + CO2 + H20

NaHCO3 → NaOH + CO2

2NaHCO3 → Na2O + 2CO2 + H20

Step 3: Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

Moles NaHCO3 = 5.0 grams / 84.0 g/mol

Moles NaHCO3 = 0.060 moles

Step 4: Calculate moles of products

For 2 moles NaHCO3 we'll have 1 mol Na2CO3

For 0.060 moles NaHCO3 we'lll have 0.060 / 2 = 0.030 moles Na2CO3

For  1 mol NaHCO3 we'll have 1 mol NaOH

For 0.060 moles NaHCO3 we'll have 0.060 moles NaOH

For 2 moles NaHCO3 we'll have 1 mol Na2O

For 0.060 moles NaHCO3 we'll have 0.030 moles Na2O

Step 5: Calculate mass of products

Mass = moles * molar mass

Mass of Na2CO3 = 0.030 moles * 105.99 g/mol = 3.18 grams

Mass of NaOH = 0.060 moles * 40.0 g/mol = 2.4 grams

Mass of Na2O = 0.030 moles *61.98 g/mol = 1.86 grams

Step 6: Calculate the percent yield

% yield = actual yield / theoretical yield

% yield Na2CO3 = (2.92 grams / 3.18 grams) *100% =  91.8 %

% yield NaOH = (2.92 grams / 2.4 grams ) *100% = 121.6 %

% yield of Na2O = (2.92 grams / 1.86 grams ) * 100% = 157 %

The product made is Na2CO3, the other reactions have a % yield greater than 100 %

7 0
3 years ago
At 500 degree C, F_2 gas is stable and does not dissociate, but at 840 degree C, some dissociation occurs: F_2 (g) 2 F(g). A fla
bazaltina [42]

Answer:

2.73 is the equilibrium constant for the dissociation of F_2 gas at 840 degree Celsius.

Explanation:

F_2(g)\rightleftharpoons 2F(g)

Initial

0.600 atm    0

Equilibrium

(0.600 atm - p)        2p

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P= 0.600 atm - p)+2p=0.984 atm

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For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

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