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Cloud [144]
4 years ago
12

Consider four different samples: aqueous LiF, molten LiF, aqueous AgF, and molten AgF. Current run through each sample produces

one of the following products at the cathode: solid lithium, solid silver, or hydrogen gas. Match each sample to its cathodic product. Drag the appropriate items to their respective bins.
Chemistry
1 answer:
meriva4 years ago
6 0

Answer:

Molten LiF forms solid lithium

Aqueous AgF forms solid silver

Aqueous LiF forms hydrogen gas.

Explanation:

Aqueous solution of LiF has water molecule, there will be competing will be two competing cations for reduction, while Li^{+}(aq) andH^{+}(aq).However the reduction potential of H^{+}(aq) is  more than that of the Li^{+}(aq) . Hence, H^{+}(aq) reduced to hydrogen gas.

MoltenLiF  has only solid form of LiF and Li^{+}(aq) reduced to form Li solid.

An aqueous form of AgF contains water molecules.there will be competing will be two competing cations for reduction, while Ag^{3+}(aq) andH^{+}(aq).However the reduction potential of Ag^{3+}(aq) is  more than that of the H^{+}(aq) . Hence, Ag^{3+}(aq) reduced to solid silver.

Therefore, Molten LiF forms solid lithium

Aqueous AgF forms solid silver

Aqueous LiF forms hydrogen gas.

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Convert 4.50 moles of Fe to atoms of Fe
olya-2409 [2.1K]
Each mole of a substance contains 6.02 x 10²³ particles
Atoms of Fe = 4.5 x 6.02 x 10²³
= 2.709 x 10²⁹ atoms
7 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
Pls help 6th grade science
lys-0071 [83]

Answer:

C.)

Explanation:

I also saw this question on Brainly! Hope this helps!

5 0
3 years ago
Stacy reads that a 50:50 mixture of methanol and water is best for keeping a car’s radiator from freezing when the temperature g
salantis [7]
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.

2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.

3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.

4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.

5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
5 0
3 years ago
Read 2 more answers
Bromine has two naturally occurring isotopes. Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Using the atomic mass
dsp73

Explanation:

Given parameters:

Mass of Br-79 = 78.918 amu

Abundance = 50.69%

Unknown:

Mass of Br - 81 = ?

Solution:

Relative Atomic mass of Br  = 80

The proportion by which each fraction of an isotope occurs in nature is called the geonormal abundance.

   RAM =  Ma Ba     +      McBc

  The formula above is used to find the average mass of the given isotopes.

Since we know the percentage abundance of Br-79, that of Br-81 = 100-50.69= 49.31%

  RAM =  Ma Ba     +      McBc

RAM = relative atomic mass

   Ma Ba = mass and abundance of Br-79

    McBc = mass and abundance of Br-81

     80 = (78.918 x \frac{50.69}{100}   )    +    ( \frac{49.31}{100} x C)

    80 =  40 + 0.49C

        0.49C = 40

                C = \frac{40}{0.49}   = 81.63amu

learn more;

Isotopes brainly.com/question/10862182

#learnwithBrainly

6 0
3 years ago
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