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Zolol [24]
3 years ago
11

Predict the species that will be oxidized first if the following mixture of molten salts undergoes electrolysis:

Chemistry
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

Explanation:

the species that will be oxidized first will depend upon their oxidation potential .  

oxidation potential are given as follows

F⁻ =  -  2.87 V   , Br⁻ = - 1.09 V , Cl⁻ = -1.36 V, Mg⁺² = -2.37 V, Cu⁺² = +.34 V

Higher the oxidation potential , higher the tendency to be oxidised .

so Cu⁺²  is easiest to be oxidised  . F⁻ is to be oxidised most difficult.

The order from easiest to most difficult as follows

Cu⁺² >  Br⁻ > Cl⁻ > Mg⁺² > F⁻

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Use the ideal-gas law to estimate the number of air molecules in your physics lab room, assuming all the air is N2. Assume a roo
stiv31 [10]

Answer:

4.13×10²⁷ molecules of N₂ are in the room

Explanation:

ideal gases Law → P . V = n . R . T

Pressure . volume = moles . Ideal Gases Constant . T° K

T°K = T°C + 273 → 20°C + 273 = 293K

Let's determine the volume of the room:

18 ft . 18 ft . 18ft = 5832 ft³

We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L

1 atm .  165045.6 L = n . 0.082 L.atm/mol.K . 293K

(1 atm .  165045.6 L) / 0.082 L.atm/mol.K . 293K = n

6869.4 moles of N₂ are in the room

If we want to find out the number of molecules we multiply the moles by NA

6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

4 0
3 years ago
Which food type will show no colour charge when it is tested with millon's reagent /Bluret solution​
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Answer:

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Explanation:

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8 0
3 years ago
Oxidation state of Nitrogen in N2O5​
balandron [24]

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

7 0
2 years ago
Which of the following hazards does not apply for methanol?
AleksAgata [21]
Answer: the answer is C oxidizing
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3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
Anestetic [448]

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

8 0
3 years ago
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