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MAVERICK [17]
3 years ago
7

Consider four different samples: aqueous LiBr, molten LiBr, aqueous AgBr, and molten AgBr. Current run through each sample produ

ces 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:
alexandr1967 [171]3 years ago
6 0

Answer:

  • aqueous LiBr → hydrogen gas
  • molten LiBr → solid lithium
  • Aqueous AgBr → solid silver
  • molten AgBr → solid silver

Explanation:

Electrolysis is a process in which an electric current is passed through an electrolyte in an electrolytic cell. The electrolyte maybe in <em>aqueous form or in molten state. </em>The electrolyte is an ionic compound, so during electrolysis, the anions move towards anode while the cations move towards the cathode. Oxidation takes place at the anode. While reduction takes place at the cathode.

In the molten form, only the constituent ions are present to be oxidized or reduced. Thus, in electrolysis of molten LiBr, lithium is reduced at the cathode and bromine is oxidized at the anode.

Li^{+}(aq) + e^{-} → Li(s)

Solid Lithium is produced at the cathode

Similarly, solid Silver is produced at the cathode during electrolysis of molten AgBr:

Ag^{+}(aq) + e^{-} → Ag(s)

However, in aqueous form, H_{2} O ions and LiBr ions compete with each other. Since H+ has a greater standard redox potential than lithium,H+ will be reduced at the cathode to produce hydrogen gas

2H^{+}(aq) + 2e^{-} → H_{2}(g)

Similarly,in aqueous AgBr, Ag will be produced at the cathode because the standard redox potential for Ag+ is higher than H+ .

Ag^{+}(aq) + e^{-} → Ag(s)

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How many particles can be stored in a 4.0 L container at room temperature (23 C) and standard atmospheric pressure
velikii [3]

Answer: 9.91×10²³ particles

Explanation:

To find the amount of particles, you will need to use the Ideal Gas Law with what we are given.

Ideal Gas Law: PV=nRT

After we find moles, we can use Avogadro's number to convert to particles.

n=\frac{PV}{RT}

P=101.3kPa=1.00 atm

V=4.0 L

T=23°C+273.15=296.15 K

R=0.08206 Latm/Kmol

n=\frac{(1.00 atm)(4.0 L)}{(0.08206Latm/Kmol)(296.15K)}

n=0.164595 mol

Now that we have moles, we can convert to particles.

0.164595mol*\frac{6.022*10^2^3 particles}{mol} =9.91*10^2^2 particles

5 0
3 years ago
8.<br> What behavior do all liquid<br> or plastic solids near a<br> heat source exhibit?<br> L
LuckyWell [14K]

Answer:

Explanation:

One of the properties of a liquid is that, <u>it's particles move freely (not tightly packed)</u> hence the reason for <u>it's free flowing (no definite shape)</u> when shaken in a container, unlike a solid whose <u>particles are tightly packed with restricted/no movement</u> and hence the reason for it's compactness and it's definite shape.

When a plastic solid (whose particle is tightly packed and have a restricted movement/no movement) is placed near a heat source, <u>it's particles gains energy in the process and starts to move (though slightly free) and become less tightly packed</u> hence the reason it is observed that plastic solids near a heat source melts.

From the above, it can be deduced that a liquid and a plastic solid near a heat source <u>have there particles move freely (and not tightly packed) hence making the two substances flow freely with no definite shape.</u>

7 0
3 years ago
You use 50.00 mL of a 12.0 M solution of HCl solution to make a 500.00 mL solution. What is the concentration of the new solutio
Alexandra [31]

Answer:

1.2 M

Explanation:

If you use the dilution equation (M1V1=M2V2), you end up with (50)(12)=(500)(M2), and when you solve for M2 you get 1.2 M.

6 0
3 years ago
Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine
Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

  • Firstly, we should write the balanced equation of the reaction:

<em>N₂ + 3H₂ → 2NH₃.</em>

<em>1) To determine the limiting reactant of the reaction:</em>

  • From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
  • This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
  • We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

  • From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>

  • As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
  • Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
  • The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
  • ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0
3 years ago
A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

5 0
3 years ago
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