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Slav-nsk [51]
3 years ago
8

A galvanic (voltaic) cell consists of an electrode composed of aluminum in a 1.0 M aluminum ion solution and another electrode c

omposed of silver in a 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C_______________.. Refer to the list of standard reduction potentials.
Chemistry
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

E° = 2.46 V

Explanation:

We have a voltaic cell of aluminum and silver in their respective solutions. To know where will take place the reduction (cathode) and where will take place the oxidation (anode), we need to compare the standard reduction potentials.

E°(Al³⁺/Al) = -1.66 V

E°(Ag⁺/Ag) = +0.80 V

Since Ag⁺/Ag has the highest reduction potential that is where reduction will take place while oxidation will take place in the Al electrode.

Reduction (cathode): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Oxidation (anode): Al(s) ⇒ Al³⁺(aq) + 3 e⁻

The standard potential of the cell (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red,cathode - E°red,anode = 0.80 V - (-1.66V) = 2.46 V

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How many sulfur atoms are generated when 9.42 moles of H2S react according to the following equation: 2H2S+SO2→3S+2H2O
8_murik_8 [283]

Answer:

A) 8.51 × 10²⁴  

Explanation:

1. Gather all the information

            2H₂S + SO₂ ⟶ 3S + 2H₂O

n/mol:   9.42

2. Calculate the moles of S atoms

The molar ratio is 3 mol S:2 mol H₂S

\text{Moles of S} = \text{9.42 mol H$_{2}$S} \times \dfrac{\text{3 mol S }}{\text{2 mol H$_{2}$S }} = \text{14.13 mol S}

3. Calculate the atoms of S

\text{Atoms of S } = \text{14.13 mol S} \times \dfrac{6.022 \times 10^{23}\text{ S atoms}}{\text{1 mol S}} = \mathbf{8.51 \times 10^{24}}\textbf{ S atoms}

 

6 0
3 years ago
Read 2 more answers
In the chemical equation below, name the reactant(s) and product(s).
Olenka [21]

Answer:Methane(CH4) and dioxygen(O2) are the reactants

Carbon dioxide(CO2) and water(H2O) are the products

Explanation:

4 0
2 years ago
A chemistry graduate student is given 500. mL of a 0.20 M chloroacetic acid (HCH2ClCO2) solution. Chloroacetic acid is a weak ac
zhenek [66]

Answer:

12 g of choloracetic acid

Explanation:

The buffer equilibrium is:

HCH₂ClCO₂ ⇄ CH₂ClCO₂⁻ + H⁺

pka= -log ka =

Ka: 1,3x10⁻³ = [CH₂ClCO₂⁻] [H⁺] / [HCH₂ClCO₂]

By Henderson-Hasselbalch equation:

pH = pka + log₁₀ [A⁻] / [HA]

3,01 = 2,89 + log₁₀ [A⁻] / [HA]

1,318 = [A⁻] / [HA]

As molar concentration of chloroacetic acid (HA) is 0,20M

[A⁻] = 0,26 mol/L

The volume is 500 mL ≡ 0,5 L

0,26mol/L × 0,5 L = 0,13 moles of chloroacetic acid. In grams:

0,13 mol × (94,5g / 1mol) = <em>12 g of choloracetic acid</em>

<em></em>

I hope it helps!

3 0
4 years ago
NH3
soldier1979 [14.2K]

Answer:

First one is: ammonia

Second one is: calcium hydroxide

Explanation:

4 0
3 years ago
Use the normal boiling points: propane, C3H8, –42.1˚C; butane, C4H10, –0.5˚C; pentane, C5H12, 36.1˚C; hexane, C6H14, 68.7˚C; hep
svlad2 [7]

Answer:

The answer to your question is 126.1°C

Explanation:

                  Boiling point         Difference of boiling points  

C₃H₈             - 42.1°C

C₄H₁₀             -   0.5°C                 41.6 °C

C₅H₁₂               36.1°C                  36.6°C                       41.6 - 36.6 = 5°C          

C₆H₁₄               68.7°C                 32.6°C                        36.6 - 32.6 =4°C

C₇H₁₆               98.4°C                 29.7°C                        32.6 - 29.7 = 2.9°C

We can observe on the table that the difference of boiling points diminishes 1°C when the hydrocarbon has one more carbon, then the difference of temperature between the hydrocarbon of 8 carbons and the hydrocarbon of 7 carbons must be 2°C.

So, this difference is 29.7°C - 2°C = 27.7°C.

And the boiling point of octane is approximately 98.4 + 27.7°C = 126.1°C

7 0
3 years ago
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