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3 years ago

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What energy moves though a conductors

1 answer:

ira [324]3 years ago

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<span> silver copper gold aluminum iron steel <span>brass</span></span>

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Given the following chemical reaction: 2 O2 + CH4 --> CO2 + 2 H2O

Elis [28]

The answer is D) 144 grams O2

5 0

4 years ago

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Can someone please help me with this ASAP??

kipiarov [429]

1.00

Explanation:

the density of water is always 1

7 0

3 years ago

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Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2

Helen [10]

Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>

<em />

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>

6 0

3 years ago

What is the independent variable in this reaction

jolli1 [7]

Answer:

Explanation:

what wheres the answer???

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3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago