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Nimfa-mama [501]
3 years ago
12

an electric heater having resistance equal to 5 ohm is connected to electric source . If it produces 180 J of heat in one second

,find the potential difference across the electric heaterl​
Chemistry
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

V = 30 volts

Explanation:

Given that,

The resistance of an electric heater, R = 5 ohm

Heat produced is 180 J in one second

We need to find the potential difference of the electric heater. The heat produced is given by :

E=I^2Rt\\\\or\\\\E=\dfrac{V^2}{R}t\\\\V=\sqrt{\dfrac{ER}{t}} \\\\V=\sqrt{\dfrac{180\times 5}{1}} \\\\V=30\ V

So, the potential difference across the heater is 30 volts.

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An iron storage tank is placed in the ocean with no cathodic protection. a. What is the electrolyte in this reaction? b. As the
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Answer:

Iron is being oxidized at the anode and water is acting as the electrolyte.

Explanation:

When iron is exposed to oxygen and water , the rusting of iron takes place.

<u>The reaction taking place at anode : Oxidation of iron.</u>

{anode:}\;\text{Fe}(s)\;{\longrightarrow}\;\text{Fe}^{2+}(aq)\;+\;2\text{e}^{-}

The reaction taking place at cathode : Reduction of oxygen in the air.

{cathode:}\;\text{O}_2(g)\;+\;4\text{H}^{+}(aq)\;+\;4\text{e}^{-}\;{\longrightarrow}\;2\text{H}_2\text{O}(l)  

The overall reaction:

{overall:}\;2\text{Fe}(s)\;+\;\text{O}_2(g)\;+\;4\text{H}^{+}(aq)\;{\longrightarrow}\;2\text{Fe}^{2+}(aq)\;+\;2\text{H}_2\text{O}(l)

The rust that is hydrated iron(III) oxide can form iron(II) ions which can react further with oxygen.

4\text{Fe}^{2+}(aq)\;+\;\text{O}_2(g)\;+\;(4\;+\;2x)\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;2\text{Fe}_2\text{O}_3{\cdot}x\text{H}_2\text{O}(s)\;+\;8\text{H}^{+}(aq)

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Iron is being oxidized at the anode and water is acting as the electrolyte.

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4 years ago
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Answer:

7  

Explanation:

Assume we have 1 L of each solution.

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\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}

Solution 2

pH = 12

pOH = 14.00 - pOH = 14.00 - 12 = 2.0

\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}

3. pH after mixing

               H⁺  +  OH⁻ ⟶ H₂O

I/mol:     10⁻²    10⁻²  

C/mol:   -10⁻²   -10⁻²

E/mol:      0        0

The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.

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Answer:

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Explicación:

Los diferentes tipos de elementos se distribuyen en 8 grupos de tabla periódica en función de sus propiedades. Los elementos que tienen propiedades similares se colocan en el mismo grupo. Por ejemplo, el primer grupo de la tabla periódica son los metales alcalinos. Todos los metales alcalinos tienen algunas propiedades similares, es decir, un electrón en su capa más externa, alta reactividad y forma metálica, mientras que, por otro lado, los ocho elementos del grupo son gases nobles que tienen una capa más externa completa y no tienen reactividad.

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