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3 years ago

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A house is built with a granite countertop. The heat capacity of the countertop is 158.5 kJ/°C. A hot pan of water is placed on

the countertop, and 5000 J of heat energy is transferred into the countertop. By how much does the temperature of the countertop change?

1 answer:

Svetlanka [38]3 years ago

8 0

The temperature of the countertop changes by 0.032 °C

The quantity of heat transferred to the countertop is given by

Q = CΔT where Q = quantity of heat transferred to the countertop = 5000 J = 5 kJ, C = heat capacity of the countertop = 158.5 kJ/°C and ΔT = temperature change of the countertop.

Since we require the temperature change of the countertop, we make ΔT subject of the formula.

So, ΔT = Q/C

So, substituting the values of the variables into the equation, we have

ΔT = Q/C

ΔT = 5 kJ/158.5 kJ/°C

ΔT = 0.032 °C

So, the temperature of the countertop changes by 0.032 °C

Learn more about temperature change here:

brainly.com/question/16384350

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Answer- The correct answer is ammonia.

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Lastly Hydrogen and chlorine combined together to form hydrochloric acid. Hence the only compound left with nitrogen is ammonia and has nitrogen and hydrogen as its basic elements.

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Which elements form ions that are usually colored in solid compounds and in solution? A. alkali metals B. alkaline earth metals

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Answer:

C. transition elements

Explanation:

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3 years ago

1. How much heat is required to vaporize 25 g of water at 100˚C?

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3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of

Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

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