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3 years ago

How much heat is released when 0.762 mol of copper cools from 81.6°C to 52.2°C?

Chemistry

1 answer:

andriy [413]3 years ago

8 0

Answer:

548.089 J

Explanation:

-The molar mass of copper is 63.546 grams.

-We obtained the mass of 0.762 moles of copper:

$mass =moles\times molar\ mass\\\\\\=0.762\times 63.546\\\\=48.422\ g$

-We apply the specific heat formula to calculate the amount of heat released:

$q=mC\bigtriangleup T\\\\\\=48.422\ g\times 0.385 \ J/g\textdegree C\times (81.6-52.2)\textdegree C\\\\\\=548.089 J$

Hence, the amount of heat released is 548.089 J

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ΔH for the reaction below is -826.0 kJ/mol. Calculate the heat change when a 69.03-g sample of iron is reacted.4Fe(s) + 3 O2(g)

Anuta_ua [19.1K]

Answer:

c. -1020.9 kJ

Explanation:

4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s) ΔH = -826.0 kJ/mol.

atomic weight of iron = 56

69.03 g = 69.03 / 56

= 1.23268 moles

Heat released by 1.23268 moles

= 1.23268 x 826.0

= -1020.9 kJ .

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jeka57 [31]

Answer:

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3 years ago

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For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

Now, we conclude that the oxidation number increases this means $KNO_{3}$ itself get oxidized to $O_{2}$ and it can act as reducing agent.

4 0

4 years ago

Read 2 more answers