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Nostrana [21]
3 years ago
8

Based on the oxidation states of the atoms in this reaction, answer the questions. Fe(0) + O2(0) → Fe2(3+)O3(2-) How many electr

ons does the iron half-reaction lose? How many electrons does the oxygen half-reaction gain?
Chemistry
2 answers:
Oliga [24]3 years ago
6 0
Half reaction 1: 2Fe° → Fe₂³⁺ + 6e⁻ /×2.
4Fe° → 2Fe₂³⁺ + 12e⁻.
Iron is oxidized from neutral charge (0) to oxidation number +3, one iron lose three electrons, two irons lose six electrons and four irons twelve electrons.
Half reaction 2: 12e⁻ + 3O₂ → 2O₃²⁻ 
Oxygen is reduced from neutral chage to oxidation number -2, one oxygen gain two electrons, six oxygens gain twelve electrons.
Balanced chemical reaction: 4Fe + 3O₂ → 2Fe₂O₃.
aalyn [17]3 years ago
6 0

Answer:

The iron half-reaction loses 12.

The oxygen half reaction gains 12.

The total number of electrons that are moved in this oxidation-reduction reaction is 12.

Explanation:

Edge 2020

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docker41 [41]

Answer:

126000J

Explanation:

From the question given, we obtained the following information:

M = 1500g

C = 4.2J/g°C

ΔT = 20°C

Q =?

Q= MCΔT

Q = 1500 x 4.2 x 20 =

Q = 126000J

7 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
3 years ago
How much heat is lost when 0.440 mol of steam condenses at 100 °C?
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Answer:

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Explanation:

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Q=n\lambda_v

where

n is the number of moles of the substance

\lambda_v is the latent heat of vaporization

The formula can be applied if the substance is at its vaporization temperature.

In this problem, we have:

n = 0.440 mol is the number of moles of steam

\lambda_v=40,660 J/mol is the latent heat of vaporization of water

And the steam is already at 100C, so we can apply the formula:

Q=(0.440)(40660)=17,890 J

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