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dem82 [27]
3 years ago
15

AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is

Chemistry
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

A- 25.9 kJ

Explanation:

ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.

ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.

Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:

5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.

If 1 mole release -826kJ, 0.0313 moles release:

0.0313 moles ₓ (-826kJ / 1 mole) =<em> -25.9kJ</em>

Thus, heat involved is:

<h3>A- 25.9 kJ </h3>

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The increase of angles between two bones is and ankle
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Describe how u would separate a mixture of salt , sand and iodine into different components. please help i will give a brainlies
almond37 [142]
<h2>answer.</h2>

to remove iodine sublime

to remove sand filter

to remove water evaporate

Explanation:

1. put the mixture in a beaker

2.Heat the mixture using a watch glass filled with cold water, the iodine will sublime at the base of the watch glass.

3.Add water to sand and salt. Stir for salt to diffuse in the water

4.Filter sand from the solution using a filter paper

5.evaporate the remaining water to remain with Salt crystals.

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-Dominant- [34]

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3 years ago
At 300K, the pressure inside a gas-filled container is 500 kPa. If the pressure inside decreases to 100 kPa, but the volume rema
alukav5142 [94]
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3 0
4 years ago
Read 2 more answers
If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, wh
Valentin [98]

Answer:

N₂

Explanation:

Sodium is a larger molecule with a much higher molecular weight. However, 20g of N₂ would be a smaller amount of molecules than 20g of sodium due to how there are multiple nitrogen molecules.

2/3 * given mass of N₂ = mass of N₃

N₃ (Azide ion) given mass = 20g

Na = 20g

Masses of chemicals are equal

Na = 22.990g/mol

20g/22.990g/mol = 0.8699mol of Na

N₃ = 20g

N₃ = N g/mol x 3

N = 14.007 g/mol

14.007 x 3 = 42.021 g/mol

N₃ = 42.021g/mol

20g/42.021g/mol = 0.4759 mol of N₃

Notice how there are fewer moles of the Azide ion than the Sodium.

0.4759 moles of NaN₃ is produced

combine molecular weights:

42.021 + 22.990 = 65.011 g/mol

multiply by amount of moles of the limiting reactant:

0.4759 mol *65.011 = 30.942 g

Also, here is the balanced equation:

3N₂ + 2Na = 2NaN₃

The results are the same as the balanced equation.

4 0
3 years ago
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