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White raven [17]
3 years ago
6

Iron (III) oxide is formed when Iron combines with oxygen in the air. How many grams of Fe₂O₃ are formed when 16.7 grams of reac

ts completely with oxygen? 4Fe + O₃ ---> 2Fe₂O₃
Chemistry
1 answer:
sladkih [1.3K]3 years ago
7 0

Answer:

23.9g of Fe₂O₃ are produced

Explanation:

<em>Are formed when 16.7g of Fe reacts completely...</em>

<em />

Based on the reaction:

4Fe + O₃ → 2Fe₂O₃

<em>4 moles of Iron react per 1 mole of O₃ producing 2 moles of Fe₂O₃.</em>

<em />

To solve this question we need to convert the mass of iron to moles. The ratio of reaction is 2:1 -That is, 2 moles of Fe produce 1 mole of Fe₂O₃-. Thus, we can find the moles of Fe₂O₃ produced and its mass:

<em>Moles Fe -Molar mass: 55.845g/mol-:</em>

16.7g Fe * (1mol / 55.845g) = 0.299 moles of Fe

<em>Moles Fe₂O₃:</em>

0.299 moles Fe * (2 mol Fe₂O₃ / 4 mol Fe) = 0.150 moles Fe₂O₃

<em>Mass Fe₂O₃ -Molar mass 159.69g/mol-:</em>

0.150 moles Fe₂O₃ * (159.69g / mol) =

<h3>23.9g of Fe₂O₃ are produced</h3>

<em />

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Answer:

See the answer below

Explanation:

<em>The carbon will have to travel in the form of CO2 from the atmosphere to a primary producer (green plant), from there to a primary consumer (herbivorous animal), and finally to a secondary consumer.</em>

The primary producer (a green plant) would fix the carbon in the CO2 to carbohydrate through a process known as photosynthesis. The equation of the process is as shown below:

6 CO_2 + 6 H_2O --> C_6H_1_2O_6 + 6 O_2

The carbon, now in the form of carbohydrate, would then be picked up by an animal (a primary consumer) that feeds on the green plant. The carbon would eventually get into a secondary consumer when the secondary consumer feeds on the primary consumer that fed on the green plant.

8 0
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At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g
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Answer:

Total pressure at equilibrium is 0.2798atm.

Explanation:

For the reaction:

H₂S(g) ⇄ H₂(g) + S(g)

Kp is defined as:

Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834

If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:

H₂S(g): 0.150atm - x

H₂(g): x

S(g): x

Replacing in Kp:

\frac{X*X}{0.150atm-X} = 0.834

X² = 0.1251 - 0.834X

X² +  0.834X - 0.1251 = 0

Solving for X:

X = -0.964 → False solution: There is no negative pressures

X = 0.1298

Thus, pressures are:

H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>

H₂(g): <em>0.1298atm</em>

S(g): <em>0.1298atm</em>

Thus, total pressure in the container at equilibrium is:

0.0202atm + 0.1298atm + 0.1298atm = <em>0.2798atm</em>

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3 years ago
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marishachu [46]

Given molecule Lithium iodide (LiI)

Heat of hydration = -793 kj/mol

Lattice energy = -730 kJ/mol

Heat of hydration = Heat of solution - Lattice energy

Heat of solution = Hydration + Lattice = -793 + (- 730) = -1523 kJ/mol

Now,

Mass of LiI = 15.0 g

molar mass of LiI = 134 g/mol

# moles of LiI = 15/134 = 0.112 moles

Heat of solution for 1 mole of LiI = -1523 KJ

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