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olga nikolaevna [1]
2 years ago
12

How many moles of iron(III) oxide are produced from 3.0 moles of oxygen and excess Iron Sulfide as described by the chemical equ

ation below? 4FeS + 502 → 2Fe2O3 + 4502​
Chemistry
1 answer:
8090 [49]2 years ago
5 0

Answer: 1.2 mol

Explanation:

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A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa
german

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

PV=Constant

That means:

P_{1}V_{1}=P_{2}V_{2}

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

P_{1}V_{1}=P_{2}V_{2}

so,

V_{2}=85\times 525\div 65

=686 mL

Volume of gas will be 686 mL.

8 0
2 years ago
Read 2 more answers
Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous iron atom.
Nikolay [14]
The ionization energy of an element is the amount of energy required to remove one mole of electrons from the element in its gaseous state. The equations for the first three are:

Fe(g) → Fe⁺(g) + e⁻

Fe⁺(g) → Fe⁺²(g) + e⁻

Fe⁺²(g) → Fe⁺³(g) + e

7 0
2 years ago
When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and
Vika [28.1K]
The correct answer is 1. Lose electrons and become positive ions.


I hope my answer was beneficial to you! c:
5 0
2 years ago
Read 2 more answers
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
2 years ago
¿Cuántos moles de aluminio (AI) se necesitan para formar 3,7 mol de Al2O3? 4Al(s) + 302 (g) -> 2A1203 (s) ​
tatuchka [14]

Answer:

3.7 mol Al2O3 x 4 mol Al = 7.4 mol Al 2 mol Al2O3

Explanation:

8 0
2 years ago
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