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ratelena [41]
3 years ago
6

Write the balanced equation for the oxidation of iron to iron (III) oxide. Will mark brainlest answer

Chemistry
1 answer:
Butoxors [25]3 years ago
5 0

Start with the unbalanced chemical equation

<span>Fe(s)+O2(g)→Fe2O3(s)</span>

The idea behind balancing chemical equations is that the number of atoms an element has on the reactants' side must be equal to the number of atoms it has on the products' side.

These atoms will become a part of different compounds once the reaction is completed, but they must always be in equal numbers on both sides.

So, look at iron first. One atom reacts, but two are produced - notice the 2 subscript iron has in Fe2O3. This means you must double the number of atoms on the reactants' side to reach an equality.2

<span>Fe(s)+<span>O2(g)</span>→Fe2<span>O3(s)</span></span>

Now look at oxygen. Two atoms react, but three are produced. The trick here is to find a common multiple that will make the number of atoms equal on both sides.

The easiest way to do this is to multiply the atoms that react by 3, which will give you 6 oxygen atoms that react, and the atoms that are produced by 2 - this will get you 6 oxygen atoms produced.

<span>2Fe(s)+3<span>O2(g)</span>→2Fe2<span>O3(s)</span></span>

However, notice that the iron atoms are unbalanced again. You have 2 that react, but 4 that are produced → multiply the atoms that react by 2 again, which will give you

4<span>F<span>e<span>(s)</span></span>+3</span><span><span>O<span>2(g)</span></span>→2</span><span>F<span>e2</span><span>O<span>3<span>(s</span><span>

</span></span></span></span>

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It is also called a finite resource

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The boiling point of methanol is 64 7°C. Its melting point is -976°C. At room temperature (-25°C), methanol is in which state?
marissa [1.9K]

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.

The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.

At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults

3 0
2 years ago
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
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