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LenKa [72]
3 years ago
11

Write a chemical equation of Dr. Jeff’s demonstration using a battery to light steel wool on fire. The battery is used to allow

a current to travel through the steel wool, thus heating it up. Steel wool is made from iron (Fe).
Chemistry
1 answer:
Elis [28]3 years ago
6 0

Answer:

2Fe(s) + 3O2(g) --------> 2FeO3(s)

Explanation:

According to the question, a  battery was used  to light the steel wool by bringing the terminals very close together. When  the battery came into contact with the steel wool, current was sent out through the thin wire.  This caused the iron to heat up quite well.

Iron reacts with oxygen under these conditions as follows;

2Fe(s) + 3O2(g) --------> 2FeO3(s)

This is the chemical reaction that occurs when the steel wool is set on fire.

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Which of the following would not be studied within the topic of stoichiometry?
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Answer:

The amount of energy required to break the ionic bonds in CaF2.

4 0
3 years ago
The mass in grams, to five sig figs, of carbon dioxide produced when 22.031 grams of C6H6 is combusted is the same as the mass o
enyata [817]

Answer:

74.566

Explanation:

C6H6 + 02 =6CO2 + 3H2O

2C6H6 + 15O2=12CO2 + 6H2O

mm of C6H6=(12x6)+(6x1)= 72+6 =78gram/mol

mass conc = 2x78=156

mm CO2= 12+(16x2)=

12+32= 44gram/mol

mass conc of CO2 = 12x44

=  528

from The equation

156g of C6H6=528g of CO2

22.031g of C6H6=Xg of CO2

therefore, Xg=

22.031 x 528/ 156

= 74.566g

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3 years ago
If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of ai
Gnom [1K]

In accordance with Dalton's Law of multiple proportions

<h3>Further explanation</h3>

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

  • air(O₂) as a limiting reactant(product=14 g)

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

  • air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions

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3 years ago
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