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1 year ago

Determine the empirical formula of a compound that was found to contain 6.412 g potassium, 2.292 g n, and 7.871 g o.

Chemistry

1 answer:

Jet001 [13]1 year ago

5 0

Answer: KNO3

Explanation: Hello! the answer to your question is KNO3

We get this by dividing each by their molar mass

Potassium: 6.412/39.10 = 0.1639

Nitrogen: 2.292/14.01 =0.1635

Oxygen: 7.871/16.00 = 0.4919

We then pick the smallest of these results, this would be Nitrogen and divide everything with this answer, 0.1635

Potassium: = 0.1639/0.1635 = 1.002, about 1

Nitrogen: 0.1635=0.1635, 1

Oxygen: = 0.4919, 3.008, about 3

Now that we have the number of moles, we put it together, making KNO3

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If 25 mol of C8H18 are available, how many mol of CO2 can be produced

Nady [450]

Answer:

200 mol

Explanation:

2C8H18+25O2=16CO2+18H2O

2 mol → 16 mol

25 mol → X mol

X=25×16÷2=200

7 0

3 years ago

Convert 99 moles of CO2 to atoms

lorasvet [3.4K]

Answer:

1.78 × 10²⁶ Atoms

Explanation:

Relation between number of molecules and moles is,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 99 mol × 6.022 × 10²³ Molecules/mol

No. of Molecules = 5.96 × 10²⁵ Molecules

Also, In CO₂ Molecule there are 3 atoms.

So,

No. of atoms = 5.96 × 10²⁵ Molecules × 3

No. of atoms = 1.78 × 10²⁶ Atoms

3 0

2 years ago

Be sure to answer all parts. For the complete redox reactions given here, write the half-reactions and identify the oxidizing an

OlgaM077 [116]

Answer : $O_2$ is the oxidizing agent and Fe is the reducing agent.

Explanation :

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

$4Fe+3O_2\rightarrow 2Fe_2O_3$

The half oxidation-reduction reactions are:

Oxidation reaction : $Fe\rightarrow Fe^{3+}+3e^-$

Reduction reaction : $O_2+4e^-\rightarrow 2O^{2-}$

In order to balance the electrons, we multiply the oxidation reaction by 4 and reduction reaction by 3 then added both equation, we get the balanced redox reaction.

Oxidation reaction : $4Fe\rightarrow 4Fe^{3+}+12e^-$

Reduction reaction : $3O_2+12e^-\rightarrow 6O^{2-}$

$4Fe+3O_2\rightarrow 2Fe_2O_3$

In this reaction, $'Fe'$ is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and $'O_2'$ is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.