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LuckyWell [14K]
3 years ago
6

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

Chemistry
1 answer:
algol133 years ago
7 0

Answer:

1) C₃H₄N

Explanation:

The empirical formula of a compound is the formula that gives the positive integer ratio of the atoms of the elements in the compound in the simplest form

The mass of carbon in the compound = 90 grams

The molar mass of carbon = 12.011 g/mol

The number of moles of carbon = 90 g/(12.011 g/mol) ≈ 7.4931313 moles

The mass of hydrogen in the compound = 11 grams

The molar mass of hydrogen = 1.00794 g/mol

The number of moles of carbon = 11 g/(1.00794 g/mol) ≈ 10.913348 moles

The mass of nitrogen in the compound = 35 grams

The molar mass of nitrogen = 14.0067 g/mol

The number of moles of carbon = 35 g/(14.0067 g/mol) ≈ 2.49880414 moles

Dividing by the smallest mole ratio gives;

The proportion of carbon, C = 7.4931313/2.49880414 = 2.9987 ≈ 3

The proportion of nitrogen, N = 10.913348 /2.49880414 = 4.367 ≈ 4

The proportion of nitrogen, N = 2.49880414 /2.49880414 = 1

Therefore, the empirical formula of the compound is C₃H₄N.

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AleksandrR [38]

Hey there!:

1 mole of  C6H12O6 ------------------ 6 moles of C

3.0 moles of C6H12O6 ------------- ??

3.0 *6 / 1 =>

18.0 moles of C


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3 years ago
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If air is a fluid, why do we not include the buoyant force caused by the displacement of air by the objects in this experiment?
uysha [10]

Archimedes' principle allows us to find that the reasons why the thrust is not written when a body is in the air is:

  • The thrust of air is about 800 times less than the thrust of a fluid
  • In general the other forces (weight, tension) are much greater than thrust

Archimedes' principle establishes that the thrust is equal to the weight of the dislodged liquid (fluid)

        B = ρ g V

Where B is the thrust, ρ and V the density and volume of the fluid, respectively, g the acceleration due to gravity.

In the attachment you have a diagram of a system in equilibrium in air and water, we can see that in the two cases for a system in equilibrium

                         B -W = 0

                         B = W

Let's find the value of the thrust in each case and compare

               B_{air} = \rho_{air} g V_{air}

               B_{water} = \rho_{water} \ g V_{water}

Used the density

               \rho_{air} = \  1.26 \  kg/m^3 \\\rho_{water} = 997 kg/m^3

               B_{air} = 1.26 \ g \ V_{air}\\B_{water} = 997 \ g \ V_{water}

Suppose that the volume of the two bodies is the same

               B_{water} = 792 B_{air}r

           

We can see that the thrust in air or other gas is about 800 times less than the thrust in liquids. This is the reason that in many problems the thrust is not written when the body is in the air.

In conclusion, using Archimedes' principle, we find that the reason why the healed thrust is not written for a body is in the air is:

  • The thrust of air is about 800 times less than the thrust of a fluid
  • In general the other forces (weight, tension) are much greater than thrust

Learn more about Archimedes' principle here:

brainly.com/question/787619

5 0
2 years ago
A 7.337 gram sample of chromium is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 9.595 g. Dete
Thepotemich [5.8K]
<h3>Answer:</h3>

Empirical formula is CrO

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of sample of Chromium as 7.337 gram
  • Mass of the metal oxide formed as 9.595 g

We are required to determine the empirical formula of the metal oxide.

<h3>Step 1 ; Determine the mass of oxygen used </h3>

Mass of oxygen = Mass of the metal oxide - mass of the metal

                          = 9.595 g - 7.337 g

                         = 2.258 g

<h3>Step 2: Determine the moles of chromium and oxygen</h3>

Moles of chromium metal

Molar mass of chromium = 51.996 g/mol

Moles of Chromium = 7.337 g ÷ 51.996 g/mol

                                 = 0.141 moles

Moles of oxygen

Molar mass of oxygen = 16.0 g/mol

Moles of Oxygen = 2.258 g ÷ 16.0 g/mol

                            = 0.141 moles

<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>

Mole ratio of Chromium to Oxygen

          Cr : O

0.141 mol : 0.141 mol

             1 : 1

Empirical formula is the simplest whole number ratio of elements in a compound.

Thus the empirical formula of the metal oxide is CrO

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