CHEMISTRY MIDDLE SCHOOL

Write the pressure equilibrium constant expression for this reaction. 2CO2 (g) + 4H2O (g) → 2CH3OH (l) + 3O2 (g)

Answers

Answer 1
Answer:

Answer:

The pressure equilibrium constant (Kp) = (P O₂)³/(P CO₂)²(P H₂O)⁴.

Explanation:

  • For the reaction:

2CO₂ (g) + 4H₂O (g) → 2CH₃OH (l) + 3O₂ (g).

The pressure equilibrium constant (Kp) = the product of the pressure of the products side components / the product of the pressure of the reactantss side components.

each one is raised to a power equal to its coefficient.

∴ The pressure equilibrium constant (Kp) = (P O₂)³/(P CO₂)²(P H₂O)⁴.

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The rise and fall of ocean Water are called

.is a pot holder a conductor or a insulator

Answers

Insulator, to be a conductor some form of heat energy would have to pass through.
Insulator because it does not let heat pass through

Which variables can affect the speed of a chemical reaction? (Choose all that apply)A.) Addition of inhibitor

B.) Change in concentration of reactant

C.) Temperature

D.) Particle size

Answers

The variables which can affect the speed of a chemical reaction are inihibitor, concentration, temperature and particle size.

What is the speed of reaction?

Speed of any reaction also known as rate of reaction and it gives idea about the duration that at how much time it will complete.

  • Addition of inhibitor slows down the speed of the reaction as inhibitor blocks the activity of the reactants.
  • Change in concentration of reactant also chnages the speed of the reaction because it completely depends on the reactant concentration.
  • Temperature changes also affect the speed of the reaction as it changes the kinetic property of the reaction.
  • Particle size provides the surface area to get react so changes in this property will also changes the speed of the reaction.

Hence all of the above will changes the speed of the reaction.

To know more about speed of reaction, visit the below link:

brainly.com/question/15841605

Your answer is B(I'm not very much sure about this one), C, and D.

I hope very much that I am correct, hope this helps you.

~Onyx~

An elements atomic number is 99. How many protons would an atom of this element have ?

Answers

A elements atomic number is is number of protons. It would have 99 protons. Because an atom by nature is neutral, it would also have 99 electrons

Least to greatest 2.62,2 2/5, 26.8%, 2.26%,271%

Answers

2.26%, 26.8%, 2 2/5, 2.62, 271%

Hope this helps! :D
2.62
2.50
26.8%  - .268
2.26%  - .0226
271%  - 2.71
Converting a Percent to a Decimal is by dividing by 100.

so your answer would be
2.26%, 26.8%, 2.50, 2.62, 271%

Define and explain what an atom is

Answers

Hi,

Atom is the second smallest form of any matter behind elementary particles. They are not visible to human eye. It consists of protons and neutrons in the atom core and electrons outside its core. All protons, neutrons and electrons are a part of atomic nuclei.

The more neutrons atom haves the bigger the mass of an atom is.

Number of electrons determines which element does atom represent. The more electrons atom haves the heavier element atom represents.

Different elements have different properties.

Hope this helped you understand what atom is better.

Hope this helps.
r3t40

Please help me!!!30 ptsShow all work and box in your answers.
1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of
sulfur. Calculate the empirical formula of this compound.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen
indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?


3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %
oxygen. Find the empirical formula of this compound.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that
4.00 grams of calcium are present. What is the empirical formula of the compound formed?

7. A 60.00 gram sample of tetraethyl lead, a gasoline additive, is found to contain 38.43grams of
lead, 17.83 grams of carbon, and 3.74 grams of hydrogen. Find its empirical formula.

8. Determine the molecular formula for the compound with empirical formula P2O5 and molar mass of
284 g/mol.

9. Determine the molecular formula for the compound with empirical formula OCNCl and molar mass
of 232.41 g/mol.

10. A compound is found to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. Its molar mass is
60.0 g/mol. What is its molecular formula?

11. A compound is 64.9%carbon, 13.5% hydrogen and 21.6% oxygen. Its molar mass is 74.0 g/mol.
What is its molecular formula?

12. A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molar mass is 88.0 g/mol.
What is its molecular formula?

Answers

Answer:

Question 7 to 12 are given in attached file because character limit is only 5000

Explanation:

1.  A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Given data:

Mass of sample = 15 g

Mass of sodium = 8.83 g

Mass of sulfur = 6.17 g

Empirical formula = ?

Solution:

Number of gram atoms of Na = 8.83 / 23 = 0.4

Number of gram atoms of S = 6.17 / 32 = 0.2

Atomic ratio:

            Na               :               S          

           0.4/0.2         :            0.2/0.2  

            2                  :               1        

Na : S  = 2 :  1

Empirical formula is Na₂S.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Given data:

Mass of phosphorus = 4.433 g

Mass of oxygen = 10.150 g - 4.433 g = 5.717 g

Empirical formula = ?

Solution:

Number of gram atoms of P = 4.433 / 30.9738 = 0.1431

Number of gram atoms of O = 5.717/ 15.999 = 0.3573

Atomic ratio:

            P                        :               O          

        0.1431/0.1431         :            0.3573/0.1431

            1                         :                  2.5        

P : O  = 2(1 : 2.5)

Empirical formula is P₂O₅.

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Given data:

Percentage of sodium = 36.48%

Percentage of sulfur = 25.41%

Percentage of oxygen = 38.11%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 36.48 / 23 = 1.6

Number of gram atoms of S = 25.41/ 32 = 0.8

Number of gram atoms of O = 38.11/ 16 = 2.4

Atomic ratio:

            Na              :               S              :      O

        1.6/0.8            :            0.8/0.8       :     2.4/0.8

            2                  :                1              :       3

Na: S : O  = 2 :  1 : 3

Empirical formula is Na₂SO₃.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Given data:

Percentage of iron = 63.52%

Percentage of sulfur = 36.48%

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 63.52 / 55.845 = 1.14

Number of gram atoms of S = 36.48 / 32 = 1.14

Atomic ratio:

            Fe                :                   S            

        1.14/1.14            :               1.14/1.14    

            1                  :                   1            

Fe : S  = 1 :  1

Empirical formula is FeS.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %  oxygen. Find the empirical formula of this compound.

Given data:

Percentage of sodium = 32.38%

Percentage of sulfur = 22.65%

Percentage of oxygen = 44.99%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 32.38 / 23 = 1.4

Number of gram atoms of S = 22.65/ 32 = 0.7

Number of gram atoms of O = 44.99/ 16 = 2.8

Atomic ratio:

            Na              :               S               :            O

        1.4/0.7            :            0.7/0.7        :           2.8/0.7

            2                  :                1             :             4

Na: S : O  = 2 :  1 : 4

Empirical formula is Na₂SO₄.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that  4.00 grams of calcium are present. What is the empirical formula of the compound formed?

Given data:

Mass of sample = 20g

Mass of bromine = 20 g - 4 g = 16 g

Mass of calcium = 4 g

Empirical formula = ?

Solution:

Number of gram atoms of bromine = 16 / 80= 0.2

Number of gram atoms of calcium =  4/ 40= 0.1

Atomic ratio:

            Ca                :               Br      

        0.1/0.1              :            0.2/0.1

            1                   :                2      

Ca: Br  = 1 :  2

Empirical formula is CaBr₂.

Answer:

1)Na2S

2)P2O5

3)Na2SO3

4)FeS

5)Na2SO4

6)CaBr2

7)C8H20Pb

8)P4O10

9)C3Cl3N3O3

10)C2H4O2

11)C4H10O

12)C4H8O2

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Moles Na = 8.83 grams / 22.98 g/mol = 0.384 moles

Moles S = 6.17 grams / 32.065 g/mol = 0.192 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 0.384  / 0.192  = 2

S: 0.192 /0.192 = 1

For each mol S we have 2 mol Na

The empirical formula is Na2S

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Mass of oxygen = 10.150 - 4.433 = 5.717 grams

Moles P = 4.433 grams / 30.97 g/mol

Moles P = 0.143 moles

Moles O = 5.717 grams / 16.0 g/mol = 0.357 moles

To find the mol ratio we divide by the smallest amount of moles

P: 0.143 moles / 0.143 moles = 1

O = 0. 357 moles / 0.143 moles = 2.5

For each P atom we have 2.5 O atoms

The empirical formula is P2O5

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Na = 36.48 grams / 22.98 g/mol = 1.587 moles

Moles S = 25.41 grams / 32.065 g/mol = 0.792 moles

Moles O = 38.11 grams / 16.0 g/mol = 2.382 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 1.587 moles / 0.792 moles = 2

S: 0.792 moles / 0.792 moles = 1

O: 2.382 moles / 0.792 = 3

The empirical formula = Na2SO3

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Fe = 63.52 grams / 55.845 g/mol = 1.137 moles

Moles S = 36.48 grams / 32.065 g/mol = 1.137 moles

The empirical formula is FeS
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