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For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

Answers

Answer 1

Answer:

Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

Answer 2

Answer:

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

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Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(SO_4) and Ba(OH)_2 At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution? What is the conducting species in this initial solution? Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker? What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker? Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

Answers

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The balanced equation is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Final answer:

The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.

Explanation:

Chemical Reaction and Metric Titration

Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:

Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)

In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.

Learn more about Metric Titration here:

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Which of the following chemical equations violates the law of conservation of mass? (1 point) Li2SO4 + 2Ca(NO3)2 yields 2LiNO3 + 2CaSO4 Ti + 2Cl2 yields TiCl4 2C2H2 + 5O2 yields 4CO2 + 2H2O MgSO4 + 2NaOH yields Mg(OH)2 + Na2SO4

Answers

i hope it hepls.......

Answer: Li2SO4 + 2Ca(NO3)2 —> 2LiNO3 + 2CaSO4

Consider the balanced equation for the following reaction:7O2(g) + 2C2H6(g) → 4CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.

Answers

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=(8g)/(32g/mol)=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $(4)/(7)* 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\n\n\text{Mass of carbon dioxide}=(0.143mol* 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}* 100\n\n\text{Experimental yield of carbon dioxide}=(90* 6.292)/(100)=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

What is the isoelectric point of proteins?

Answers

Isoelectric point. The isoelectric point (pI, pH(I), IEP), is the pH at which a particular molecule carries no net electrical charge in the statistical mean. The standard nomenclature to represent the isoelectric point is pH(I), although pI is also commonly seen, and is used in this article for brevity.

The isoelectric point (pI, pH(I),IEP), is the pH at which a particular molecule carries no net electrical charge in the statistical mean.

5. What is the speed (Velocity) of a cyclist who covers 10 km (convert to metersfirst!) in 14 minutes and 30 seconds? Remember time must be in seconds first!

Answers

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Kinematics.

Here, we know that, Velocity = Distance / Time,

So here, Distance = 10km = 10×1000 = 10000 metres.

, Time = 14 min 30 sec = 870 seconds,

so now, we get velocity as,

=> V = 10000 ÷ 870 => 11.49 m/s .

Hence, Velocity is 11.49 m/s.

Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for acetic acid = 1.8 x10-5 ).

Answers

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these rare metals and nonmetals.

The correct answer is 5.59.

What is PH?

pH, historically denoting "potential of hydrogen" is a scale used to specify the acidity or basicity of an aqueous solution.
Acidic solutions are measured to have lower pH values than basic or alkaline solutions

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH= pka+log(salt)/(acid)$