Answer:
0.294 M
Explanation:
The computation of the final molarity of acetate anion is shown below:-
Lead acetate = Pb(OAc)2
Lead acetate involves two acetate ion.
14.3 gm lead acetate = Mass ÷ Molar mass
= 14.3 g ÷ 325.29 g/mol
= 0.044 mole
Volume of solution = 300 ml.
then
Molarity of lead is
= 0.044 × 1,000 ÷ 300
= 0.147 M
Therefore the molarity of acetate anion is
= 2 × 0.147
= 0.294 M
To calculate the final molarity of acetate anion in the solution, consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. When ammonium sulfate is added, it reacts with the lead(II) cations, leaving only the acetate anions in the solution. The final concentration of acetate anions is therefore the same as the initial concentration.
To calculate the final molarity of acetate anion in the solution, we need to consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. Lead(II) acetate will dissociate into lead(II) cations (Pb2+) and acetate anions (CH3COO-) in solution. However, when ammonium sulfate is added, the sulfate anions (SO42-) react with the lead(II) cations, forming lead(II) sulfate and removing them from solution. This leaves us with only the acetate anions.
First, calculate the concentration of the acetate anions in the lead(II) acetate solution. Then subtract the concentration of the acetate anions that reacted with the lead(II) cations to form lead(II) sulfate. This will give us the final concentration of acetate anions in the solution.
Let's assume we have an initial concentration of lead(II) acetate of X M. The dissociation of lead(II) acetate can be represented as:
Pb(CH3COO)2(s) ⇌ Pb2+(aq) + 2CH3COO-(aq)
Since we assume the volume of the solution doesn't change when the lead(II) acetate is dissolved, the initial concentration of acetate anions is also X M.
When ammonium sulfate is added, it reacts with the lead(II) cations according to the reaction:
Pb2+(aq) + SO4^2-(aq) ⇌ PbSO4(s)
Since the concentration of lead(II) sulfate is negligible, we can assume that all the lead(II) cations react with the sulfate anions. This removes the lead(II) cations from solution, leaving us with only the acetate anions.
Therefore, the final concentration of acetate anions is still X M.
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Answer:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
α =2
β = 19
γ = 12
δ = 14
53.2moles of O₂
Explanation:
Proper equation of the reaction:
αC₆H₁₄ + βO₂ → γCO₂ + δH₂O
This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:
CₓHₙ + (x + )O₂ → xCO₂ + H₂O
From the given combustion equation, x = 6 and n = 14
Therefore:
β = x + = 6 + = 6 + 3.5 = 9
γ = 6
δ = = = 7
The complete reaction equation is therefore given as:
C₆H₁₄ + 9O₂ → 6CO₂ + 7H₂O
To express as whole number integers, we multiply the coefficients through by 2:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
Problem 2
From the reaction:
2 moles of hexane are required to completely react with 19 moles of O₂
∴ 5.6 moles of hexane would react with k moles of O₂
This gives: 5.6 x 19 = 2k
k =
k = 53.2moles of O₂
Answer:
Boron
Explanation:
Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.
Among boron, carbon, aluminum, and silicon, aluminum has the lowest first ionization energy due to its position on the periodic table, which is further to the left and in a higher period than the other elements.
Ionization energy refers to the energy necessary to remove an electron from an atom in its gaseous state. The element with the lowest first ionization energy among boron, carbon, aluminum, and silicon is aluminum. Ionization energy increases from left to right across a period in the periodic table and from bottom to top in a group. Thus, aluminum, being to the left of boron, carbon, and silicon, has the lowest first ionization energy. Furthermore, aluminum is in the third period, which is below boron and carbon's second period.
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6. What is the activation energy for this reaction? _____
7. What is the change in free energy for the reaction? _____
8. How many intermediates are involved in this reaction? _____
9. How many transition states are involved in the reaction mechanism? _____
10. Is the reaction endergonic or exergonic overall?
6. Activation energy is the difference in energy between the reactant and the transition state. The energy of the reactant and the transition state are 5 kJ/mol and 15 kJ/mol respectively. The activation energy for this reaction is 10 kJ/mol.
7. The change in free energy for the reaction is the difference in energy between the reactant and the product. The energy of the reactant and the product are 5 kJ/mol and 10 kJ/mol respectively. The change in free energy for the reaction is 5 kJ/mol.
8. There are no intermediates involved in this reaction.
9. There is only one transition state involved in the reaction mechanism. The transition state is indicated by the highest point of the graph.
10. The reaction is endergonic overall. The energy of the product is higher than the energy of the reactants, which is only possible if energy is absorbed by the reaction.
A property that will NOT change if temperature changes
A property that changes if the amount of substance changes
A property that does NOT change if the amount of substance changes
Help :( pls
Answer:
A property that changes if the amount of substance changes
Explanation:
An extensive property is a property that depends on the amount of matter in a sample.
An extensive property changes if the amount of substance changes. For instance, mass and volume are extensive properties as they would vary depending on the amount of substance.
An extensive property is a property that changes if the amount of substance changes. For example, mass and volume are extensive properties. If you have two separate samples of a substance, each with a different amount, their mass and volume would be different. On the other hand, the melting point or boiling point of the substance, which are examples of intensive properties, would not change regardless of the amount of substance.
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Answer:
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