Answer:
A) involves changes in temperature
Explanation:
The figure is missing, but I assume that the region marked X represents the region in common between Gay-Lussac's law and Charle's Law.
Gay-Lussac's law states that:
"For an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"
Mathematically, it can be written as
where p is the pressure of the gas and T its absolute temperature.
Charle's Law states that:
"For an ideal gas kept at constant pressure, the volume of the gas is directly proportional to its absolute temperature"
Mathematically, it can be written as
where V is the volume of the gas and T its absolute temperature.
By looking at the two descriptions of the law, we see immediately that the property that they have in common is
A) involves changes in temperature
Since the temperature is NOT kept constant in the two laws.
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
Answer:
1. Na2SO4 + Ba(NO3)2 → NaNO3 + BaSO4
2. NaNO3 + NH4Cl → NaCl + NH4NO3
4. Sorry, I don't know
5. Refer to the attachment..
6. NaC2H3O2(aq) + AgNO3(aq) = NaNO3(aq) + AgC2H3O2
<h3>Hope it can help you and please mark me as a brainlist... </h3>
Answer:
pH = 3.23
Explanation:
Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:
HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)
The Ka of this reaction is:
Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]
<em>Where [] are concentrations in equilibrium of each species</em>
As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:
[HC₂H₃O₂] = 0.020M - X
[C₂H₃O₂⁻] = X
[H₃O⁺] = X
<em>Where X is reaction coordinate.</em>
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Repalcing in Ka expression:
1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]
1.8x10⁻⁵ = [X] [X] / [0.020M - X]
3.6x10⁻⁷ - 1.8x10⁻⁵X = X²
3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0
Solving for X:
X = -0.0006M → False solution. There is no negative concentrations
X = 0.000591M → Right solution.
As:
[H₃O⁺] = X
[H₃O⁺] = 0.000591M
As pH = -log[H₃O⁺]
<h3>pH = 3.23</h3>
