Answer:
reproduction
Explanation:
reproduction, process by which organisms replicate themselves
Answer : The mole fraction and partial pressure of 
and 
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of 
= 1.79 mole
Moles of 
= 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and gases.
and,
and,
Thus, the mole fraction of and gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and gases are, 1.54, 1.03 and 3.20 atm respectively.
Answer:
pH = 11.05
Explanation:
It is possible to answer this question using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A] / [HA⁺]
Where A in this case is weak base (dimethylamine) and conjugate acid (HA⁺) is dimethylamine hydrochloride.
As Ka= Kw / Kb = 1x10⁻¹⁴ / 7.4x10⁻⁴ = 1.35x10⁻¹¹ And pKa is -log Ka = <em>10.87 </em> pH of the solution is:
pH = 10.87 + log₁₀ [0.600] / [0.400]
<em>pH = 11.05</em>
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I hope it helps!
Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
