Answer:
his is an example of a first-year chemistry question where you must first convert two of the pressures to the units of the third and add them up, per Dalton’s law of additive pressures. There are three possible answers, one for each of the three pressure units.
1 atm = 760 torr …… torr and mm Hg are the same
1 atm = 101.3 kPa
Dalton’s law:
P(total) = P(O2) + P(N2) + P(CO2)
Explanation:
Gases will assume whatever pressure depending on the equation of state of the mixture (in this case) and the volume htey are contained in. That could be the ideal gas law and simple mixing law, If you are quoting the partial pressures which you call simply “the pressure” of each gas, and that these refer to their values in the present mixture, then yes, we would add them up. The pressures are low enough for the ideal gas law to apply provided the temperature is not extremely low as well .
Answer:Effect of Catalysts on the Activation Energy. Catalysts provide a new reaction pathway in which a lower Activation energy is offered. A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.
Explanation:
Your answer is in this
Answer:
Bottom left corner of the periodic table
Explanation:
The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water, for example.
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.
