Photosynthesis is the process where plants create energy. It requires water, carbon dioxide and sunlight. The end result is glucose, which the plants consume, and oxygen. Cellular respiration requires oxygen and glucose. The end result is carbon dioxide, ATP, and water.
They cause most of their damage when they reach land
OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml
</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100
</span>therefore:
V1 = 20 ml
Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.
Answer:
Decreasing the temperature will shift the equilibrium leftwards towards reactants.
Explanation:
Hello!
In this case, since the reaction between chromate anions and hydrogen ions yields dichromate anions, water and heat, we can infer this is an exothermic reaction by which heat is released (remember in endothermic reactions heat is absorbed as a reactant), it means that considering the LeChatelier’s which states that increasing the temperature of an exothermic reaction shifts the equilibrium leftwards since heat is a product, otherwise (decreasing the temperature) the equilibrium will be shifted rightwards.
Therefore, decreasing the temperature is the perturbation that will shift the equilibrium leftwards towards the reactants.
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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.
