We are given the pOH of the solution of 10.75. pOH is the property of the solution that is related to the OH ion concentration of the solution. THe formula to be followed is pOH = -log (OH); OH- = 10^-pOH. In this case, OH- = 10^-10.75 equal to B. 1.778 x 10^-11 M
Answer:
Option 2= Glucose
Explanation:
Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.
Facilitated diffusion:
it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.
Primary active transport:
The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.
Secondary active transport:
It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.
PH is defined as the negative log of Hydrogen ion concentration. Mathematically we can write this as:
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
We are given the concentration of
. Using the value in formula, we get:
![pH=-log[1.8*10^{-4}]=3.745](https://tex.z-dn.net/?f=pH%3D-log%5B1.8%2A10%5E%7B-4%7D%5D%3D3.745%20)
Therefore, the pH of the solution will be 3.745
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