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.
Answer:
pH = 8.1
Explanation:
Assuming that we are at 25 degrees Celsius, pH + pOH = 14.
We can then plug in the given pOH and solve for pH:
pH + pOH = 14
pH + 5.9 = 14
pH = 14 - 5.9 = 8.1
Field or discipline
hope it helped
Explanation:
It is given that volume of 
is 300 mL and molarity is 0.200 M.
Volume of NaCl is 200 mL and molarity is 0.050 M.
The chemical reaction will be as follows.
for 
is given as 
.
As, molarity is number of moles present in liter of solution.
Hence, moles of 
(aq) will be calculated as follows.
moles of (aq) = 
= 0.06 mol
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
= 0.120 M
Mole of 
= 
= 0.010 M
Now, Q = ![[Pb^{2+}][Cl^{-}]^{2}](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D)
= 
= 
As, Q < hence, there will be no formation of precipitate.
