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.
Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.
5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
4 x 6.02 x 10²³ = 2.41 x 10²⁴
