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:
Less expensive than natural fibers
Strong and long-lasting
Easily available
Easy to maintain
Dry up quickly
It will help burn more calories and build muscle in your legs because of the extra weight you are carrying when walking or running
Answer:
89,4%
Explanation:
If you have a solutio of 2,5g of acetanilide in 50mL of water and you warm this solution to 100°C you will dissolve all acetanilide because the maximum solubility in 50mL will be:
5,5g / 100mL → 2,75g / 50mL.
Then, if you cold the water to 0°C the solubility in 50mL will be:
0,53g / 100mL → 0,265g / 50mL.
That means you will precipitate:
2,5g - 0,265g = <em>2,235g of acetanilide</em>
The theoretical percent recovery will be:
2,2365g / 2,5g ×100 = <em>89,4%</em>
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I hope it helps!
Answer is: the % ionization of hypochlorous acid is 0.14.
Balanced chemical
reaction (dissociation) of an aqueous solution of hypochlorous acid:
HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).
Ka = [H⁺] · [ClO⁻] / [HClO].
[H⁺] is equilibrium concentration of hydrogen cations or protons.
[ClO⁻] is equilibrium concentration of hypochlorite anions.
[HClO]
is equilibrium concentration of hypochlorous acid.
Ka is the acid
dissociation constant.
Ka(HClO) = 3.0·10⁻⁸.
c(HClO) = 0.015 M.
Ka(HClO) = α² · c(HClO).
α = √(3.0·10⁻⁸ ÷ 0.015).
α = 0.0014 · 100% = 0.14%.
