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Which of the following would NOT diffuse through the plasma membrane by means of simple diffusion?1 oxygen2 glucose3 a steroid h

Mumz [18]

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.

5 0

3 years ago

I need help in this :(

alexandr402 [8]

Answer:

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Explanation:

6 0

2 years ago

What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?

anzhelika [568]

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

5 0

3 years ago

Read 2 more answers

Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY

Anton [14]

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

$K_{sp}$ = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

$K_{sp}$ = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

$K_{sp}$ value describes the solubility of a particular ionic compound. The higher the $K_{sp}$ value, the higher the Solubility will be.

6.

This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility

4 0

3 years ago

What is the percent yield for the reaction of nitrogen and hydrogen to produce ammonia, if the theoretical yield was 12.5 g but

koban [17]

Answer:

81.6%

Explanation:

10.2/12.5= 0.816

0.816=81.6%

4 0

2 years ago