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Korvikt [17]
4 years ago
11

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

ormone4 a lipid soluble vitam
Chemistry
1 answer:
Mumz [18]4 years ago
5 0

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.

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Write a balanced chemical equation, complete ionic equation and a net ionic equation for Copper lll sulfate and zinc.
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Answer: CuSO4 (aq)+ Zn (s) → Cu (s) + ZnSO4(aq)

Explanation: do you mean copper II sulphate CuSO4?

4 0
3 years ago
What keeps galaxies togther? How?
aleksklad [387]
Tough question, I would suggest using Google maybe, or just plain out asking your science teacher.
8 0
3 years ago
Which part of the experiment is not touched by the independent variable or is the normal/comparison
kobusy [5.1K]
The part of the experiment that’s is not touched by the independent variable and is for comparison is called the :
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8 0
3 years ago
Write the balanced half equation of iron 2 and permanganate in a solution of acid. Show all of your work.
Aleks [24]

Answer:

5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

Explanation:

Fe⁺² + MnO₄⁻ +  H⁺ => Mn⁺² + Fe⁺³ +  H₂O

5(Fe⁺² => Fe⁺³ + 1e⁻)      =>                        5Fe⁺² => 5Fe⁺³ + 5e⁻

<u>MnO₄⁻ + 5e⁻ => Mn⁺²    =>   MnO₄⁻ + 8H⁺ + 5e⁻ => Mn⁺² + 4H₂O</u>

                                      =>  5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

6 0
3 years ago
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0
4 years ago
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