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CaHeK987 [17]
3 years ago
8

Use the following scenario to answer the question: A cell has an antiport protein on its apical surface. The cell is placed in a

solution that contains a high concentration of Na+ relative to the inside of the cell, and a high concentration of Ca⁺⁺ relative to the inside of the cell. The antiporter will move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. This is an example of what type of transport?
Chemistry
1 answer:
balu736 [363]3 years ago
6 0

Answer:

The correct answer is "Secondary active transport".

Explanation:

Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.

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The equilibrium expression for kp for the reaction below is __________. n2 (g) o2 (g) 2no (g)
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Read 2 more answers
If 50.0 g of KCl reacts with 50.0 g of O2 to produce KClO3 according to the following equation, how many grams of KClO3 will be
Sliva [168]

Answer:

A. 82.2g of KClO3

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3

C. Formula equation:

2KCl + 3O2 —> 2KClO3

Explanation:

The balanced equation for the reaction. This is given below:

2KCl + 3O2 —> 2KClO3

Next, we shall determine the masses of KCl and O2 that reacted and the mass of KClO3 produced from the balanced equation. This is illustrated below:

Molar Mass of KCl = 39 + 35.5 = 74.5g/mol

Mass of KCl from the balanced equation = 2 x 74.5 = 149g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 122.5g/mol

Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g

Summary:

From the balanced equation above:

149g of KCl reacted.

96g of O2 reacted.

245g of KCl were produced.

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

149g of KCl reacted with 96g of O2.

Therefore, 50g of KCl will react with = (50 x 96)/149 = 32.21g of O2.

Since a lesser mass of O2 ( i.e 32.21g) than what was given (i.e 50g) is needed to react completely with 50g of KCl, therefore, KCl is the limiting reactant and O2 is the excess reactant.

A. Determination of the mass of KClO3 produced from the reaction.

In this case the limiting reactant will be used.

From the balanced equation above,

149g of KCl reacted To produce 245g of KClO3.

Therefore, 50g of KCl will react to produce = (50 x 245)/149 = 82.2g of KClO3.

Therefore, 82.2g of KClO3 is produced from the reaction.

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3.

C. Formula equation:

2KCl + 3O2 —> 2KClO3

4 0
4 years ago
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