1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^4
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.
Answer:
Acid are those substances which release H + ions when dissolved in water.
Get that hundooo!
Answer:
(a)
(b)
(c)
(d)
(e)
Explanation:
To calculate de pH of an acid solution the formula is:
were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.
(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.
(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:
clearing the
(d) is a weak base so it is necessary to solve the equilibrium first, knowing
The reaction is → so the equilibrium is
clearing the <em>x</em>