Question:
Which of the following statements correctly describe(s) the driving forces for diffusion of Na+ and K+ ions through their respective channels? Select all that apply.
A)The diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the plasma membrane.
B)The diffusion of Na+ ions into the cell is impeded by the electrical gradient across the plasma membrane.
C)The diffusion of K+ ions out of the cell is impeded by the K+ concentration gradient across the plasma membrane.
D)The diffusion of K+ ions out of the cell is impeded by the electrical gradient across the plasma membrane. The electrochemical gradient is larger for Na+ than for K+.
Answer:
"The concentration gradient and the electro-chemical gradient" describes the driving forces for diffusion of Na+ and K+ ions through their respective channels
Explanation:
The Na ions diffusion inside the cell is facilitated by the concentration gradient of the Na ions which is present across the plasma membrane. Hence, the diffusion of the K ions which is present outside the cell and will be impeded due to the electrical gradient which is present near the plasma membrane. Thus, the electro-chemical gradient is greater as compared to the Na ion than that of the K ion.
Answer:
Mass of proton is
<h2>
1.6726219 × 10-27 Kg</h2>
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Answer:
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Explanation:
Answer:
9.9652g of water
Explanation:
The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:
n = PV / RT
Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)
Replacing:
n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K
n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:
1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>
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As the initial mass of water was 10g, the mass of water that remains in liquid phase is:
10g - 0.0348g = <em>9.9652g of water</em>
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I hope it helps!
