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.
Well, 2 litres is 2 kilograms. 1000 grams = 1 kilogram
So that means that 2 litres of water = 2000 grams
The amount left after 20 years = 154.15 mg
<h3>Further explanation
</h3>
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles
The decay formula for isotope :
Then for t=20 years, the amount left :
Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
