Answer:
Active transport by the Na+-K+ pump
Explanation:
Active transport by the Na+-K+ pump
Maintenance (and restoration) of the resting ion concentrations depends on the Na+-K+ pump. Once gated ion channels are closed, the combined action of the pump and ion leakage (particularly that of K+) establishes a resting membrane potential in a typical neuron of around âˆ'70 mV.
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
6.4 g BaSO₄
Explanation:
You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.
Molarity (mol/L) = moles / volume (L)
(Step 1)
55 mL / 1,000 = 0.055 L
Molarity = moles / volume <----- Molarity ratio
0.5 (mol/L) = moles / 0.055 L <----- Insert values
0.0275 = moles <----- Multiply both sides by 0.055
(Step 2)
Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (BaSO₄): 233.387 g/mol
0.0275 moles BaSO₄ 233.387 g
--------------------------------- x ------------------- = 6.4 g BaSO₄
1 mole
there are 8 planets in our solar system 1.Mercury 2.venus 3.earth 4.mars 5.jupiter 6.saturn 7.uranus 8. neptune
Answer:
involuntary, attached to the eyeball, nonstriated.
Explanation: