Electrochemical gradients Because ions carry a charge (positive or negative) their transport across a membrane is governed not only by concentration gradients across the membrane but also by differences in charge across the membrane (also referred to as membrane potontia) Together, the conoentration (chemical) gradient and the charge difforence (electrical gradient) across the plasma membrane make up the electrochemical gradient. consider the plasma membrane of an animal cell that oontains a sodium potassium pump as wel as two non-gated (always open) ion channelse a Nat channel and a K channel. The effect of the sodium potassium pump on the concentrations of Na and K as well as the distribution of charge across the plasma membrane is indicated in the figure below. Outside cell channel (Nat) high K 1 high inside cel which of the following statements correctly describes) the driving forces for diffusion of Na and K ions through their respective channels? 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 gradiont across the plasma membrane c. The diffusion of K ions out of the cel is impeded by the KT concentration gradient across the plasma membrane.

Answerhund's rule;

Answer: temperature? i hope this and have a wonderful

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³