Answer:
The correct answer is
B.action potential; sodium
Explanation:
Action potential
Nerve action potentials are initiated and propagated in an all or nothing fashion. The all-or nothing action potential is initiated when the membrane potential threshold value is reached
Sodium
Voltage-gated Na⁺ ions membrane channels open enabling an influx of sodium ions to rapidly enter the cell, raising the membrane potential up to +40 mV
Depolarization in the neuron is based ion exchange channels opening to allow the rapid influx of sodium ions (Na⁺) and the efflux of potassium ions (K⁺)
The neuronal responds rapidly and for short periods to the activation of an ion channel receptor by a drug or natural neurotransmitter is rapid rel
H2O, one Oxygen in the middle, surrounded by 2 hydrogen, usually close to each-other, which gives it polarity
Answer:
6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5
Explanation:
For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).
Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.
HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴
Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).
For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)
= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.
Check using the Henderson - Hasselbalch Equation...
pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18
Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.
pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.
One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.
Answer:
Some formulas for calculating mole are
Mole = Mass/ Molar mass
Mole = no of particles / avogadros constant
NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23
so we will be using the second formula
Mole = no of particles / avogadros constant
Mole = 5.03 x 10 ^23/6.02 x10^23
Mole = 8.355x10^45
hope it helps :)
Explanation:
Answer:
0.414 mole (3 sig. figs.)
Explanation:
Given grams, moles = mass/formula weight
moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)
≅ 0.414 mole (3 sig. figs.)
