During action potential, positively charged sodium ions move inside the cell.
So option D is correct one.
The sodium ion moves inside the cell during a action potential. The stage of action potential is called depolarization . This open voltage gated sodium channel.
Action potentials ( those electrical impulse that send signals around body ) is nothing but more than temporary shift ( from negative to positive ) in the neuron's membrane potential caused by ions suddenly flowing in and out of the neuron.
It consists of phases:
- Depolarization
- overshoot
- repolarization
An active potential propagates along the cell membrane of an axon until it reaches the terminal button.
to known more about action potential
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Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻)
= 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) =
5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
Answer:
1. NaN₃(s) → Na(s) + 1.5 N₂(g)
2. 79.3g
Explanation:
<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>
NaN₃(s) → Na(s) + 1.5 N₂(g)
<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>
First, we have to calculate the moles of N₂ from the ideal gas equation.
The moles of NaN₃ are:
The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:
and 
.
Assuming complete decomposition of both samples,
First compound:
; 
of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a 
ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; 
of the first compound would contain
and therefore the empirical formula
.
This is a tricky question. All that matters are ratios of percentages, not percentages themselves. So no one should directly compare 27.2 with 42.9. We must and shall compare the ratios (27.2 to 72.8) and (42.9 to 57.1).
Take them both down to 1 to and see what happens.
Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.
A
Carbon Oxygen
27.2g 72.8g (100-27.2)
Moles 27.2/12 72.8/16
2.27 4.55
Ratio 1 2
Do the same with the other
