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.
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Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
The element that gains electrons, becomes reduced.
While the one which loses electrons, becomes oxidized.
In this equation,
CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.
By balancing the equation, we will get:
3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O
Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.
Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.
So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .