Answer: 44g
Explanation: The formular for finding Moles is ;
Moles = Mass / Molar Mass or Formular Mass.
Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?
Making 'Formular Mass', subject of the formular; we thus have;
Formular mass = Mass / Moles = 440/ 10 = 44g
In contrast to an inhibitory transmitter, an excitatory transmitter encourages the development of an electrical signal known as an action potential in the receiving neuron.
Depolarization is brought on by excitatory neurotransmitters (decrease in membrane potential). Hyperpolarization is brought on by inhibitory neurotransmitters (increase in membrane potential).
Neurotransmitters fall into two categories: excitatory and inhibitory. While inhibitory neurotransmitters work to stop an action potential, excitatory neurotransmitters function to activate receptors on the postsynaptic membrane and enhance the effects of the action potential.
While inhibitory neurons release neurotransmitters that prevent action potential firing, excitatory neurons release neurotransmitters that cause an action potential to occur in the postsynaptic neuron.
Let's know more about Excitatory & Inhibitory
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Answer:
Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.
Explanation:
Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.
Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.
No. of moles =
= 
= 0.2 mol
Molarity of acetic acid is calculated as follows.
Density = 
1 g/ml = 
volume = 100 ml
Hence, molarity = 
= 
= 2 mol/l
As reaction equation for the given reaction is as follows.

So, moles of NaOH = moles of acetic acid
Let us suppose that moles of NaOH are "x".
(as 1 L = 1000 ml)
x = 20 L
Thus, we can conclude that volume of NaOH required is 20 ml.