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
brainly.com/question/13021637
Answer:
Decomposition
Explanation:
If we look at the process;
2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
We can see that NaHCO3 was broken down into Na2CO3, H2O and CO2.
The breakdown of one compound to yield other chemical compounds is known as decomposition.
Hence the NaHCO3 was decomposed in the process above.
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Democritus, a Greek philosopher, first developed the idea of atoms (around 460 B.C., I believe).
Hope this helps!