Plants combine water and Carbon dioxide from the air to make glucose for themselves, and giving off oxygen in the process. So CO2 and H2O would be on ther reactant side (left) with O2 and C6H12O6 being on the product side (right).
Answer:
Option A. 0
Explanation:
To know the number of lone pair on each H in NH3, we shall determine how NH3 is formed. This is illustrated below:
3H + N —> NH3
Three atoms of Hydrogen, H reacted with 1 atom of nitrogen to produce ammonia, NH3. Each atoms contribute one electron each to form the covalent bond in NH3.
Hydrogen has only one electron and it will share it with the nitrogen atom to produce ammonia.
Further details can be seen in the attached photo.
Further more, in the attached photo, we can see that there is no lone pair of electron in the hydrogen atom as all it's electron has been used to form bond with the nitrogen atom. Only the nitrogen has a lone pair of electron.
Therefore, there are zero lone pair of electron on each hydrogen, H atom in ammonia, NH3.
Simple!
Atomic number= number of protons AND number of electrons (in an atom)
Mass number= number of neutrons
Answer:
Two objects are in thermal equilibrium if they are in close contact that allows either to gain energy from the other.
Imidazole's cyclic structure is aromatic and the lone pair present at the N₁ atom available for donation. hence, according to question N₁ is the most basic atom in the structure.
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What is resonating structure?</h3>
- Two π bonds (between C₂-C₃ and N₁-C₅), as well as one lone pair on N₄, can interact with one another to generate a delocalized π system in the cyclic structure.
- This delocalization is intriguing since it has the same number of delocalized electrons as benzene—six.
- As a result, imidazole, like benzene, has a closed, delocalized ring with six π electrons. So, like benzene, it is regarded as an aromatic chemical with resonance stability.
- N₄ is neutral since it cannot be donated because it needs to use its lone pair to be aromatic.
- On the other hand, N₁ already forms a π connection, which helps the system become delocalized.
- N₁ is sp² hybridized and has a trigonal planar basic form. Its lone pair cannot communicate with the delocalized π system since it is pointed away from the cyclic structure.
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