HI(aq)+NaOH(aq)→ what the final balanced chemical equation with the phases included
1 answer:
Answer:
Explanation:
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In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:
Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:
Which is also balanced as the number of atoms of all the elements is the same at both sides.
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Answer:
1.56 mol H₂
Explanation:
Mg₃(Si₂O₅)₂(OH)₂
<em>There are 4 Si moles per Mg₃(Si₂O₅)₂(OH)₂ mol</em>. With that in mind we can <u>calculate how many Mg₃(Si₂O₅)₂(OH)₂ moles are there in the sample</u>, using the <em>given number of silicon moles</em>:
- 3.120 mol Si *
= 0.78 mol Mg₃(Si₂O₅)₂(OH)₂
Then we can <u>convert Mg₃(Si₂O₅)₂(OH)₂ moles into hydrogen moles</u>, keeping in mind that <em>there are 2 hydrogen moles per Mg₃(Si₂O₅)₂(OH)₂ mol</em>:
- 0.78 mol Mg₃(Si₂O₅)₂(OH)₂ * 2 = 1.56 mol H₂
The number of sigma and pi bonds are,
Sigma Bonds = 16
Pi Bonds = 3
Explanation:
Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.
Sigma Bonds = 16
Pi Bonds = 3
Explanation:
Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.