Answer:
2KClO3 —> 2KCl + 3O2
The coefficients are 2, 2, 3
Explanation:
From the question given above, we obtained the following equation:
KClO3 —> 2KCl + 3O2
The above equation can be balance as follow:
There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:
2KClO3 —> 2KCl + 3O2
Now, the equation is balanced.
Thus, the coefficients are 2, 2, 3
The number of sigma and pi bonds are,
Sigma Bonds =
16 Pi Bonds =
3Explanation: 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.
Answer:
Zero
Explanation:
Recall that;
E = q + w
Where;
q = heat, w = work done
When heat is absorbed by the system q is positive
When heat is evolved by the system q is negative
When the system does work, w is negative
When work is done on the system w is positive
Step 1
ΔE1= 60 KJ + 40 KJ = 100KJ
Step 2
ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ
ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have
as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
Answer:
As blood travels through the body, oxygen is used up, and the blood becomes oxygen poor. Oxygen-poor blood returns from the body to the heart through the superior vena cava (SVC) and inferior vena cava (IVC), the two main veins that bring blood back to the heart.
Explanation: