Another product: CO₂
<h3>Further explanation</h3>
Given
Reaction
2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O
Required
product compound
Solution
In the combustion of hydrocarbons there can be 2 kinds of products
If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product
If Oxygen is low, you'll get Carbon monoxide(CO) and water
Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products
if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left
2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O
When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.
<u>Explanation</u>:
- Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
- The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.
- The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
