Answer:
The correct answer:
8)- e)2, 2 dymetilpropane
9)- b) 2-chloropropane
10)- b) hydroxyl
See the explanation below, please.
Explanation:
In the cases of exercises 8 and 9:
Correspond to alkanes, having 3 carbons are named with the prefix prop and suffix anus. In 8 it has 2, 2 methyl groups in carbon and 9 in a 2-carbon chlorine group.
In the case of 10, it corresponds to a 3-carbon alcohol: suffix prop, and prefix ol: 2- propanol; in group 2 it has an OH group corresponding to alcohol (hydroxyl).
Answer:
When water vapor in the air comes into contact with something cool, its molecules slow down and get closer together.
Explanation:
I hate it when you accidently drop your drink haha. Have a good day!!
We are given with two amounts of the reactants and is asked to determine which of the two is the limiting reactant. In this case, we just have to convert the mass of HCL first and divide by 4. The answer is 0.33. Hence, HCl is used up first. The grams of Cl2 produced is 23.43 grams
Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.
<span> 2 NaHCO</span>₂<span> </span> →<span> Na</span>₂<span>CO</span>₃<span> (s) </span>+ <span> CO</span>₂<span> (g) + H</span>₂<span>O (g)
</span>
Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.
Answer:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
Explanation:
To balance the equation, you should apply the law of conservation of mass for the equations.
The law of conservation of mass states that the no. of each atom is equal in both sides (reactants and products).
Hydrocarbon is burned in oxygen to produce CO₂ and H₂O.
So, the given equation is balanced as:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
that 1.0 mole of C₃H₈ is burned in 5.0 moles of O₂ to produce 3.0 moles of CO₂ and 4.0 moles of H₂O.
The no. of all atoms is the same in both of reactants and products side.
C (3), O (10), and H (8).
