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).
Yes. its a chemichal change.
<h3>
Answer</h3>
HF
<h3>Explanation</h3>
A buffer solution contains <em>a weak acid</em> and<em> its conjugate base</em>. The two species shall have a similar concentration in the solution. It's also possible for <em>a weak base</em> and <em>its conjugate acid</em> to form a buffer solution.
The KF solution already contains large number of 
ions. The objective is to thus find its conjugate acid or base.
contains no proton 
and is unlikely to be a conjugate acid. Assuming that is a conjugate base. Adding one proton to would produce its conjugate acid.
Therefore 
is the conjugate acid of . happens to be a weak acid. As a result, combining with would produce a solution with large number of both the weak acid and its conjugate base, which is a buffer solution by definition.
Answer:
here's the set up for the right side
BaSO4 + HCl
Ba 1 0
Cl 0 1
H 0 1
S 1 0
O 4 0
Explanation:
You only have to answer whats in the chemical compound so since there is no hydrogen in BaSO4, you can put 0 or leave it blank :)
