Answer:
0.1472 mol/L is the concentration of the barium hydroxide solution.
Explanation:

Mass of potassium hydrogen phthalate = 2.050 g
Molar mass of potassium hydrogen phthalate = 
According to reaction , 2 moles of potassium hydrogen phthalate reacts with 1 mole of barium hydroxide, then 0.01004 moles of potassium hydrogen phthalate will :
of barium hydroxide
Moles of barium hydroxide = 0.005020 mol
Volume of the barium hydroxide solution = 34.10 mL = 0.03410 L
1 mL = 0.001 L

Molarity of the barium hydroxiude silution :

0.1472 mol/L is the concentration of the barium hydroxide solution.
Answer:
<h2>118,438.5 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula

where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have

We have the final answer as
<h3>118,438.5 moles</h3>
Hope this helps you
Answer: (Structure attached).
Explanation:
This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.
The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.
Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).
The reaction occurs in three steps:
- The attack on the electrophile forms the sigma complex.
- The loss of a proton regenerates an aromatic ring.
- The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).
Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Answer: O2+6H12O6=CO2+ENERGY(ATP)
I DON'T THINK SHE IS CORRECT
Explanation: