Answer:
case1.
The addition of acid and base leads to a change in pH of the water when adding to deionized water due to fact that acid and bases dissociated in dissolving in water. If the H+ ion increases in the water as acid addition hikes it, it will result in decreasing the pH value. The intensity of the acid also affects the dissociation of the ions.
case2
Buffers are normally formed by weak acid and its conjugate base, and adding acid to the buffer it absorbs the H+ ions so the pH will be lower and adding base or increase of OH- conjugate base resists the pH value to increase.
Answer:
sorry to waste ur time but im getting points bc i have a question and i need more points to say ig
Answer:
Explanation:
1. Calculate the moles of copper(II) hydroxide
2. Calculate the molecules of copper(II) hydroxide
<h3><u>Answer;</u></h3>
Polar: IF, PCl3, IF5
Nonpolar: CS2, SO3, SF6
<h3><u>Explanation:</u></h3>
- Polar molecules form when two atoms do not share electrons equally in a covalent bond.
- A molecule is classified as a polar molecule when the arrangement of the atoms is such that one end of the molecule has a positive electrical charge and the other end has a negative charge.
- A non-polar molecule does not have electrical poles.The electrons are distributed more equally.
- Therefore, a non-polar molecule does not have a profusion of charges at the opposite ends. The majority of hydrocarbon liquids are non-polar molecules.
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl * = 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L