Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
Answer:
31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .
Explanation:
Given data:
Number of moles of CaCl₂ = 2.6 mol
Number of Cl₂ ions = ?
Solution:
CaCl₂ → Ca²⁺ + 2Cl⁻
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
In one mole of CaCl₂ there are two moles of chloride ions present.
In 2.6 mol:
2.6×2 = 5.2 moles
1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions
5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol
31.31× 10²³ number of Cl⁻
Answer:
a bird sitting on a branch
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
