Answer: 3d
Explantion:
1) Period 4 contains the elements with atomic numbers 19 through 36.
2) The elements with atomic numbers 19 (K) and 20 (Ca) fill the orbital 4s.
3) After that, as Aufbau's rule may help you to remember, the energy of the orbitals 3d is lower than the energy of the orbtitals 4p. So, the element 21 (Sc) start fillind the orbital 3d.
There are ten 3d orbitals, so the elements 21 through 30 fill the 3d orbitals.
Those elements, called transition metals are: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn.
When the 3d orbitals are full, the next elements in the same period 4, fill the six 4p orbitals.
Answer:
28%
Explanation:
Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.
It's a pretty straightforward question.
At the final step that's missing, I Did
(y)C3H8 = 2.8 / ( 2.8 + 7.1)
(y)C3H8 = 0.28
Answer:
6.22 × 10⁻⁵
Explanation:
Step 1: Write the dissociation reaction
HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺
Step 2: Calculate the concentration of H⁺
The pH of the solution is 2.78.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M
Step 3: Calculate the molar concentration of the benzoic acid
We will use the following expression.
Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution
Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M
Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid
We will use the following expression.
Ka = [H⁺]²/Ca
Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵
Answer:
0.6743 M
Explanation:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:
- Molarity = moles / volume
- moles = Molarity * volume
- 0.4293 M * 39.27 mL = 16.86 mmol NaOH
<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.
Finally we <u>calculate the concentration (molarity) of acetic acid</u>:
- 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
