Iron rusts when in contact with water.
For an element whose third shell contains six electrons, the appropriate electron configuration is; 1s2 2s2 2p6 3s2 3p4.
The electron configuration shows the distribution of electrons in the shells of an atom and in orbitals.
We have been told that the six electrons are found in the third shell. This shell has n=3 and the configuration of this shell must ns2 np4.
The only electron configuration that meets this standard is 1s2 2s2 2p6 3s2 3p4.
Learn more: brainly.com/question/18704022
Positive ion with a radius smaller than the radius of the atom
Answer:
(A) 0.129 M
(B) 0.237 M
Explanation:
(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:
- 2HA + Ba(OH)₂ → BaA₂ + 2H₂O
Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).
We <u>convert mass of phthalate to moles</u>, using its molar mass:
- 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol
Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:
- 9.27 mmol HA *
= 6.64 mmol Ba(OH)₂
Finally we calculate the molarity of the Ba(OH)₂ solution:
- 6.64 mmol / 35.8 mL = 0.129 M
(B) The reaction between Ba(OH)₂ and HCl is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
So<u> the moles of HCl that reacted </u>are:
- 17.1 mL * 0.129 M *
= 4.41 mmol HCl
And the <u>molarity of the HCl solution is</u>:
- 4.41 mmol / 18.6 mL = 0.237 M
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
