Unit 1 Practice: Medium

Answer: 3.01 × 10²³ O atoms

Work:

Each CO₂ contains 2 oxygen atoms.

0.250 mol CO₂ × 2 = 0.500 mol O atoms

Then:

0.500 × 6.022 × 10²³

= 3.01 × 10²³ O atoms

Explanation: Convert molecular moles into moles of the specific atom first, then use Avogadro's number.

Answer: 3.01 × 10²³ molecules

Work:

First convert grams to moles:

9.00 / 18.02 ≈ 0.499 mol

Then:

0.499 × 6.022 × 10²³

≈ 3.01 × 10²³ molecules

Answer: 10.8 amu

Work:

Convert percentages to decimals:

20.0% = .200 80.0% = .800

Then:

(10.0)(.200) + (11.0)(.800)

= 2.0 + 8.8

= 10.8 amu

Explanation: Average atomic mass is a weighted average, so the more abundant isotope contributes more.

Answer: CH₂O

Work:

Assume 100 g:

C: 40.0 g H: 6.7 g O: 53.3 g

Convert to moles:

C:

40.0/12.01 ≈ 3.33

H:

6.7/1.008 ≈ 6.65

O:

53.3/16.00 ≈ 3.33

Divide all by 3.33:

C ≈ 1 H ≈ 2 O ≈ 1

Therefore:

CH₂O

Answer: C₆H₁₂O₆

Work:

Empirical formula mass:

12.01 + 2(1.008) + 16.00 ≈ 30.03 g/mol

Find multiplier:

180/30.03 ≈ 6

Multiply every subscript by 6:

C₆H₁₂O₆

Answer: Potassium has one more occupied principal energy level.

Explanation: Na's valence electrons are mainly in n=3.

K's valence electron is in n=4.

That extra shell places the outer electron farther from the nucleus and increases shielding, so K is larger.

Answer: Na > Mg > Al > Cl

Explanation: All are in Period 3.

Across a period, effective nuclear charge increases and pulls electrons inward, so radius generally decreases from left to right.

Answer: Na < Mg < Cl

Explanation: Ionization energy generally increases across Period 3 because valence electrons experience stronger attraction to the nucleus.

Answer: The second electron would have to be removed from a stable core shell.

Explanation: Na is:

[Ne]3s¹

The first ionization removes the 3s electron.

Na⁺ becomes:

[Ne]

Removing another electron means breaking into the filled neon core, which requires much more energy.

Answer: [Ar]3d⁵

Explanation: Neutral Fe:

[Ar]4s²3d⁶

Remove electrons from 4s first:

Fe²⁺:

[Ar]3d⁶

Remove one more from 3d:

Fe³⁺:

[Ar]3d⁵

Answer: 2

Explanation: Oxygen ends in:

2p⁴

There are three p orbitals.

Hund's rule says electrons fill them singly first.

The arrangement gives one paired orbital and two orbitals with one unpaired electron each.

Answer: Electrons occupy orbitals of equal energy singly before pairing.

Explanation: For example, the three p orbitals receive one electron each before any orbital receives a second electron.

This reduces electron-electron repulsion and produces a lower-energy arrangement.

Answer: Al loses a higher-energy 3p electron, while Mg loses a 3s electron.

Explanation: Mg:

3s²

Al:

3s²3p¹

The 3p electron is higher in energy and slightly more shielded, so it is easier to remove.

This creates an exception to the general trend.

Answer: 25.0%

Work:

Total mass:

25 + 75 = 100 g

Then:

(25/100)(100%) = 25%

Answer: 24.0%

Work:

% X = (12.0/50.0)(100)

= 24.0%

Answer: H₂O

Work:

H:

4.00/1.008 ≈ 3.97 mol

O:

32.0/16.00 = 2.00 mol

Divide by 2.00:

H ≈ 1.985 ≈ 2 O = 1

So:

H₂O

Answer: Core electrons closest to the nucleus.

Explanation: These electrons experience strong electrostatic attraction to the positively charged nucleus and are difficult to remove.

Answer: They are closer to the nucleus and experience stronger nuclear attraction.

Explanation: Valence electrons are farther away and experience more shielding from inner electrons.

Core electrons therefore require more energy to eject.

Answer: Na

Explanation: Na loses its entire outer 3s shell when forming Na⁺.

Na⁺ has fewer occupied energy levels and less electron-electron repulsion, so it is significantly smaller.

Answer: Cl⁻

Explanation: Cl⁻ has one more electron than neutral Cl but the same number of protons.

The additional electron increases electron-electron repulsion and spreads the electron cloud out.