VSEPR
Intermolecular Forces
Gas Laws
Equations
Reactions and pH
100

What does VSEPR theory predict about molecules?

It predicts the 3D shape of molecules based on electron pair repulsion.

100

What is the strongest intermolecular force present in H₂O?

Hydrogen bonding.

100

State Boyle’s Law.

At constant temperature, pressure and volume are inversely proportional (P₁V₁ = P₂V₂).

100

State the ideal gas law equation.

PV = nRT.

100

A solution has a hydrogen ion concentration of [H+]=3.2×10−4 M Calculate the pH of the solution.

pH = 3.49 

200

What is the molecular shape of CO₂ according to VSEPR?

Linear.

200

Rank F₂, Cl₂, Br₂, I₂ in terms of increasing boiling points.

F₂ < Cl₂ < Br₂ < I₂ (increasing dispersion forces with molar mass).

200

A balloon has a pressure of 1.00 atm and a volume of 2.00 L. If pressure increases to 2.00 atm, what is the new volume (T constant)?

Answer: 1.00 L.

200

A gas sample has n = 0.500 mol, T = 298 K, P = 1.00 atm. Find the volume.

V = nRT/P = (0.500 × 0.0821 × 298) / 1.00 = 12.2 L.

200

Explain how increasing temperature affects the rate of reaction, referring to collision theory.

Higher temperature increases kinetic energy of particles, leading to more frequent and energetic collisions, increasing the proportion of particles above activation energy, thus faster reaction.

300

Predict the shape of NH₃ and state the bond angle.

Trigonal pyramidal

300

Explain why CH₄ is nonpolar while CHCl₃ is polar.

CH₄ has symmetrical bonds that cancel out dipoles; CHCl₃ has polar C–Cl bonds that do not cancel due to asymmetry.

300

A 2.50 L balloon at 1.25 atm is taken to a pressure of 95 atm. What is the new volume (T constant)?

V₂ = (P₁V₁)/P₂ = (1.25 × 2.50)/95 = 0.0329 L.

300

A 0.85 mol sample of an element has mass 67.1 g. Identify the element.

M = 67.1 ÷ 0.85 = 79 g/mol → Bromine (Br).

300

Blood has [OH⁻] = 7.25 × 10⁻⁴ M. Calculate the pH.

pOH = 3.14 → pH = 10.86

400

Explain why H₂O has a bent shape while CO₂ is linear, despite both having central atoms bonded to two others.

H₂O has two lone pairs on oxygen, causing repulsion and a bent geometry; CO₂ has no lone pairs on the central atom, making it linear.

400

Justify why NH₃ has a higher boiling point than PH₃.

NH₃ has hydrogen bonding; PH₃ has only weaker dipole-dipole forces.

400

A gas sample has volume 5.00 L at 300 K. What will its volume be at 450 K, assuming constant pressure?

V₂ = V₁ × (T₂/T₁) = 5.00 × (450/300) = 7.50 L.

400

How many atoms are in 23.4 moles of bromine atoms? (Avogadro’s number = 6.02 × 10²³)

23.4 × 6.02 × 10²³ = 1.41 × 10²⁵ atoms.

400

A solution has a pH of 5.2. Calculate the hydrogen ion concentration [H+] in the solution.

[H+] = 6.31×10−6M

500

Compare and contrast the geometry of SiO₂ and H₂S using VSEPR theory.

SiO₂ has a tetrahedral arrangement, whereas H₂S is a discrete bent molecule due to two lone pairs on Sulfur. 

500

Explain whether CH4 or H2O will form a stronger intermolecular force with water and why?  

Water (H2O) will form stronger intermolecular interactions with other water molecules than with methane (CH4). This is because the O-H bond in H2O is highly polar, creating partial positive and negative charges that allow hydrogen bonding, which is a strong type of intermolecular force. In contrast, the C-H bond in CH4 is only weakly polar, so methane can only interact with water through very weak London dispersion forces. Therefore, water interacts much more strongly with itself than with CH4.

500

A balloon has V = 2.50 L, P = 1.25 atm, T = 285 K. Calculate the moles of gas in the balloon.

n = PV/RT  

n = 0.134 mol 

500

A 2.5 L container holds oxygen gas at a pressure of 150 kPa and a temperature of 27 °C. Calculate the number of moles of oxygen gas in the container.

0.15 moles of oxygen gas.

500

Determine the products to this chemical equation to determine if a precipitate will form: NaCl(aq) + AgNO₃(aq) → 

NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s) 

  • NaNO₃ is soluble in water → stays in solution.

  • AgCl is insoluble in water → forms a precipitate.

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