Q: A mixture contains 3.0 moles of He and 2.0 moles of Ne at a total pressure of 2.50 atm. What is the partial pressure of Ne?

$$P_{\mathrm{Ne}} = X_{\mathrm{Ne}} \times P_{\mathrm{total}} = \frac{2.0}{5.0} \times 2.50 = 1.00\ \mathrm{atm}$$

Q: A participant in a medical study exhales, reducing lung volume from 6.15 L (0.254 mol air) to 2.55 L. How many moles of air remain in the lungs after exhalation? (Assume constant temperature and pressure.)

$$n_2$$ = $$n_1$$ \times \frac{$$V_2$$}{$$V_1$$} = 0.254 \times \frac{2.55}{6.15} = 0.105\ \mathrm{mol}$$

Question 1

Which of the following molecules is nonpolar, despite having polar bonds, due to its molecular geometry?

Accepted Answer

CO\_2

Question 2

A sample of gas occupies 2.50 L at 1.00 atm and 300 K. If the pressure is increased to 2.00 atm and the temperature is decreased to 150 K, what will be the new volume? (Assume the amount of gas is constant.)

Accepted Answer

$$V_2$$ = \frac{$$P_1$$ $$V_1$$ $$T_2$$}{$$P_2$$ $$T_1$$} = \frac{(1.00)(2.50)(150)}{(2.00)(300)} = 0.625\ \mathrm{L}$$

Question 3

Which statement best explains why CH\_4 (methane) has four equivalent C-H bonds?

Accepted Answer

The carbon atom undergoes sp\^3 hybridization, forming four equivalent orbitals.

Question 4

A mixture contains 3.0 moles of He and 2.0 moles of Ne at a total pressure of 2.50 atm. What is the partial pressure of Ne?

Accepted Answer

$$P_{\mathrm{Ne}} = X_{\mathrm{Ne}} 	imes P_{\mathrm{total}} = \frac{2.0}{5.0} 	imes 2.50 = 1.00\ \mathrm{atm}$$

Question 5

Which of the following molecules has a square planar geometry according to VSEPR theory?

Accepted Answer

XeF\_4

Question 6

A 0.50 g sample of dry ice (CO\_2) is placed in a 4.6 L vessel at 30.0°C. What is the pressure in atm after all the dry ice sublimes? (R = 0.08206 L·atm/K·mol, molar mass CO\_2 = 44.01 g/mol)

Accepted Answer

$$n = \frac{0.50}{44.01} = 0.0114\ \mathrm{mol}$$; $$T = 30.0 + 273.15 = 303.15\ \mathrm{K}$$; $$P = \frac{nRT}{V} = \frac{(0.0114)(0.08206)(303.15)}{4.6} = 0.061\ \mathrm{atm}$$

Question 7

Which of the following molecules is expected to be polar due to the presence of a lone pair on the central atom?

Accepted Answer

NH\_3

Question 8

What is the hybridization of the central atom in BeF\_2?

Accepted Answer

sp

Question 9

Which of the following best describes the difference between sigma (σ) and pi (π) bonds?

Accepted Answer

Sigma bonds result from end-to-end overlap of orbitals, while pi bonds result from side-to-side overlap.

Question 10

A gas sample has a density of 1.43 g/L at 296 K and 0.789 atm. Using $$PM = dRT$$, what is the molar mass of the gas?

Accepted Answer

$$M = \frac{dRT}{P} = \frac{(1.43)(0.08206)(296)}{0.789} = 44.3\ \mathrm{g/mol}$$

Question 11

Which of the following molecules has a trigonal planar electron domain geometry and is nonpolar?

Accepted Answer

BF\_3

Question 12

Which of the following statements about the kinetic molecular theory of gases is correct?

Accepted Answer

At the same temperature, different gases have the same average kinetic energy.

Question 13

What is the bond angle in a molecule with tetrahedral electron domain geometry?

Accepted Answer

$$109.5^\circ$$

Question 14

Which molecule below contains both sigma and pi bonds?

Accepted Answer

C\_2H\_4 (ethene)

Question 15

A participant in a medical study exhales, reducing lung volume from 6.15 L (0.254 mol air) to 2.55 L. How many moles of air remain in the lungs after exhalation? (Assume constant temperature and pressure.)

Accepted Answer

$$n_2$$ = $$n_1$$ 	imes \frac{$$V_2$$}{$$V_1$$} = 0.254 	imes \frac{2.55}{6.15} = 0.105\ \mathrm{mol}$$

Question 16

Which of the following conditions causes real gases to deviate most from ideal behavior?

Accepted Answer

Low temperature and high pressure

Question 17

What is the electron domain geometry and hybridization of the central atom in HCN?

Accepted Answer

Linear geometry, sp hybridization

Question 18

Which molecule below is expected to have a net dipole moment?

Accepted Answer

CH\_3Cl

Question 19

A fixed amount of gas at room temperature occupies a volume of 1.800 L at a pressure of 775.0 mmHg. What will be the pressure if the gas is compressed to 0.8500 L at the same temperature?

Accepted Answer

$$P_2$$ = \frac{$$P_1$$ $$V_1$$}{$$V_2$$} = \frac{775.0 	imes 1.800}{0.8500} = 1641\ \mathrm{mmHg}$$

VSEPR Theory, Molecular Polarity, Valence Bond Theory, and Gas Laws: Study Notes for General Chemistry

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