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Question 1 (Review): Make the following conversions (4 pts). A. Grams of oxygen in 0.158 mol...

Question 1 (Review): Make the following conversions (4 pts).

A. Grams of oxygen in 0.158 mol dinitrogen tetroxide (1 pt)

B. TOTAL ions in 32.7 g of Copper (I) carbonate (1 pt)

C. Grams of nickel (II) nitrate in a 0.0233 M solution with a volume of 55.0 mL (1 pt)

D. Atoms of gold in a gold nanoparticle (density of gold = 19.3 g/cm3 , volume of a gold nanoparticle, 33.5 nm3 ) (1 pt)

Question 2 (Review): Balance and classify the following reactions. States of matter should be included for full credit. (6 pts)

A. Combustion of ethane (C2H6) (2 pts)

B. Aqueous solutions of lead (II) nitrate and potassium chloride are mixed (2 pts)

C. Stoichiometry: 62.1 mL of 0.338 M potassium chloride and 45.2 mL of 0.219 M lead (II) nitrate are mixed. What mass of lead chloride is produced (2 pts)?

Question 3, Making Connections: Based on your knowledge from Chapter 6/7, predict the behavior of the following atoms when bonding with other elements. Explain your answers for full credit (6 pts).

A. Is fluorine more likely to form polar covalent, nonpolar covalent, or ionic bonds (2 pts)?

B. Is sodium more likely to form polar covalent, nonpolar covalent, or ionic bonds (2 pts)?

C. Is carbon more likely to form polar covalent, nonpolar covalent, or ionic bonds (2 pts)?

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Answer #1

Solution:

3) Questions

Part A) Number of moles of N2O3 = 0.158 mol

Mass = Number of mol x molar mass

= 0.158 mol x 76.0 g mol-1 = 12.0 g

Since, 76.0 g contains 48 g O

Hence, 12 g contains = 12 g x 48 /76.0 = 7.58 g

Part B) Molar mass of Cu2CO3 = 187.0 g mol-1

Thus, number of moles in 32.7g = mass / molar mass

= 32.7 g / 187.0 g mol-1 = 0.175 mol

Thus,

Number of ions = number of mol x avagadro number

= 0.175 mol x 6.023 x 10^23 = 1.05 x 10^23 ions

Part C)

Molarity = mass / molar mass x volume in L

Thus,

0.0233 M = Mass / 182.7 g mol-1 x 0.055 L

Mass = 0.0223 M x 182.7 g mol-1 x 0.055 L

= 0.224 g

Part D)

Density (D) = n M / NA x V

n = NA x V x D / M

n = 6.023 x 10^23 x 33.5 x 10^-21 cm3 x 19.3 g cm-3 / 197 gmol-1

n = 1976.73 atoms

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