Chem 120 Graded Worksheet Name

Chem 120 Graded Worksheet Name

According to estimates, 1.0L of seawater contains on the average 4.0×10^-9 g of gold. If gold is worth $390 per troy ounce and a troy ounce weighs 31.103 g, how many cubic meters of seawater would have a million dollars worth of gold in it?

Gold is measured in troy ounces, one of the units of apothecaries’ weights. In this system of weights, 1 scruple is 20 grains (exactly), 1 ounce is 480 grains (exactly), 1 ounce is 31.103 g, and 12 ounces equals 1 (troy) pound. How many scruples of gold are needed to make a ring weighing 3.77 g?

The LD50 for a substance in the dose that would be lethal for 50% of the population that receive that dose. The LD50 for aspirin in rats is 1.75 grams per kilogram of body weight. How many aspirin tablets, each of which contains 325 mg of aspirin, would you have to feed to a rat weighing 290 g to give him this dose?

At sea level, sound travels at about 740 miles per hour. How far will it travel in 5.75 seconds? If a shot could be heard around the world, how long would it take the sound of a shot to go around the world at the equator? The circumference of the earth at the equator is about 4.01×10^7 m.

If a person weighing 165 pounds contains 5.37 g of magnesium. How much magnesium will a person weighing 90 kg contain? (1 kg = 2.2 pounds)

Convert 65 miles per hour to meters per second. (1 mile = 5280 ft; 1 ft = 12 in; 1 in = 2.54 cm)

A lead sphere has a diameter of 1.88 cm. How many lead (Pb) atoms are in the sphere? The density of lead is 11.4 g/mL. Volume of a sphere = (4/3)πr^3. A student was given a bag of lead shot (small spheres of lead all the same size) and asked to find the volume of a single average pellet. The student first weighed a beaker and found it weighed 65.78 g. He then counted 50 of the pellets into the beaker and found it now weighed 103.11 g. The density of lead is 11.4 g/mL. What is the volume of a single shot?

What is the mass of a single atom of gold in kilograms?

Concentrated sulfuric acid has a density of 1.84 g/mL. What is the volume of 50.0 g of the acid? What is the mass of 50.0 mL of the acid?

How many atoms of gold are in a ring with a mass of 2.03 g?

Paper For Above instruction

The following comprehensive report addresses a selection of quantitative and qualitative chemistry problems designed to deepen understanding of fundamental concepts such as molar conversions, atomic structure, density calculations, and physical properties of substances. These problems offer practical applications reflecting real-world scenarios ranging from environmental quantifications to biological assessments and physical measurements.

1. Gold in Seawater and Value Estimation

The concentration of gold in seawater is approximately 4.0×10^-9 grams per liter. To determine the volume of seawater needed to accumulate one million dollars worth of gold, we start by calculating the total mass of gold corresponding to this value. Given the price of gold at $390 per troy ounce and the weight of one troy ounce at 31.103 grams, the value per gram is computed as:

Value per gram = $390 / 31.103 g ≈ $12.53/g.

Thus, the total gold mass for one million dollars is:

Mass = $1,000,000 / $12.53 ≈ 79,808 grams.

Since 1 liter of seawater contains 4.0×10^-9 grams of gold, the volume of seawater required is:

Volume = total gold / gold per liter = 79,808 g / 4.0×10^-9 g/L ≈ 2.00×10^13 liters.

Converting liters to cubic meters (1 m^3 = 1000 L):

Volume ≈ 2.00×10^10 m^3.

Therefore, approximately 20 billion cubic meters of seawater are required to yield a million dollars worth of gold at current concentrations.

2. Calculating Scruples for Gold Jewelry

To determine how many scruples are needed to make a 3.77 g gold ring, we first convert grams to grains using the conversion factor: 1 ounce = 480 grains, and 1 ounce = 31.103 g. Therefore, 1 grain = 31.103 g / 480 ≈ 0.06476 g.

Number of grains in 3.77 g:

= 3.77 g / 0.06476 g ≈ 58.2 grains.

Since 1 scruple equals 20 grains, the total scruples needed are:

58.2 grains / 20 ≈ 2.91 scruples.

Hence, approximately 2.91 scruples of gold are required to make the ring.

3. Aspirin Dosage in Rats

The LD50 of aspirin for rats is 1.75 g/kg. For a rat weighing 290 g (0.290 kg), the lethal dose is:

LD50 dose = 1.75 g/kg × 0.290 kg ≈ 0.5075 g.

Each aspirin tablet contains 325 mg (0.325 g). The number of tablets needed is:

Number = 0.5075 g / 0.325 g ≈ 1.56 tablets.

Therefore, feeding approximately 1.56 tablets would provide the LD50 dose for this rat.

4. Sound Travel Distance and Time

At 740 miles/hour, the distance traveled in 5.75 seconds is calculated by converting miles/hour to miles/second:

Speed = 740 miles/hour ÷ 3600 seconds/hour ≈ 0.2056 miles/sec.

Distance = speed × time = 0.2056 miles/sec × 5.75 sec ≈ 1.183 miles.

To find how long a sound would take to circumnavigate the Earth at the equator (circumference = 4.01×10^7 m),:

Speed in m/sec = 740 miles/hour × 1609.34 m/mile / 3600 sec/hour ≈ 330.5 m/sec.

Time = distance / speed = 4.01×10^7 m / 330.5 m/sec ≈ 121,300 seconds.

Converting to hours: ≈ 33.7 hours.

5. Magnesium Content Scaling

A person weighing 165 pounds contains 5.37 g of magnesium. To find the magnesium content in a 90 kg individual, first convert pounds to kilograms:

165 pounds / 2.2 ≈ 75 kg.

Magnesium per kilogram = 5.37 g / 75 kg ≈ 0.0716 g/kg.

Thus, a 90 kg person contains:

= 0.0716 g/kg × 90 kg ≈ 6.44 g of magnesium.

6. Conversion of Speed Units

Speed in miles/hour = 65 mph.

First, convert miles to meters: 1 mile = 1,609.34 meters.

Speed in m/sec = (65 miles × 1,609.34 m/mile) / 3600 sec ≈ 29.1 m/sec.

7. Calculating Number of Lead Atoms in a Sphere

Radius of sphere r = 1.88 cm / 2 = 0.94 cm = 0.0094 m.

Volume of sphere = (4/3)πr^3 ≈ (4/3) × 3.1416 × (0.0094)^3 ≈ 3.5×10^-6 m^3.

Mass of sphere = density × volume = 11.4 g/mL × 3.5×10^-6 L (since 1 mL = 1 cm^3) = 11.4 g/mL × 3.5×10^-6 mL ≈ 4.0×10^-5 g.

Number of atoms = total mass / atomic mass of lead (207.2 g/mol) × Avogadro's number (6.022×10^23 mol^-1):

Mass per atom = 207.2 g / 6.022×10^23 ≈ 3.44×10^-22 g.

Number of atoms ≈ 4.0×10^-5 g / 3.44×10^-22 g ≈ 1.16×10^17 atoms.

8. Volume and Mass of Lead Shot Pellets

The mass difference in the beaker is 103.11 g - 65.78 g = 37.33 g for 50 pellets, so mass per pellet is:

≈ 0.7466 g.

Volume per pellet = mass / density = 0.7466 g / 11.4 g/mL ≈ 0.0655 mL.

9. Atomic Mass of Gold in Kilograms

The atomic mass of gold is 197 g/mol. The mass of a single atom is:

Mass per atom = 197 g / 6.022×10^23 ≈ 3.27×10^-22 g.

In kilograms: 3.27×10^-22 g × (1 kg / 1×10^3 g) = 3.27×10^-25 kg.

10. Volume and Mass of Concentrated Sulfuric Acid

Mass of 50.0 g corresponds to a volume of:

volume = mass / density = 50.0 g / 1.84 g/mL ≈ 27.17 mL.

Similarly, 50.0 mL of sulfuric acid has a mass of:

mass = volume × density = 50.0 mL × 1.84 g/mL = 92.0 g.

11. Atoms of Gold in a Ring

Mass of gold in ring = 2.03 g. Number of atoms = 2.03 g / (197 g/mol) × 6.022×10^23 ≈ 7.78×10^21 atoms.

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