According To Estimates, 10L Of Seawater Contains On Average

According To Estimates 10l Of Seawater Contains On the Average 40

According to estimates, 1.0L of seawater contains on average 4.0×10^-9 g of gold. If gold is valued at $390 per troy ounce and one troy ounce weighs 31.103 g, this paper calculates how many cubic meters of seawater would contain gold worth one million dollars. Additionally, the paper explores various related calculations involving weights, conversions, and physical properties, including the measurement of gold in scruples, the amount of aspirin needed for a lethal dose, the distance sound travels in a given time, magnesium content in humans, unit conversions, the number of atoms in a lead sphere, and the volume and number of atoms in gold samples.

Paper For Above instruction

Understanding the concentration of precious metals in seawater and its implications for economic and scientific purposes involves complex calculations across chemistry, physics, and environmental science. This paper explores these calculations step by step, starting with the estimation of the volume of seawater needed to contain gold worth a specific amount of money, and moving through related problems involving mass conversions, physical properties, and atomic scales.

Gold in Seawater and Economic Valuation

The initial calculation centers on determining how much seawater one must process to obtain a certain monetary value of gold. Given that 1.0L of seawater contains 4.0×10^-9 g of gold, and gold’s market value is $390 per troy ounce, with a troy ounce weighing 31.103 g, we first convert the gold content into a dollar value per liter. This involves converting grams to troy ounces and then multiplying by the price per ounce to get the dollar value per liter.

Calculating the dollar value per liter:

Gold per liter = 4.0×10^-9 g

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

Value per liter = 4.0×10^-9 g × $12.53/g ≈ 5.012×10^-8 dollars.

To find the volume of seawater containing a million dollars’ worth of gold, divide one million dollars by the value per liter:

Volume (L) = 1,000,000 / 5.012×10^-8 ≈ 1.995×10¹³ liters.

Converting liters to cubic meters (since 1 m³ = 1000 L):

Volume (m³) ≈ 1.995×10¹⁰ cubic meters.

This immense volume underscores the scarcity of gold in seawater and the impracticality of extraction for economic gain at current market prices.

Measuring Gold in Scruples

The apothecaries’ system defines 1 scruple as 20 grains, and 1 ounce as 480 grains, with 1 ounce equivalent to 31.103 g. To determine how many scruples are needed to make a ring weighing 3.77 g, first convert the ring’s weight to grains:

Weight in grains = 3.77 g × 480 grains/ounce / 31.103 g ≈ 58.20 grains.

Number of scruples = 58.20 grains / 20 grains per scruple ≈ 2.91 scruples.

This illustrates the close relationship between traditional weight units and modern mass measurement in grams.

Aspirin Dosage Calculation

The LD50 for aspirin in rats is 1.75 g/kg. For a rat weighing 290 g (0.29 kg), the lethal dose equates to:

Lethal dose = 1.75 g/kg × 0.29 kg ≈ 0.5075 g.

Each aspirin tablet contains 325 mg (0.325 g), so the number of tablets needed to reach this dose is:

Number of tablets = 0.5075 g / 0.325 g ≈ 1.56 tablets.

Since it’s unsafe to administer a fraction of a tablet, approximately 2 tablets would represent the lethal dose for this rat, emphasizing the importance of dosage regulation in pharmacology.

Sound Travel Distance and Speed

At sea level, sound travels at about 740 miles per hour. To find the distance traveled in 5.75 seconds, convert speed to miles per second:

Speed (miles/sec) = 740 mi/hr ÷ 3600 ≈ 0.2056 mi/sec.

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

The question also explores how long a shot sound would take to go around the world at the equator with a circumference of 4.01×10⁷ m. First, convert the speed of sound to meters per second:

740 mph ≈ 332.77 m/sec (since 1 mile ≈ 1609.34 m).

Time to travel around the Earth:

Time = circumference / speed = 4.01×10⁷ m / 332.77 m/sec ≈ 120,600 seconds ≈ 33.5 hours.

This illustrates the vast scale of the Earth and the limitations of sound propagation over such distances.

Magnesium Content in Humans

A person weighing 165 pounds contains 5.37 g of magnesium. To find how much magnesium a 90 kg person contains, convert 90 kg to pounds:

90 kg × 2.2 pounds/kg = 198 pounds.

Assuming magnesium content scales proportionally:

Magnesium in 90 kg person = (5.37 g / 165 pounds) × 198 ≈ 6.44 g.

Thus, a person weighing 90 kg would contain approximately 6.44 grams of magnesium, reflecting the proportional distribution of minerals in the human body.

Unit Conversions and Volume Calculations

Converting 65 miles per hour to meters per second involves multiple steps:

Miles to feet: 1 mile = 5280 ft

Feet to inches: 1 ft = 12 in

Inches to centimeters: 1 in = 2.54 cm

Then, centimeters to meters (divide by 100):

Speed in m/sec = (65 miles/hr) × (5280 ft/mile) × (12 in/ft) × (2.54 cm/in) / (100 cm/m) / (3600 sec/hr) ≈ 29.06 m/sec.

Regarding the lead sphere, with a diameter of 1.88 cm, calculate its volume:

Radius r = 0.94 cm.

Volume = (4/3) × π × r³ ≈ (4/3) × 3.1416 × (0.94)^3 ≈ 3.48 cm³.

The mass of the sphere, given the density of lead (11.4 g/mL or 11.4 g/cm³), is:

Mass = volume × density ≈ 3.48 cm³ × 11.4 g/cm³ ≈ 39.67 g.

Atomic Scale Calculations

The number of gold atoms in a 2.03 g ring involves calculating moles and atoms:

Moles of gold = mass / molar mass = 2.03 g / 197 g/mol ≈ 0.0103 mol.

Atoms = moles × Avogadro’s number (6.022×10²³) ≈ 0.0103 mol × 6.022×10²³ ≈ 6.2×10²¹ atoms.

The mass of a single gold atom in kilograms is:

Mass per atom = molar mass / Avogadro’s number = 197 g / 6.022×10²³ ≈ 3.27×10^-22 g ≈ 3.27×10^-25 kg.

Sulfuric Acid Volume and Mass Calculations

Given the density of concentrated sulfuric acid (1.84 g/mL), the volume of 50 g of acid is:

Volume = mass / density = 50 g / 1.84 g/mL ≈ 27.17 mL.

Conversely, the mass for 50.0 mL of acid: 50.0 mL × 1.84 g/mL = 92 g.

Atoms of Gold in a Ring

This calculation is similar to the previous atom count: in a 2.03 g gold ring, the number of atoms is approximately 6.2×10²¹, illustrating the immense scale of atomic quantities in tangible objects.

Conclusion

These diverse calculations reveal the interconnectedness of chemistry, physics, and biology, and demonstrate how fundamental units and constants underpin our understanding of matter, energy, and environmental resources. Whether estimating the concentration of valuable metals in seawater, converting units across systems, or calculating atomic-scale quantities, these computations showcase the importance of precise measurements and conversions in scientific inquiry. Such analyses are vital for informing resource extraction, pharmacology, environmental science, and understanding our physical world.

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