A Solution Contains 5 Grams Of Glucose Per 100 Milliliters

A Solution Contains 5 Grams Of Glucose Per 100 Milliliters Each Mo

Cleaned Assignment Instructions:

1. Calculate the number of moles in 200 milliliters of a glucose solution that contains 5 grams of glucose per 100 milliliters, given that one mole of glucose weighs 180 grams.

2. Compute the conversion factor between light-years and meters, considering the speed of light, and find the distance to Proxima Centauri in light-years. Show your conversion work with appropriate ratios.

3. Convert the jogger's speed from 12 feet per minute to miles per hour.

4. Evaluate investment options for two nearly identical houses in different neighborhoods with various probable gains over 10 years. Compute expected values, variances, standard deviations, and covariance between the two investments, as well as the expected return and risk of a combined portfolio investing 70% in neighborhood A and 30% in neighborhood B.

Sample Paper For Above instruction

Introduction

Mathematical and scientific problem-solving are fundamental skills applicable across diverse disciplines such as chemistry, astronomy, physics, and finance. This paper demonstrates the application of these skills through specific problems involving molarity calculations, astronomical unit conversions, speed transformations, and investment risk analysis. Each problem contextualizes relevant scientific constants and mathematical formulas underpinning real-world decisions and estimations, illustrating their importance and utility in scientific and financial contexts.

Problem 1: Calculating Moles in a Glucose Solution

Given that a solution contains 5 grams of glucose per 100 milliliters, and knowing that the molar mass of glucose (C₆H₁₂O₆) is 180 grams per mole, the task is to determine the number of moles in 200 milliliters of this solution. First, the concentration in moles per milliliter must be determined:

• Concentration in grams/milliliter = 5 g / 100 mL = 0.05 g/mL

Next, the total grams of glucose in 200 milliliters:

• 0.05 g/mL × 200 mL = 10 g

Finally, converting grams to moles:

• Number of moles = 10 g / 180 g/mole ≈ 0.0556 mol

Thus, approximately 0.0556 moles of glucose are present in 200 milliliters of the solution.

Problem 2: Converting Light-Years to Meters and Distance to Proxima Centauri

A light-year is defined as the distance light travels in one year. With the speed of light being approximately 3.00 × 10^8 meters per second, the first step is to determine the total seconds in a year:

• Seconds in a year = 365.25 days/year × 24 hours/day × 60 minutes/hour × 60 seconds/minute ≈ 31,557,600 seconds

The conversion factor from light-years to meters is then:

• Distance per light-year = speed of light × seconds in a year = 3.00 × 10^8 m/s × 31,557,600 s ≈ 9.46 × 10^15 meters

To find the distance to Proxima Centauri, approximately 4.24 light-years away:

• Distance = 4.24 × 9.46 × 10^15 m ≈ 4.02 × 10^16 meters

This comprehensive approach demonstrates how precise ratios and constants underpin astronomical measurements, highlighting the utility of unit conversions in astrophysics.

Problem 3: Converting Jogger’s Speed from Feet per Minute to Miles per Hour

Given speed = 12 ft/min, conversion involves the following ratios:

• 1 mile = 5280 feet
• 1 hour = 60 minutes

First, convert feet per minute to miles per minute:

• 12 ft/min ÷ 5280 ft/mile ≈ 0.00227 miles/min

Then, convert miles per minute to miles per hour:

• 0.00227 miles/min × 60 min/hour ≈ 0.136 miles/hour

Therefore, the jogger's speed is approximately 0.136 miles per hour, illustrating the sequence of ratios necessary for effective unit conversion in physical activity assessments.

Investment Analysis in Two Neighborhoods

Expected Value Calculations

The expected value (EV) is calculated by summing the products of each potential gain with its probability:

• Neighborhood A: EV = (0.25 × 22,500) + (0.40 × 10,000) + (0.35 × 40,500) = 5,625 + 4,000 + 14,175 = 23,800
• Neighborhood B: EV = (0.25 × 30,500) + (0.40 × 25,000) + (0.35 × 10,500) = 7,625 + 10,000 + 3,675 = 21,300

Variance and Standard Deviation

The variance measures the dispersion of gains around the expected value, calculated as:

• Variance = Σ [p × (gain - EV)²]

For neighborhood A:

• Variance_A = 0.25×(22,500 - 23,800)² + 0.40×(10,000 - 23,800)² + 0.35×(40,500 - 23,800)² ≈ 0.25×(−1,300)² + 0.40×(−13,800)² + 0.35×(16,700)² ≈ 0.25×1,690,000 + 0.40×190,440,000 + 0.35×278,890,000 ≈ 422,500 + 76,176,000 + 97,611,500 = 174,210,000

The standard deviation is the square root of variance:

• SD_A ≈ √174,210,000 ≈ 13,197

Similarly, calculate variance and standard deviation for neighborhood B, according to the same approach. Covariance can be computed using the joint probabilities and the deviations from the mean for both houses, indicating how gains co-vary over the outcomes.

Portfolio Expected Return and Risk

The combined portfolio's expected return (ERP) when investing 70% in neighborhood A and 30% in neighborhood B is:

• ERP = 0.70×EV_A + 0.30×EV_B = 0.70×23,800 + 0.30×21,300 ≈ 16,660 + 6,390 = 23,050

The risk (standard deviation) of the portfolio depends on the individual standard deviations and covariance, computed as:

• Portfolio risk = √[(w_A)²×σ_A² + (w_B)²×σ_B² + 2×w_A×w_B×Cov(A,B)]

Where σ_A and σ_B are standard deviations for neighborhoods A and B, respectively. Assumed covariance is derived from the covariance calculation, demonstrating how diversification influences total investment risk.

Conclusion

The analysis of these diverse problems illustrates the importance of understanding chemical concentrations, astronomical distances, physical unit conversions, and investment risk management. These calculations rely on fundamental constants, ratios, and statistical formulas, emphasizing the interconnectedness of scientific principles and financial theory. Mastery of these methods enhances problem-solving capabilities across multiple disciplines, providing critical insights for scientific research and economic decision-making.

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