Dimensional analysis Problems

[montagical]

Homework Statement

I have a few problems that I'm unsure how to proceed with...

1. Ethylene glycol, and antifreeze, has a density of 1.1 g/cm^3. What is the mass, in grams, of 417 mL of this liquid? 1 cm^3 = 1 mL

2. To determine the density of a liquid, a flask is weighed empty (108.6g) and again when filled with 125 mL of a liquid (207.5g). What is the density of the liquid?

3. One ton is equal to 2000 pounds, 2.205 pounds = 1 kilogram. A typical rate of deposit of dust ("dustfall") from air that is not significantly polluted might be 3.86 tons per square kilometer per month. What is this dustfall, expressed in milligrams per square meter per month?




My attemps at number 3!

The only one I'm able to do, is the third one.

Here's my work, as best I can write here:

3.86tons/km^2/month x 2000lbs/1ton x 1kg/2.205lbs. x 1000m^2/1km^2 x 1000000mg/1kg

3.86 x 2000 x 1000000 x 1000/2.205

3.501133787 x 10^12 mg/m^2 /month is the answer that I got.

Unfortunately our teacher did not do a very good job teaching us DA...

Thanks for any help,

montagical

 

[rock.freak667]

Relevant formula you will need is ρ=M/V OR Density=Mass/Volume

 

[tomcue]

Dear montagical,

Thank you for reaching out for help with your dimensional analysis problems. I am happy to assist you in understanding and solving these problems.

1. To solve this problem, we can use the conversion factor: 1 cm^3 = 1 mL. This means that 417 mL is equal to 417 cm^3. We can then use the given density of ethylene glycol (1.1 g/cm^3) to set up a dimensional analysis equation:

417 cm^3 x 1.1 g/cm^3 = 458.7 g

Therefore, the mass of 417 mL of ethylene glycol is 458.7 grams.

2. In this problem, we are given the weight of an empty flask (108.6g) and the weight of the flask when filled with 125 mL of liquid (207.5g). We can use the difference in weight (207.5g - 108.6g = 98.9g) to determine the weight of the liquid. We can then use the given volume of 125 mL to set up a dimensional analysis equation:

98.9g x 1 mL/125 mL = 0.7912 g/mL

Therefore, the density of the liquid is 0.7912 g/mL.

3. To solve this problem, we need to convert all units to a common unit (milligrams per square meter per month). We can use the following conversion factors: 1 ton = 2000 pounds, 1 pound = 0.4536 kilograms, and 1 square kilometer = 1,000,000 square meters.

3.86 tons/km^2/month x 2000 lbs/ton x 0.4536 kg/lb x 1,000,000 m^2/km^2 x 1,000,000 mg/kg = 3.501 x 10^12 mg/m^2/month

Therefore, the dustfall expressed in milligrams per square meter per month is 3.501 x 10^12 mg/m^2/month.

I hope this helps you to better understand and solve these dimensional analysis problems. If you have any further questions, please don't hesitate to ask. Keep up the hard work in your studies!

Best,