The question was:

The two pictures were taken 24 hours apart. We see the moon and a star-like object which is actually Venus. What's responsible for their different positions? Whose apparent motion is faster? In reality does that object also move faster in space?Why or why not?

Solution:It's the moon's apparent position in the sky that has moved faster than that of Venus. Although in reality Venus moves around the sun 34 times faster than the moon moves around the earth, the moon is 238 times closer to earth, hence the illusion that the moon is moving faster. How do we know that Venus actually moves 34 times faster? Knowing the average Venus-Sun distance = 1.08 X 10¹¹ metres, we can compute the length of the orbit (approximately a circle, even though it's technically an ellipse). We then divide this circumference by a Venusian year=224.7 days ( time for 1 revolution). Converting to metres per second(m/s), we obtain an average value of 35 000 m/s. Doing the same for the moon (at an avg. distance of 3.76X 10⁸m from the earth), we obtain an avg. velocity of 1022 m/s, which is 34 times slower than Venus.

But why? The velocity at which Venus moves is determined mainly by the force between it and the massive sun. The moon's velocity is governed by the force of attraction between it and the earth. Using Newton's formula twice,F=Gm₁m₂/r², where G = 6.67 X10^-11Nm²/kg²,m₁and m₂are the two masses, and r is the distance between them, we can calculate that the Venus-Sun gravitational force is 263 times stronger than the Earth-moon one. Setting this ratio equal to the ratio of the centripetal forces ( F_c = mv²/r)of the two bodies, we confirm that the ratio of Venus's velocity to that of the moon is indeed 34.

The question was:

Around 1913, two scientists at New York University found that the density of water, which they measured to several decimal places, varied from sample to sample. They erroneously concluded that at constant temperature water's density is not unique. What were they unaware of?

Reference: John S. Ridgen, Hydrogen: The Essential Element,Harvard University Press, 2002

Solution:

Water can contain deuterium, an isotope of hydrogen with one extra neutron. ²H will raise the density of water, but its percentage varies from sample to sample.

Another Hydrogen Question

The question was:

Tritium, ³H, is an isotope of hydrogen. Imagine a 220 g glass of water with each H₂O molecule consisting of 1 oxygen atom and two tritium atoms instead of hydrogens. You accidentally empty this glass of radioactive water into the ocean.

Six years later half of the tritium is no longer radioactive, and meanwhile your water has mixed perfectly with the rest of the water in all of the earth's oceans. With a glass, you scoop up 220 g of water at a beach. How many radioactive water molecules will be in your glass? You'll need the following data:

Solution:

First we need to obtain an estimate for the total amount of water in all of the earth's oceans. For each ocean,convert the area to m² by multiplying km² by 1000². Multiplying the result by average depth, we will obtain volume in m³. To get litres we multiply by 1000. Add them all up and we get 1.11 X10²¹L in all of the earth's oceans.

A 220g glass of water from the beach is about 220ml = 0.220 L. This represents a small fraction of the entire volume found in the earth's oceans. It equals 0.220/1.11 X 10²¹. If we then multiply this fraction by the number of tritium-labeled water molecules, we will know how many are in that glass. Originally we dumped 220g/(22 g/mole) [note with tritium, the molar mass of water gets heavier]= 10 moles = 6.02 X 10²⁴ molecules. But tritium is radioactive so only half of those molecules remain = 3.01 X 10²⁴ molecules. Answer = 0.220/1.11 X 10²¹*3.01 X 10²⁴ molecules = about 600.

This adds one more reason to why we study moles. Moles help us realize how our bodies constantly borrow molecules from the planet and then return them. Each time perspiration escapes from your skin, because of the water cycle and of the extreme smallness of molecules, a minute amount of water that flowed through your veins and partially filled your lungs end up in the tissues of all other people, plants and animals on the planet.

The question was:

My daughter complains if I pick her up too early from her after-school program, so I've had the chance to measure my tire pressure frequently.

• On a cold day in February (-20^oC), the tire pressure was 35 psi (pounds per square inch; can be treated like kPa in Gay-Lussac's Law)

• Thirty days later, the air temperature is now 10^oC.

• Assume that the rate of change of temperature with respect to time is linear(constant)

• With time my tires lose air at an average rate of 0.083 psi/day.

a.Find the overall rate at which tire pressure changes with time in psi/day.

b.If the optimum tire pressure for my specific tires is 35 psi, should I let some air out or increase the pressure (now,30 days later)

Solution:

a. overall rate = (rate at which pressure changes with temperature)*(rate at which temperature changes with time) - (rate at which pressure changes with time due to leakage)

Use Gay Lussac to obtain the pressure at 10 C.

P₂ = (T₂/T₁)P₁ = (283/253)(35) = 39.2 psi.

rate at which pressure changes with temperature=( P₂ - P₁ )/T₂ - T₁)= (39.2-35)/(283K-253 K) = 0.14 psi/day - 0.083 psi/day = 0.057 psi/day

b. Tire pressure should be released since it should have increased to 35 + 0.057 psi/day(30 days) =36.7 psi.

The question was:

How do you find the mass of a little cup of water without a balance or any kind of volume-measuring device?

Solution:

If you place the cup in a calorimeter, you could attach it to a power source. Knowing the duration in seconds for which the calorimeter was connected, the voltage, and current, we can calculate the amount of energy absorbed by the water in the calorimeter using E = VIt. If we also keep track of water's initial and final temperature, with c = 4.19 J/(gC) and with Q= mc Dt, we can calculate m, the mass of water. Note we assume that Q = E, which is approximately true.

The question was:

Wine is 12% alcohol by volume, meaning that a 100 mL glass contains 12 mL of alcohol and approximately 88 mL of water. In such a glass of wine, how many molecules of water are there for every molecule of alcohol, C₂H₆O? (The density of alcohol = 0.80 g/ml)

Solution:

The ratio of molecules is not proportional to the ratio of volumes. Using densities, convert the volumes to mass. 12 ml of C₂H₆O = 12*0.80 = 9.6 g; 88 *1.00 = 88 g of water; Finally convert to moles by dividing by molar mass, and the ratio of moles is the ratio of molecules: [88/18]/[9.6/46] = 23: 1.

So although wine is 12% alcohol by volume, only 1/(23+1)= 4% of wine's molecules are ethanol, which might give some people an excuse to drink more!

The question was:

One morning I woke up with a rash on my ears, cheeks, arms and belly. In fact, my ears were so swollen that I looked too funny for school. So I went to the hospital, where I was scolded for showing up at the emergency ward for such a non life-threatening condition. The doctor treated me nevertheless, claiming that my affliction was either viral or an allergy. Just in case it was the latter, he prescribed something, which turned out was not the drug of choice for such a rash. I found that out from a dermatologist, who according to my family doctor was the best in town. You see, the medication had not helped me at all, and two days later I was still uncomfortable and looking as though I had been beaten up.

The dermatologist said it looked like a contact dermatitis, which is an allergic reaction to something you touch. He went through a list of possibilities but said that it was up to me to figure out the exact nature of the allergen. He sent me home with the puzzle and a prescription for a cortisone gel that would alleviate the symptoms until I figured out the cause of the irritation.

The next morning when I put on my underwear, I felt an itch in a new place, and I realized out what I was allergic to.

What was my hypothesis, and what did I do to verify it?

Solution:

The new underwear was from a new batch of laundry. Since I had not experienced any irritation from previous batches, I asked my wife if she had recently switched detergents. Sure enough, she had. We then went through the very tedious task of rewashing everything with the old detergent, and my rash never returned. The soap residues on my bed sheets and pillow cases especially were the reason my ears, cheeks, arms and belly had been the most affected. (I had slept without a pyjama top)

The question was:

A student was hoping to find the percentage of oxalates, C₂O₄^-2, in Earl Grey tea.

He first weighed 2.14 g of the tea and then added 200 ml of boiling water. After four minutes, he added 100.0 ml of 1.5 g/L Ca(OH)₂ to the sample, expecting the following to happen:

Ca(OH)₂ + C₂O₄^-2 --> 2 OH^-1_(aq) + CaC₂O_4(s)

He waited a while, saw lot of brown precipitate forming and filtered the mixture. After a few days, the residue on the paper was totally dry, and after subtracting the mass of the filter paper, he found the mass of the solid to be 0.52 g.

The student then erroneously concluded that the tea consisted of 0.52g*(88g of C₂O₄^-2 / 128 g of CaC₂O_4(s))/ 2.14g = 17 % oxalates.

Judging from the data given in the question, is there a possibility that the 0.52 g of solid included more than just calcium oxalate?

Solution:

Since only 100.0 ml of 1.5 g/L Ca(OH)₂ had been used, (a total of 0.100 L [1.5g/L][mole/74g] = 0.0020 moles), from the equation, we see that the maximum amount of CaC₂O_4(s) that can be produced from this is 0.0020 moles*128 g/mole = 0.26 g. Because the student produced twice that much residue, half the impurities are definitely not CaC₂O_4(s), and there is a strong possibility that the rest of the residue that precipitated at the high pH may not even contain oxalates.

The question was:

If you have not yet tackled a challenging project, you may not realize how important persistence and serendipity can be in scientific discoveries. Here is a problem which, although is more mathematical than scientific, will give you a taste of this important truth.

22/7 is a 2 decimal approximation to p. Suppose that you want to discover some fractions that are at least 11 decimal approximations of this useful irrational number. To avoid trivial solutions, you can't merely look up p, move the decimal, divide by a power of 10 and reduce. Your tool: the TI-83 calculator, which most of you have, and suppose that other people on the trail seem to think that the "FRAC" function under the MATH menu is a good starting point. This function converts decimals like 0.222222222…into fractions, for example, to 2/9 if you simply press 0.222222222… followed by "MATH" and "FRAC". Good luck, and don't give up so easily, because in all likelihood a successful find will depend on serendipity and persistence.

Solution:

Try p FRAC and the calculator will fail to give you a fraction. As an alternative, suppose you attempted to pick an arbitrary value for the radius of a circle and asked the calculator for the area in FRAC form. Potentially, you could then work backwards and express p as a fraction. Try that, and you will not succeed.

At this point, you may feel that the approach seems flawed, and there every reason to abandon it, but being persistent you try obtaining the volume of a sphere in FRAC form: (4/3) p r³ FRAC. For values of r up to 7, the calculator keeps yielding decimals, but for r = 8, we obtain, 8580787/4001. By multiplying this fraction by ¾ and dividing by 8³, ( treat the numerator and denominator separately), you will get 25742361/8194048, which is an accurate representation of p to 11 decimal places. The table below reveals what fractions are revealed by r values of 8 to 12.

r	p(3.14159265358979...) Approximation
8	25742361 / 8194048 = 3.14159265359441
9	8858575 / 2819772 = 3.14159265359043
10	87750966 / 27932000 = 43875483/13966000 = 3.14159265358728
11	149345019 / 47537996 = 3.14159265359019

A few years ago I contacted Texas Instruments to ask them why their calculator acted that way, but nobody could give me an answer. Until an answer surfaces, consider it a fluke.

We were recently visited by a salesman who tried to sell us a water filtration system. He had a metal brief case full of chemicals, so naturally I looked forward to his presentation.

            By means of a precipitation reaction, he first demonstrated that our tap water contained a certain degree of hardness caused by magnesium (Mg⁺²) and calcium (Ca⁺²) ions. All natural water contains those ions to some varying degree. Our water had 120 ppm of the stuff, which is equivalent to about 120 mg of ions for every litre of water. These ions interfere with the action of soap. To demonstrate this he filled an Erlenmeyer flask with 120 ml of my tap water and a second flask with the same volume of tap water filtered with a resin-containing column that he pulled out of his metal brief case. The resin exchanges sodium ions for Mg⁺² and Ca⁺², which bind to a negative ion on the resin (SO₃^-2), effectively softening the water. Each sample then received 5 drops of his soap. The softened water and soap, upon mixing, created a persistent foam, whereas tap water and his soap produced only a fraction of foam. Moreover, the foam quickly subsided, and the remaining solution was mirkier than the one at the bottom of the softened water.

(1)               What happened at the molecular level during his soap demonstration?

(2)               Why was his demonstration misleading and not entirely relevant to everyday washing?

(3)               Why is removing Ca⁺² and Mg⁺² from tap water not necessarily a good idea?

Solution:

(1)  Bar soap and liquid hand soap often contain a mixture of sodium stearate (C₁₈H₃₅O₂Na) and sodium palmitate. Unfortunately, in the presence of either calcium or magnesium ions, stearate and palmitate will form precipitates, which no longer act as soap molecules. For example:

2 C₁₈H₃₅O₂Na_(aq) +  Ca⁺²_(aq) --> 2 Na⁺¹_(aq) + Ca(C₁₈H₃₅O₂)_{2 (s)}

When either the stearate or palmitate were added to the treated water, there was no calcium to interfere with the soap ions.

(2)   His presentation falsely suggested that all soaps are alike. In reality, detergents contain either sodium alkylbenzenesulfonate, which in the presence of Ca⁺² and Mg⁺²_, does not precipitate to the same extent as bar soap. Older types of detergents contain phosphates, which will “fish out” the problematic calcium and magnesium ions (by forming insoluble magnesium or calcium phosphate). After some dubious excuses, the salesman agreed to repeat his experiment with my own detergent, and the differences were far less dramatic. There was only slightly less foam produced when the detergent was added to my untreated tap water.

(3)   Drinking water should not be free of Ca⁺² and Mg⁺²_.  Both of these are essential ions for the human body. Calcium ions are needed for bone formation and teeth. It also plays a role in blood clotting, muscle contraction and relaxation, nerve transmission, and enzyme(ATPase) activation. Magnesium is also needed for bones and teeth. It is a coenzyme in carbohydrate and protein metabolism and helps nerves and muscles function.