Assingment 04

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course PHY 232

004. `query 4

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Question: query problem 15 introductory problem sets temperature and volume information find final temperature.

When temperature and volume remain constant what ratio remains constant?

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Your Solution:

When temperature and volume remain constant then n and p will be constant as well.

confidence rating #$&*:e!

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Given Solution:

** PV = n R T so n R / P = V / T

Since T and V remain constant, V / T remains constant.

· Therefore n R / P remain constant.

· Since R is constant it follows that n / P remains constant. **

STUDENT QUESTION:

I don’t understand why P is in the denominator when nR was moved to the left side of the equation

INSTRUCTOR RESPONSE:

The given equation was obtained by dividing both sides by P and by T, then reversing the sides.

We could equally well have divided both sides by v and by n R to obtain

P / (n R) = T / V,

and would have concluded that P / n is constant.

To say that P / n is constant is equivalent to saying the n / P is constant.

Your Self-Critique:

The solution is very similar and requires understanding the equation.

Your Self-Critique Rating:

@&

If temperature and volume remain constant then P / n = R T / V remains constant. n and P don't remain constant, their ratio remains constant. Thus the proportional changes in n and P will be equal.

*@

Ok

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Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?

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Your Solution:

For T/V to be constant there has to be a change in volume as well as a change in temperature.

@&

Your statement isn't correct.

T / V can remain constant if neither T nor V changes.

The condition is that only T and V change, which implies that n and T remain constant so that

T / V = P / (n R)

must be constant.

*@

confidence rating #$&*:

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Given Solution:

** STUDENT ANSWER AND INSTRUCTOR RESPONSE:

They are inversely proportional. They must change together to maintain that proportion.

INSTRUCTOR RESPONSE:

You haven't justified your answer in terms of the ideal gas law:

PV = n R T so V / T = n R / P.

If only T and V change, n and P don't change so n R / P is constant.

Therefore V / T is constant, and so therefore is T / V.

You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law, using a similar strategy. **

Your Self-Critique:

The given solution is well detailed

Your Self-Critique Rating: Ok

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Question: prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost?

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Your Solution:

Since 1cal=4.184Joules then (2500cal)(4.184joules)=10460Joules

We first have to convert the joules to kilowatt then multiply by

10cents to get the cost as shown below;

1kW/hr= 0.10cents = (0.002905556)(0.10)=2.9*10^-4 approximately 29cents

0.002905556kw/hr

confidence rating #$&*:

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Given Solution:

One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 10,500,000 Joules.

A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules.

10,500,000 Joules / (3,600,000 Joules / kwh) = 3 kwh, rounded to the nearest whole kwh.

This is about 30 cents worth of electricity, and a dime per kilowatt-hour.

Relating this to your physiology:

· You require daily food energy equivalent to 30 cents’ worth of electricity.

· It's worth noting that you use 85% of the energy your metabolism produces just keeping yourself warm.

· It follows that the total amount of physical work you can produce in a day is worth less than a dime.

Your Self-Critique:

The given solution breaks it down really easy and the convertion are much straight forward to follow.

Your Self-Critique Rating:

Ok

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Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr?

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Your Solution:

Well for this case we have been given the following

Mass=1200Kg

speed=100Km/hr

so we start by converting the speed to the correct units of m/s as shown below

(100)(1000)/(3600)=27.8m/s

Since K.E=0.5*m*v^2

we substitute what we have in the equation above as shown below

KE=(0.5)(1200)(27.8)^2=463704joules

Since 1kcal=4184Joules

then ?? = 463704joules

(463704*1)/(4184)=111kcal

confidence rating #$&*:

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Given Solution:

NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities).

**STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s:

The book tells that according to energy conservation

· initial KE = final KE + heat or (Q)

· 100km/hr *3600*1/1000 = 360 m/s

INSTRUCTOR COMMENT:

100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2), not 360 m/s.

The correct conversion would be 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so.

STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS

Ke=0.5(1000Kg)(100Km)^2 = 5MJ

1Kcal=4186J

5MJ/4186J==1194Kcal

INSTRUCTOR COMMENT:

Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules.

100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx.

so

KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). **

STUDENT QUESTION:

The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly.

Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr =

1hr/3600s = 27.78 ~28m/s.

KE = 1/2mv^2

= ½(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J

470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal

INSTRUCTOR RESPONSE:

I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here.

In any case your solution is good.

Your Self-Critique:

My solution has a similar approach as the given solution.

Your Self-Critique Rating:

OK

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Question: query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, final temp.

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Your Solution:

Using the equation of (MCT)1+(MCT)2+(MCT)3=0, Where Mass= M, specific heat capacity= C and Temperature = T

We substitute the given info above into the equation and solve for the final temp as follows:

((0.4kg)*(450J/kg*c)*(25c-T))1+((0.3kg)*(450J/kg*c)*(25C-20C))2+((1.35kg)*(4186J/kg*c)*(25C-20C))=0

Therefore, 4500-180T+28930.5

33430.5=180T

T=final temp=186C

@&

Your solution is good but you need to be careful with notation.

You've used the symbol T where you intend, and apply, a change in temperature, which should be denoted `dT.

M C T implies a single value of each of the quantities M, C and T.

This notation could be misleading.

M C `dT is correct and much less likely to mislead.

*@

confidence rating #$&*:

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Given Solution:

** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters):

M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used.

0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0

Solve for T1, T1 = 214.8 Celsius

Solution below is 189.8 C.

GOOD STUDENT SOLUTION:

This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg.

For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe.

Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg.

1.35kg of water is heated by 5 degrees

· The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J of energy is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too.

mass of bucket = 0.30 kg

· specific heat of iron = 450 J/kg/degrees

· 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket

So it takes

· 675 J to heat bucket to 25 degrees celsius

· 28255 J to heat water to 25 degrees celsius

so the horse shoe transferred 675+28255 = 28930 J of energy.

Mass of horse shoe = 0.40 kg

· horse shoe is also iron

· specific heat of iron = 450 J/kg/degree

· energy transferred / mass = 28930 J / 0.40kg =72,326 J / kg

· 72 330 J / kg, at 450 (J / kg) / C, implies `dT = 72,330 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe.

A symbolic solution:

m1 c1 `dT1 + m2 c2 `dT2 + m3 c3 `dT3 = 0.

Let object 1 be the water, object 2 the pot and object 3 the horseshoe. Then `dT1 = `dT2 = + 5 C, and `dT3 = 25 C - T_03, where T_03 is the initial temperature of the horseshoe.

We easily solve for `dT3:

`dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) so

`dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) = - (1.35 kg * 4200 J / (kg C) * 5 C + .3 kg * 450 J / (kg C) ) / ((.4 kg * 450 J / (kg C) ) = -160 C, approx. so

25 C - T_03 = -160 C and

T_03 = 160 C + 25 C = 185 C, approx..

STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere.

INSTRUCTOR RESPONSE: Each of the following should be common knowledge:

· 1 liter = 1000 mL or 1000 cm^3.

· Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg.

· Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. **

Your Self-Critique:

I think the given solution has the same approach as well as the same data although there might be a few errors with conversion i think the idea

is the same.

Your Self-Critique Rating:Fair

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Question: query univ problem 18.61 / 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool.

What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem.

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Your Solution:

Since we are given some of the data will try and use this formula below

PV=nRT

First we solve for n by rearraing the equation as shown below

PV=nRT

n=PV/RT=(1.03*10^5 Pa)(1.5*10^-3 m^3)/(8.31 J/mol k)(380K)=0

therefore, 154.5/3157.8=0.049 mol

Hence, if we refer back to chemistry the atomic masses of C2H6=

C=10*2=20

H=1*6=6

Making the total to be 26g/mol

Therefore the total mass of gas will be obtained by n*Atomic mass of the whole compound

=0.049*26=1.3g

Since upon heating we expect there is to be a change in volume therefore

V=(1.5L)(380k)/(300k)=1.9L

So, (1.5L)/(1.9L)(1.3g)=1.03g

since temperature returns to 300K there will be a drop of pressure to

(1atm)(300K)/(380K)=0.79atm

confidence rating #$&*:

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Given Solution:

** use pV = nRT and solve for n.

· n = p V / (R T) = (1.03 *10^5 Pa )(1.5 * 10^-3 m^3 ) / [ (8.31 J / (mol K) )(380 K) ] = .048 mol, approx..

If the given quantities are accurate to 2 significant figures, then calculations may be done to 2 significant figures and more accurate values of the constants are not required.

The atomic masses of 2 C and 6 H add up to 30.1, meaning 30.1 grams / mol.

So total mass of the gas is initially

· m(tot) = (.048 mol)(30.1 g/mol)

· m(tot) = 1.4 g

Now if the system is heated to 380 K while open to the atmosphere, pressure will remain constant so volume will be proportional to temperature. Therefore the volume of the gas will increase to

· V2 = 1.5 liters * 380 K / (300 K) = 1.9 liters.

Only 1.5 liters, with mass 1.5 liters / (1.9 liters) * 1.4 grams = 1.1 grams, will stay in the flask.

· The pressure of the 1.1 grams of ethane is 1 atmosphere when the system is closed, and is at 380 K.

As the temperature returns to 300 K volume and quantity of gas will remain constant so pressure will be proportional to temperature.

· Thus the pressure will drop to P3 = 1 atm * 300 K / (380 K) = .79 atm, approx.. **

Your Self-Critique:

The given solution is well presented but the idea is the same as to how i approached the problem.

Your Self-Critique Rating: Ok

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Question: univ phy query problem (publisher has omitted this problem from the 12th edition) 18.62 (16.48 10th edition)

A uniform cylinder is .9 meters high, and contains air at atmospheric pressure. It is fitted at the top with a tightly sealed piston.

A little bit of mercury (density 13600 kg / m^3) is poured on top of the piston, which increases the force exerted by the piston. The piston therefore descends, compressing the confined air until the pressures equalize. Mercury continues to be added, further lowering the piston and compressing the air.

If this continues long enough, mercury will spill over the top of the cylinder. How high is the piston above the bottom of the cylinder when this occurs?

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Your Solution:

Using the PV=nRT

We use what we have to obtain what we need. H=height P=Pressure

So, (P*H)1=(P*H)2

Therefore H1=0.9m will be used in the formula below as follows

P2=Patm +rho+g*h

Patm*H1=(Patm+rho*g*y)*(H1-H)

Solving for H=0.140m

confidence rating #$&*:

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Given Solution:

** Let y be the height of the mercury column.

Since

· T and n for the gas in the cylinder remain constant we have P V = constant, and

· cross-sectional area remains constant V = A * h, where h is the height of the air column,

we have P * h = constant. Thus

· P1 h1 = P2 h2, with

P1 = atmospheric pressure = Patm and

h1 = .9 m, P2 = Patm + rho g y.

Mercury spills over when the depth of the mercury plus that of the air column is .9 m, at which point h2 = h1 - y. So the equation becomes

· Patm * h1 = (Patm + rho g y) * (h1 - y).

We can solve this equation for y (the equation is quadratic).

We obtain two solutions:

· one solution is y = 0; this tells us what when there is no mercury (y = 0) there is no deflection below the .9 m level.

· The other solution is

y = (g·h1·rho - Pa)/(g·rho) = .140 m,

which tells us that .140 m of mercury will again bring us to .9 m level.

We might assume that this level corresponds to the level at which mercury begins spilling over. To completely validate this assumption we need to show that the level of the top of the column will be increasing at this point (if the height is not increasing the mercury will reach this level but won’t spill over).

· The level of the top of the mercury column above the bottom of the cylinder can be regarded as a function f (y) of the depth of the mercury.

· If mercury depth is y then the pressure in the cylinder is Patm + rho g y and the height of the gas in the cylinder is Patm / (Patm + rho g y ) * h1. The level of the mercury is therefore

f(y) = Patm / (Patm + rho g y) * h1 + y

The derivative of this function is f ' ( y ) = 1 - Patm·g·h1·rho/(g·rho·y + Patm)^2, which is a quadratic function of y.

Multiplying both sides by (rho g y + Patm)^2 we solve for y to find that y = sqrt(Patm)·(sqrt(g* h1 * rho) - sqrt(Patm) )/ (g·rho) = .067 m approx., is a critical point of f(y).

The second derivative f '' (y) is 2 Patm·g^2·h1·rho^2/(g·rho·y + Patm)^3, which is positive for y > 0.

This tells us that any critical point of f(y) for which y > 0 will be a relative minimum.

So for y = .0635 m we have the minimum possible total altitude of the air and mercury columns, and for any y > .0635 m the total altitude is increasing with increasing y.

This proves that at y = .140 m the total height of the column is increasing and additional mercury will spill over.

To check that y = .140 m results in a total level of .9 m:

· We note that the air column would then be .9 m - .140 m = .760 m, resulting in air pressure .9 / .760 * 101300 Pa = 120,000 Pa.

· The pressure due to the .140 m mercury column is 19,000 Pa, which when added to the 101,300 Pa of atmospheric pressure gives us 120,000 Pa, accurate to 3 significant figures.

The gauge pressure will be 19,000 Pa.

A more direct but less rigorous solution:

The cylinder is originally at STP. The volume of the air in the tube is inversely proportional to the pressure and the altitude of the air column is proportional to the volume, so the altitude of the air column is inversely proportional to the pressure.

If you pour mercury to depth y then the mercury will exert pressure rho g y = 13,600 Kg/m^3 * 9.8 m/s^2 * y = 133,000 N / m^3 * y.

Thus the pressure in the tube will thus be atmospheric pressure + mercury pressure = 101,000 N/m^2 + 133,000 N/m^3 * y. As a result the altitude of the air column will be the altitude of the air column when y cm of mercury are supported:

· altitude of air column = atmospheric pressure / (atmospheric pressure + mercury pressure) * .9 m =101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m.

At the point where mercury spills over the altitude of the air column will be .9 m - y. Thus at this point

· 101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m = .9 m - y.

This equation can be solved for y. The result is y = .14 m, approx.

The pressure will be 101,000 N/m^2 + 133,000 N/m^3 * .14 m = 120,000 N/m^2.

The gauge pressure will therefore be 120,000 N/m^2 - 101,000 N/m^2 = 19,000 N/m^2. **

Your Self-Critique:

The first given solution was alittle bit confusing and not easy to follow but the direct solutions makes

a lot of sense.

Your Self-Critique Rating:Fair

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Question: query univ phy 18.79 was 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible?

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Your Solution:

Since KE=0.5mv^2=3/2kT we can rearrange the equation to solve for m=3KT/V^2

and the volume=4/3*pi*r^3 rearranging to solve for r=3sqrt(3V/4pi)

@&

You appear to be using V for both velocity and volume.

This is likely a source of confusion.

Using lower-case v for velocity and upper-case V for volume would help you keep the symbols straight.

*@

therefore r=3sqrt(9/4(0.1)^2*1000)=0.608m

@&

@&

You appear to be using an expression of the form

r = 3 sqrt(9 / (4 r^2 rho) ).

However you don't indicate how you got this expression.

You have no units in the expression

3sqrt(9/4(0.1)^2*1000)

but rather appear to simply append the anticipated units on the resulting number .608.

Had you used units in this step you would have realized that the units of your calculation do not yield the unit m, but rather the unit m / sqrt(kg).

If by 3 sqrt you mean the cube root, then you need to use the 1/3 power to indicate the cube root. 3 sqrt means 3 multiplied the the square root of whatever follows the square root sign. In any case, if you intend

r = cube root (9 / (4 r^2 rho) ),

which would be written

r = (9 / (4 r^2 rho) )^(1/3)

the units still don't work out; you will get units of m^(2/3) / kg^(1/3).

The first few lines of the given solution demonstrate a derivation of the expression for r. You should work the value of that expression out, using units throughout your calculation to verify that the result is inded in meters, and that the number is not .608.

Note also that an ice ball of this radius would weight over 1000 pounds.

*@

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and since KE=3/2KT=5.5*10^-21J

M=5.5*10^-21j/(0.5(0.1M/S^2)=1.1*10^.20kg

V=1.1*10^-20/(1000kg/m^2)=1.1*10^-23m^3

From this point i couldnt figure the rest

confidence rating #$&*:sure

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Given Solution:

** We can solve this problem knowing that ave KE per particle is 3/2 k T so the .5 m v^2 = 3/2 k T, where v is RMS velocity. Thus

· m = 3 k T / v^2.

From the density of water and the mass of the particle we can determine its volume, which is equal to 4/3 pi r^3. From this we find r.

· We obtain volume m / rho = 3 k T / (v^2 rho), where rho is the density of water.

· Setting this equal to 4/3 pi r^3 we get the equation 4/3 pi r^3 = 3 k T / (v^2 rho). The solution is

r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3).

From the mass, Avogadro's number and the mass of a mole of water we determine the number of molecules.

The following analysis shows the intermediate quantities we obtain in the process. Some of the calculations, which were done mentally, might be in error so you should redo them using precise values of the constants.

At 273 Kelvin we have ave KE = 3/2 k T = 5.5 * 10^-21 Joules.

mass is found by solving .5 m v^2 = 3/2 k T for m, obtaining m = 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.2 * 10^-14 kg.

The volume of the sphere is therefore 1.2 * 10^-14 kg / (1000 kg / m^2) = 1.2 * 10^-17 m^3.

Setting this equal to 4/3 pi r^3 we obtain radius r = ( 1.2 * 10^-17 m^3 / 4.2)^(1/3) = ( 2.8 * 10^-18 m^3)^(1/3) = 1.4 * 10^-6 m. Diameter is double this, about 2.8 * 10^-6 m. This is only 3 microns, and is not visible to the naked eye, though it could easily be viewed using a miscroscope.

A water molecule contains 2 hydrogen and 1 oxygen molecule with total molar mass 18 grams = .018 kg.

The 1.2 * 10^-14 kg mass of particle therefore consists of 1.2 * 10^-14 / (.018 kg / mole) = 6.7 * 10^-13 moles. With about 6 * 10^23 particles in a mole this consists of

6.7 * 10^-13 moles * 6 * 10^23 particles / mole = 4* 10^11 particles (about 400 billion water molecules). **

STUDENT COMMENT:

I'm still not sure about the 'visible' thing.

INSTRUCTOR COMMENT:

In any case, visible light has a wavelength between about .4 microns and .7 microns. Nothing smaller than this is visible even in principle, in the sense that its image can't be resolved by visible light.

If we mean 'visible to the naked eye', that limit occurs between 10 and 100 microns.

So this object is in principle visible (wouldn't be hard to resolve with a microscope), but not to the naked eye.

Your Self-Critique:

Prof, if you can work out this problem for me that would be great. I dont think i understand the whole idea as for mass and volume that is okay the rest seems to have been thrown

all over the place.

Your Self-Critique Rating:NOt sure

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Check my notes. I recommend that you submit a revision, according to the guidelines below:

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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