Here's a prob on boyle's law?
827779
An intubated puffin or pelican using a helix might form an opinion depending on the relative humidity of the biosphere. Elastic limits could be reached, in which case garters would be up for grabs. The issue of effervescence is something that lends itself to elision, but backwardation, triangulation, and differential calculus all offer salutary remedial solutions.
702914
Actually there's no good relationship because Boyle's Law assumes constant temperature.  Unless the air is 98.7 degrees Farenheit, breathing is going to involve the air changing temperature as well as expanding / experiencing pressure change.\n\nHowever, at 98.7 degrees farenheit, one could surmise that temperature is (mostly) constant, so then Boyle's law is clearly shown:  pressure inside a full lung is quite high, but when that air is exhaled it expands into a much larger space around the breathing person and the pressure (should) drop.  Assuming you aren't 10,000 leagues under the sea.
645900
Boyles Law PV/T I think. Anyway it has to do with the relationship of Pressure to Volume. If one increases the other decreases. Its useful in industry in making liquid Nitrogen and has endless applications in dealing with gasses in Chemisrty and Physics.
820189
Because the volume, pressure, and temperature are all related by Boyles Law and the Ideal Gas Law.\n\npv=nRt\n\nPressure times Volume equals the Number of molecules times the Gas Constant time the temperature.
248683
Sounds like its something related to 'entropy' otherwise known as the 2nd Law of Thermodynamics. "What can go wrong, must go wrong" as Murphy would say.
687306
Check the Michaelis-Menten equation.It's too long to write it all down for you ;)
1187434
I think that's dalton's law.  Whoever came up with it, the pressure is additive.  So convert so everything is in the same units, multiply by the percent, add, and then compare to the answers.  I bet it's more fun than you predict.  Whooo!  Conversion factors!  I don't want to take that kind of enjoyment away, so get cracking!
741934
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
561732
water displacement and boyles law of bouyancy
534799
Yes.  \n\nBoyle's Law states that the volume of a fixed amount of gas at constant temperature is inversely proportional to the pressure exerted on the gas.  In other words, as pressure rises volume decreases and as pressure decreases volume increases.
1076679
According to Boyle's Gas Law, the product of the pressure and volume of an ideal gas is a constant.\n\nTherefore, P1V1 = P2V2\nHere, P1 = 5 psi, V1 = 20 cubic inches, P2 = 10 psi.\n\nPutting into the equation,\n5*20 = 10*?\n\nAnswer is 10 cubic inches - B in the list of choices.
56363
Dalton's law states that the total pressure is equal to the sum of the individual partial pressures of the gases in the mixture: therefore if the vapor pressure of water @ 21.5 degrees C is 2.6kPa and the total pressure in the flask is 99.4kPa then your answer should show 99.4-2.6=96.8kPa\n\nHope that helps...
794147
You might work up some Q&A while showing this old demo of the effect of atm pressure...thereby adding some visual interest to the subject.\n\nAdd a few oz of water to a square-sided, metal, one-gallon can having a tightly fitted cap...bring to a vigorous boil (cap now off, please :-))..) and continue boiling until all air is expelled...screw cap in place and allow to cool...they will see the can collapse.\n\nWork up questions about the vacuum thus created (by the condensation of the water vapor) allowing atm pressure to exert its force.
302322
I won't do your homework for you. But I will tell you this:\n\n\n1a&b. These should be common sense. Go outside and turn on the garden hose and tell your teacher what you've observed.\n2a. Look up BERNOULLI'S LAW as it pertains to fluid pressure through a tube. There is a formula that related pressure to hose diameter and volume.\nb. The answer is: high pressure is bad for vessels weakened by atherosclerosis.
903371
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
963774
Use the fact that PV/T = a constant, we find the constant by pluging in the values we have, T = 20C, but we must convert that to K first so 20C = (273 + 20) = 293K.  And 1 atm should be converted to physicall units of Nt/m^2 for an easier solution.  1 Atm = 101325 Nt/m^2.    So our constant is (101325)(4.5)/293 = 1.336 x 10^7 so now you have one equation as\n\nPV/T = 1.336 x 10^7 setting T = 260C or 533K gives\n1)  PV = 7.121 x 10^10\n\nThe next equation that we need is the force on the cylinder end given the pressure, which would be\n\n2)  .01P = F\n\nThe spring will stretch a distance 3)  d = F / (2.00 x 10^3)  \n\nThe volume changes by  4)  V = 4.5 + d * .010   \n\nThis gives you 4 equations which you can now solve to find the solution for V, d, P and F
933014
All decent explanations so far; I hope I can add to what has already been posted.\n\nLiquid and Ice most definitely can exist at the same temperature and Ice can be and often does get colder than 0C.\n \nFirst of all the Triple Point of water is the pressure and temperature at which Ice, Liquid Water and Water Vapor exist at the same time in equilibrium.  Vapor is the right word, not steam although you can argue that the only time that Water Vapor and Steam are the same thing is at a very low pressure.  The Triple point of water only happens at a near perfect vacuum, 0.0063 ATM to be precise.\n\nAdding Salt to water will lower the freezing point, but then it isn't water anymore, it's salt water.  Even so, Liquid Salt Water and Ice Salt Water can also exist at the same place and time at the same temperature, slightly below 0C.  But let's stick to water.\n\nIce can easily be lower than 0 degrees.  Consider a bucket of water, liquid water sitting outside and the air temperature is -10C.  Little by Little, as the liquid reaches 0.0098 C, liquid starts to become ice.  The ice crystals form in the bucket, around the walls of the bucket and on the surface first.  The air is still -10C so the bucket, the liquid and the ice will keep getting colder.  The ice on the surface will drop to slightly below 0C but very slowly since it is still absorbing the heat of the not-yet-frozen water.  Eventually, the entire bucket of water will be frozen and will be just below 0C.  First the walls of the bucket will reach -10C (same as the air tempertaure).  The ice near the walls of the bucket and on the surface will reach -10C first, continually absorbing the heat energy from the relatively warmer ice near the center of the bucket.  Until finally, the entire bucket of ice is steady at -10C.  That is until the sun comes out or the day gets warmer.\n\n\nAnd by the Way, dry ice is not water, it's frozen CO2 (Carbon Dioxide)\n\nOne more thing, at a near perfect vacuum, you can actually cool liquid water to less than 0C without it freezing.  Never tried that but I did accidently freeze a beer last night...