How to calculate 90 parts per million for a swimming pool w/13,500 gals?
546438
DO NOT add sodium bicarbonate! The pH is already too high. Bicarb will only COMPOUND the problem, not fix it. What you need is acid. Bicarb has an alkaline pH.
109042
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
327092
www.poolandspa.com\n\nGood Resource
315339
It's basically a swimming pool with a slab over, though you have the opportunity of using stub walls or columns to reduce the slab spans. If the tank is small the sizes will be determined by practical rather than structural considerations. I'm hoping to build such a tank later this year as part of extentions to my house.
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^
734516
Easy.  \n\nYou have q = mass * specific heat * change in temp.\n\nFind the specific heat of water and your set.  Q is joules so you need to convert.
358814
width of pool = 32\nwidth of fence therefore = 32 + 7 + 7 = 46\nlength of pool = 18\nlength of fence therefore = 18 + 7 + 7 = 32\n\nperimeter = 2(width) + 2(length)\np = 2(46) + 2(32)\np = 92 + 64\np = 156\n\n156 feet of fencing is needed.
581128
In order to achieve 600 ppm, \n600/120 pound= 5 pound product should be added to 1000 gal water.\n∴ In 10 gallon water, 5 pound * 10/1000 = 0.05 pound of product should be added. This is to be added in 30 seconds.\n\nHence, the flow rate of product is: 0.05/(1/2) = 0.1 lb/minute = 0.1 *60*24 lb/day = 144 lb/day * 1 gal/10 lb = 14.4 gal/day \n  \n\n⇒ %output = (14.4 gal/day)/(22 gal / day) = 65.45.. ←Answer
1161237
One cubic feet of water is 7.48 gallons.\nFind the volume of water in your swimming pool.\nSuppose it is V cubic feet.\n(If your pool has the same depth d feet every where, then \nV = (10)(15)(d) = 150 d)\nThen the number of gallons of water is V/7.48.\n\n(If the depth is not the same every where, you can make an estimate of the number of gallons, by taking d = average depth)
983864
First find the surface area of the tank:\n\nS.A. = 2x(lb+bh+hl) = 2(12x9 + 9x4 + 4x12) = 384 meter square\n\nSince it says 5 per meter and it is 2 meters wide, the Rs. 5 applies to 2x1 = 2 meter square of the sheet\n\nSo if ur paying Rs 5 for every 2m.sq., then u pay Rs. 2.5 for 1m.sq.\n\nTherefore, total cost = 384 x 2.5 = Rs 960
184529
ppm = parts per million = a/1000000\n% = parts per hundred = b/100\n\nLet's say there are n ppm(parts per million)\nn ppm\n= n/1,000,000\n\nTo convert to percent, the denominator should be 100\nWhat will you multiply to n/1000000 to make it n/100?\nThe answer is 10000.
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
864626
1) Calculate the initial concentration of the H+.\npH = - log [H+] = 2, therefore  [H+] = 10 ^ -2 mol / liter\n\n2) Calculate the dilution factor.\nWe want to change the concentration from 10 ^ -2 mol / liter to\n10 ^ -4 mol / liter. Therefore the dilution factor is 100.\n\n3) Calculate the amount of water to add.\nThe dilution factor is 100. We therefore want to have 100 x 14.5 ml = 1450 ml after dilution. We therefore have to add 1450 - 14.5 = 1435 ml of water
869360
depends on how much you want, see the link\n\nhttp://www.poolandgarden.com/aqclearstabc6.html
1044477
the smaller tank is 900L and the bigger tank is 1200L\n\nTank 1 = T\nTank 2 = T+300\n\n(2/3) T = (1/2) (T+300)\n(2/3) T = (1/2) T + 150\n(4/6) T = (3/6) T + 150\n(1/6) T = 150\nT = 900
652740
Girl Chemists are soO SexY!  We need more of them. =)\n\nThe level of active chlorine is called the free chlorine residual and should measure 1.5 - 3.0 ppm, regardless of size or volume of the pool.  How to achieve the correct ppm depends on the chlorine product you choose to use.  Once you pick your pool chlorinating product it should have a detailed instruction to make the correct concentration of pool water.  There are kits with paper indicators that you can dip into your pool water after you add Chlorine solution or tablets to see if you have the correct amount of chlorine dissolved in the water.  If you still need help, you know where to come back and ask.
198515
Let's see, if the circumference of the pool is 60 feet and you want to surround the pool (with a gate i'm assuming...not that it matters) so that there is an area of 6 feet between the edge of the pool and the gate (or whatever it is) then you need to first find the diameter. (diameter=distance from one side of the circle to the other which also intersects the center of the circle) You would model this equation as 3(pi)r, (3(pi)r=the formula for the circumference of a circle) 3(pi)r=60. divide both sides by 3 so that the equation reads (pi)r=20.  then divide both sides by (pi) so that the equation reads r=6.366197723675814ft.  You multiply that by 2 so that you get the diameter of the pool so that the equation now reads d=12.732395447351628ft or d=12ft 9in.  You then add 12 feet to that number (6ft extension on either side of the pool) so that the diameter is now 24.732395447351628ft or 24ft 9in.  you divide that by 2 because the formula for circumference of a circle deals with the radius (one-half the diameter) of the circle, not the diameter.  so your model for this situation reads as follows: 3(pi)12.366197723675814ft(or 12ft 4.5 in.)=116.5486677646163 (or 116ft 6.5in) which is how many feet of gate or whatever it would take to surround the pool 6 feet away from the edge of the pool.