A steel cylinder contains 32 dm^3 of hydrogen at 4.0 x 10^6 Pa and 39 degree C. Calculate:?
100414
Sorry but uses the information above for more infor.\n\n\n STP = 1106.2 liters\n\n and\n\n 99.5g of H2
270860
You will need to know the mass of the material inside.\n\n1.  Subtract that from the overall mass.  The remaining mass will be the steel cylinder itself.\n\n2.  Find the volume of the steel by multiplying the mass by the density factor.\n\n3.  Volume of the cylinder itself is:\n\n  [((r1^2) * pi)  - ((r2^2) * pi)] * L  = Vol of Cyl.\n\nWhere: \n\n   r1 is the outside radius \n    (known, this is half the measured diameter)\n   \n   r2 is the inside diameter (unknown)\n    \n   L  is the length of the cylinder\n\n   Vol of Cyl is the mass times the density\n\n\n4. Solve for r2 with the previous equation:\n\n  [((r1^2) * pi)  - ((r2^2) * pi)] * L  = Vol of Cyl. \n\n\n  r2 = SQRT {[((r1^2) * pi) - (Vol of Cyl * L)] / pi}\n\n5.  r1 - r2 is your cylinder wall thickness.
730844
OK...You can start with the equation of DG (I'm using D instead of writing "delta").\n\nDG = DH -TDS  You've seen that equation, right?  OK..Since the boiling point of water is 100 degreed C at 1 atm pressure, we are really talking about an equilibrium situation.  So, DG= 0.\n\nConsequently, DH = TDS.  \n\nSince you know DH and you know T (in Kelvin), you can calculate DS.  Make sure that you convert DH into Joules (instead of kilojoules) before you finish the problem.
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
899235
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
187220
W=F*d,  d=W/F,  d=600J/400N = 1.5m
358957
Volume = pi * r^2 * h\n\nr is 1/2 the radius of the drilled hole = (D'/2)\nh is equivalent to the diameter of the original cylinder = D
844835
H2          +        I2=>      2HI\n1mol                 1mol     2mol\n1,75                  1,75      3,5mol\n\nK=[HI]^2/([H2][I2])\n\n[H2]=2mol/dm^3-1,75mol/dm^3=\n=0,25mol/dm^3\n[I2]=2mol/dm^3-1,75mol/dm^3=\n=0,25mol/dm^3\n\nK=[3,5]^2/(0,25 x 0.25)=12,25/0.0625=196\n\nSo it is D (196)
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
1010084
Propane  = C3H8. The molar mass is\n(3 x 12 + 8 x 1) gram = 44 gram.\nYou give 44.1 g. That a little bit better.\n\nThe volume of the cylinder is\nAh = pi r2 h = 3.14 x (0.050 m)^2 x 1.30 = 0.212 m3.\n\nBefore: pV = nRT\n1.4×10^6 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(begin) = 119 mol.\n\nAfter: pV = nRT\n2.4×10^5 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(end) = 20 mol.\n\nUsed is (119 - 20) mol propane, being 99 x 44.1 gram = 4.4 kg
412854
I don't know what you plan on using this information for but I'm sure it it very similar to other steels. For many calculations small errors in the number have little or no practical effect on the answer.\n\nMy heat transfer book gives the following values:\n\nMild steel 36 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\nMild steel 33 Btu/(hr)(sqft)(degF/ft) @ 212 degs F\n\n1% Carbon steel 26.2 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\n1% Carbon steel 25.9 Btu/(hr)(sqft)(degF/ft) @ 212 degs F
466950
Volume = Area x height\nArea = r x r x pi\npi is approximated by the fraction 22/7\nTherefore Volume of cylinder = 4 x4 x 22/7 x7 = 352 units^3
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
750654
Find the percentahe of hydrogen first, by subtracting from 100.\nDivide each percentage by the Ar of that element, and then divide by the smaller of those answers to get a whole-number ratio. That is the empirical formula.\n\nNow work out n = pV/RT for the gas, remembering to convert to the correct units.\n\nThen calculate the Mr, using the mass given. You can now compare the Mr and the empirical formula to get the molecular formula.
742342
29.444..°C\n\nC = (F - 32) * (5/9)\nC = (85 - 32) * (5/9)\nC = 53 * 5/9\nC = 265 / 9\nC = 29 4/9° C
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
625417
pH= -log(6.0 x10^-3)\npH is about 2.22