How do you work out the percentage of combustion from 2 figures?
954537
weight of original material... weigh the ash, take one from teh other, equals the percentage of combustion... or how much was left after you burned it...
924197
PCO2 = 35-45\nPO2 = 75-100
764912
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.
553400
take the mole fraction and multiply by the atomic weight of each which you can find on the periodic table. For carbon, it would be 12 x .04 = .48 Once you have all three, sum them and divide each by this sum. So, say you find Ch and I for the other two. Then the percentage by weight of Carbon would be the number you got for Carbon (.48) divided by (.48 + Ch + I).\n\nTo double check, your weight percentages should all add up to 100%.
102536
First use the ideal gas law to find out how much oxygen gas reacts with the butane.\n\nPV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nn=PV/RT=[(101325 Pa)(0.015 m^3)]/[(8.314 J/mol K)(298 K)]\nn=0.613 moles O2\n\nNow write the balanced equation for butane combustion.\n\n2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O\n\nNow use stoichiometry to find the moles of butane.\n\n(0.613 mol O2)(2 mol C4H10/13 mol O2)= 0.094 moles C4H10.\n\nNow convert moles C4H10 to grams by multiplying by molar mass. The molar mass for butane is 58 g/mol.\n\nMass butane = (0.094 moles)(58 g/mol) = 5.47 grams butane
369810
Find the fraction/decimal/percent/whatever. Then, find how big this is for 360 degrees. For example, say you have 41%. Just find 41% of 360. If this is a problem, then just multiply the fractin/decimal/percent/whatever by 360.
258044
divide the exponent by two
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
918301
.06 bal2 + .0501 mgso4 = x BaSO4\n2.150 L x (square root of .06) = .5266401\n1.980 L X  .0501 = .099198\n.099198 / .5266401=.1883601\nTHEREFORE YOUR % PERCENTAGE YEILD IS 18.83%
855371
Look at like this. Whatever the amount of the first component\nis, consider it 100%, then for the second component, it would be 80% of the first.\nExample:\nYou have 500 gm of component 1,\nyou put 80%of that or 400 gm of component 2.\nThe total gms would be 900 gm, 100 parts component 1 to\n80 parts component 2.\nHope this helps.
250535
Assuming the reaction is elementary:\n\nK = [03]^2/[02]^3\n\n1.8*10^-7 = [03]^2/0.0012^3\n\nSolving for [03], O3 concentration is 1.76*10^-8 M.  (Actually plus or minus, but the negative concentration is meaningless).
672347
you can work that in your head just by moving the point:\n10% is 443.7\n1% is 44.37\nso 443.7 + 44.37 + 44.37 = 532.44 !
891286
Percentage composition means calculating the percentage (by mass) of an atom in its molecule. For instance, we would say:\n\n"Calcium Hydroxide (Ca(OH)2) is x% Calcium, y% Oxygen, and z% Hydrogen"\n\nTo find the percentage composition, first find the total mass of 1 mole of the molecule:\n\nFor Ca(OH)2, there is 1 Ca atom, 2 O atoms, and 2 H atoms. The atomic masses are as follows:\nCa = 40.08 g/mol. \nO = 16.00 g/mol\nH = 1.01 g/mol\n\nThe atomic mass of 1 mol of Ca(OH)2:\nCa + 2*O + 2*H = 40.08 + 2*16.00 + 2*1.01 = 74.1 g/mol\n\nTo find the percentage of Ca, divide the mass of Ca in the molecule by the total mass of the molecule:\n\n%Ca = 40.08 / 74.1 = 0.54 = 54%\n\nDo the same thing with O and H, but remember that there are TWO molecules of O and H in each molecule of Ca(OH)2:\n\n%O = 32 / 74.1 = 0.43 = 43%\n\n%H = 2.02 / 74.1 = 3%\n\n\nNow, repeat this process with the other four:\n\nNaHCO3 = Na + H + C + 3*O = 22.97 + 1.01 + 12.01 + 3*16 = 83.99 g/mol\n%Na = 22.97 / 83.99 =  0.27 = 27%\n%H = 1.01 / 83.99 = 0.01 = 1%\n%C = 12.01 / 83.99 = 0.14 = 14%\n%O = 48 / 83.99 = 0.57 = 57%\n\nCaO = Ca + O = 40.08 + 16.00 = 56.08 g/mol\n%Ca = 40.08 / 56.08 = 0.71 = 71%\n%O = 16.00 / 56.08 = 0.29 = 29%\n\nCaCO3 = Ca + C + 3*O = 40.08 + 12.01 + 3*16.00 = 100.09 g/mol\n%Ca = 40.08 / 100.09 = 0.40 = 40%\n%C = 12.01 / 100.09 = 0.12 = 12%\n%O = 48.00 / 100.09 = 0.48 = 48%\n\nAl2(SO4)3 = 2*Al + 3*S + 12*O = 2*26.98 + 3*32.07 + 12*16.00 = 342.17 g/mol\n%Al = 53.96 / 342.17 = 0.16 = 16%\n%S = 96.21 / 342.17 = 0.28 = 28%\n%O = 192.00 / 342.17 = 0.56 = 56%\n\nI hope this helps!
741999
whats the equation or formula u need help with?
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
987563
Using proportions you can do the math later to find out how to get to the numbers, but basically it goes something like this;\n\nA = 4.00 - 3.50 = 100% - 90%\nB = 3.49 - 3.00 = 89% - 80%\nC = 2.99 - 2.50 = 79% - 70%\nD = 2.49 - 2.00 = 69% - 60%\nF = 1.99  -  0    = 59%  -  0%