What is the weight percent of C in CO2?
404126
Wt percent:\n\nC atomic weight: 12\nCO3 molecular Wt : 44\n% wt = 12X100/44 = 27.3%
431250
Three ways to interpret this question.\n\n1. As a percentage of all the carbon dioxide in the atmosphere, which is ~380ppm or 600 x 10^15 g C. In this case we add <1% per year. The atmospheric carbon dioxide is then projected to double, at present rates of increase, in just over 100 years.\n\n2. As a percentage of the net input flux of carbon of which we are responsible for about 95%.\n\n3. As a percentage of the net total flux of carbon, of which we are closer to 15-20%, which I think somebody already mentioned.
88535
Ca(OH)2 \n0.185 g/100cc at 0deg C\n        0.077 g/100cc at 100deg C
909109
Carbon Dioxide and Water
891726
atomic mass unit, 1/12 of the mass the carbon 12, the most common carbon isotope. Try a periodic table.\n\n\n**** \n1 atomic mass unit = 1.66053886 × 10-27 kilograms
949367
First you have to get the molecular weight of those.\n\nits\n180 for C6H12O6\n192 for 6O2\n264 for 6CO2\n108 for H2O\n\na. Since you have the same amount of C6H12O6 and O2, your limiting reagent in this case is the O2.\n\nSo you use it to determine the CO2 produced.\n\nb. if 192 grs of O2 produce 264 grs of CO2\nthen 100 grs will do: (100 * 264)/180 = 137.5 grs of CO2.\n\nc. Now, to calculate the percent yield, in ideal conditions you would have gotten 137.5 grs (which is your 100% theorically) but practically you got 97.1 grs. so\n\n137.5 ___ 100%\n97.1______ ?\n\n(97.1*100)/137.5 = 70.62% is the perecnt yield of CO2.\n\nd. You do the same thing backwards to determine how many excess of the non-limiting reagent (C6H12O6) you have.\n\nto produce 264 grs of CO2 you need 180 grs of C6H12O6\nfor 137.5...\n\n(137.5*180)/264 = 93.75 grs of C6H12O6\n\nso the excess is 100 - 93.75 = 6.25 grs of C6H12O6
365485
I assume you mean molecular weight.\n\nSince carbon weights 12 grams, and oxygen weighs 16:\n12+(16*2) = 44 grams.\n\nYou have to give the amount of CO^2, in ML to get Moles.\n\nSo generically speaking, 1 mol of CO^2 weighs 44 grams, 2 mol weights 88 grams, and so on.\n\nTo calculate the amount of atoms in it, you would have to times it with Avogadro's number. (6.0221415 × 10)
150708
Oxygen and carbon dioxide, symbolized O2 and CO2 respectively, are colorless, odorless gases. The atmosphere, or air around us, contains approximately 21 percent oxygen and 78 percent nitrogen. There is almost no CO2 in air (about 0.03 percent)
685823
Strictly speaking, it is not true.  CO2, being heavier than air, WILL accumulate on ground level, if the air is perfectly still for a long enough period of time.  Or if the CO2 gas is colder than the air.  \n\nKeep in mind, though, that since the molecular weights of CO and of air are pretty close, the relative rate of diffusion of CO in air will be fairly slow.  That means if you have something making CO and putting it into the air, the worst place to be is right next to it.  Especially if the air is still.
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
924197
PCO2 = 35-45\nPO2 = 75-100
1100315
It is carbon dioxide molecules produced devided by oxygen molecules consumed. In humans it is roughly 0.8
897570
Dimensional analysis is easier than it first seems.\n\nFirst you need to know the molecular weight of CO2.\n\nCarbon +2(Oxygen) = 12 + 2(16) = 44 grams per mole\n\n[The above step is not dimensional analysis.]\n\nLet's say somebody asks you how many moles 440 grams of CO2 is.\n\nThis is where we do dimensional analysis.\n\n(1 mole CO2 / 44 grams CO2) * (440 grams CO2 / ? CO2)\n\nThe / sign indicates "divided by", so imagine the parts over each other.  That makes it easier but I cannot do it here in text.\n\nThe first part of the above equation is stating that for every mole of CO2, there are 44 grams of CO2.  The second part says that we have 440 grams of CO2, and want to find out how many moles we have.\n\nThe trick here is to have the terms, one above the other, so that the units cancel out and you end up with the desired unit.  Our desired unit here is the mole - we want to know how many moles are in 440 grams of CO2.  So, that is why in the above, (1 mole CO2) is on top of (44 grams CO2).  If we allow 440 grams to be on top in the next term, then grams cancels out, and moles is the only unit left.  This is what we want.\n\nSo if we solve the above\n\n(1/44)*(440/1) = 440/44 = 10 moles of CO2.\n\nSo, 440 grams of CO2 is the same as 10 moles of CO2.\n\nhope this helps\n\nyay
255550
You can't convert volume (cubic centimeters) to mass (pounds) unless you know a value for density. \n\n5800 cc equals 6.12 US quarts or 12.25 US pints, or 353.9 cubic inches.
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
912377
A has a higher percent carbon than B.\n\nThe molecular mass of A is 194 g/mol and it contains 96 g/mol of carbon.  96 g/mol carbon /194 g/mol molecule A = 0.495 = 49.5% carbon\n\nThe molecular mass of B is 342 g/mol and it contains 144 g/mol of carbon.  144 g/mol carbon/342 g/mol molecule B = 0.421 = 42.1% carbon.
116688
from the equation: 2CO + O2 -----> 2CO2,\n\n0.69 g CO2 x (1 mol CO2/ 44 g CO2) x (2 mol CO/ 2 mol CO2) x (28 g CO/ 1 mol CO)\n\n\nthe answer is 0.4391 g of CO needs to react with O2 to produce 0.69 g of CO2