how many beta particles are emitted in 45.5 sec. by a 2.44 mg sample of Co?
546696
Ok, So you know the Time, the Starting mass we'll say Ao, and the Half life.  So by the equation:\nA/Ao = (1/2)^n,\n\nwhere A is the amount remaining, Ao is the starting amount and n is an integer, which needs to be determined.\n\nn = T/(t)\n\nWhere t is the half life and T is the time elapsed.  \n\nSo you first solve for n, and then derive the expression for A.  you should get the equation,\n\nA = ((1/2)^n)*(Ao)\n\nThat'll give you the new mass and from the mass you can use Avogadros number to determine the particles emitted.
851404
At room temperature and pressure one mole (6 x 10^23 particles/atoms/molecules) of a gas fills 24 dm^3 (Litres).\n\nThis means that in every cubic centimeter (one millilitre) of air there are 2.5 x 10^19 molecules.\n\nIf we forget about the rapid movement and think how many static molecules this means in each dimension we find that each 1-dimensional 1-cm long line contains 2,924,017 molecules (let's say 3 million to keep things simpler).\n\nOver 70% of air is Nitrogen and the length of an N2 molecule is about (keeping things simple) 100 pm (picometers; 10^-12 m) so 3 million molecules take up 300 million picometers or 300 micrometers (following conversion)\n\nThere are 10,000 micrometers in a centimeter so the molecules only 3% of the available distance.\n\nIf all 3 million molecules were equally spaced they would be approximately 3.3 nm (nanometers; 10^-9 m) apart.\n\nA division of the molecular length (100 pm) by the separation (3.3 nm) shows that they are about 33 molecules length apart.\n\nNow I've made plenty of assumptions here. I'm more than happy for people to point out where I've gone wrong (if I have). Please PM me.
795154
2CO(g) + O2(g) --> 2CO2(g)
605736
1. Automobiles = carbon monoxide\n\n2. Industry = lead, carbon monoxide, respirable particulate matter\n\n3. Agriculture = oxides of nitrogen\n\n4. Power Generation = sulfur dioxide\n\n\nHere are the pollutants.\n\nSulfur dioxide (SO2)\nAnnual average = 60 µg/m3\n24 hour = 80 µg/m3\n\nOxides of Nitrogen (NO2)\nAnnual average = 60 µg /m3\n24 hour = 80 µg /m3\n          \nSuspended Particulate Matter (SPM)\nAnnual average = 140 µg/m3\n24 hour = 200 µg/m3\n\nLead\nAnnual average = 0.75 µg/m3\n24 hour = 1.0 µg/m3\n\nCarbon Monoxide\nAnnual average = 2.0 µg/m3\n24 hour = 4.0 µg/m3\n        \nRespirable Particulate Matter (RPM)\nAnnual average = 60 µg/m3\n24 hour = 100 µg/m3
705042
No, radioactive decay is pseudo-random process. Even if you had two samples of the same material with the same mass, there is nothing to guarantee that the same number of particles would be ejected. If you decay time is small enough then statistically the two sample will yield similar results. \n\nIn your case if you assume the two samples are the same isotope, then you would expect the larger sample to emit twice as many particles as the smaller sample. The number of decays is proportional to the mass.
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
286343
The effects of carbon monoxide in parts per million are listed below:\n\n    100 ppm (0.01%) Slight headache in two to three hours\n    200 ppm (0.02%) Slight headache within two to three hours\n    400 ppm (0.04%) Frontal headache within one to two hours\n    800 ppm (0.08%) Dizziness, nausea, and convulsions within 45 minutes. Insensible within two hours.\n    1,600 ppm (0.16%) Headache, dizziness, and nausea within 20 minutes. Death in less than two hours.\n    3,200 ppm (0.32%) Headache, dizziness and nausea in five to ten minutes. Death within 30 minutes.\n    6,400 ppm (0.64%) Headache and dizziness in one to two minutes. Death in less than 20 minutes.\n    12,800 ppm (1.28%) Death in less than three minutes.
798483
Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
816363
4 Fe(s) + 3 O2(g) ----> 2 Fe2O3(s)\n\nFirst, you have to check if the chemical equation is balanced. In this case, it is. So, what it tells you is that for every 4 mole of iron that is reacted, 1652 kJ will be released to the surrounding. Now, since O2 is in excess, iron will then be your limiting reagent. If you have 3.34 mole of iron, you can use ratio to determine how much energy will be released.\n\nE = (-1652/4) *3.34 = -1379.42 kJ\n\nNote that the "-ve" sign indicates energy is released instead of being absorbed.
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
743335
First you need to balance the equation.  In order to do this, there must be the same number of each atom  on both sides.\n\nFeO(s) + C(s) --> Fe(l) + CO2(g)\n\nBalanced, the equation looks like this:\n\n2FeO + C -> 2FE + CO2\n\nNotice how there are the same number of each type of atom on both sides of the equation?  In order to make CO2, you have to have 2FeO because each FeO molecule has only one oxygen atom to provide for CO2.  This leaves 2 Fe left over after those oxygens are popped off and put on the carbon molecule.\n\nWhat the equation means is that for every 2 of FeO you have, you need 1 carbon in order to convert it into CO2 and you get 2 Fe ions left over from the reaction.\n\nIf you start out with .135 moles of FeO, you need 1/2 that amount in carbon for the reaction, because for every 2 FeO, you only need 1 carbon for the conversion.  You get the same amount of Fe as you had FeO and half that as CO2.\n\nIn other words:\n\n.135 moles of FeO plus .0675 (this is half of .135) moles of C\n\nwill be converted into:\n\n.135 moles of Fe plus .0675 moles of CO2
930311
Two per molecule of glucose
14216
Know-it-all... is wrong.\nBeing radioactive, carbon-14 eventually decays to form nitrogen-14 and a  beta particle (high speed electron). \nThe process is called beta decay.\nThe HowStuffWorks link explains it adequately
254702
Well...for CO2 I have:\n\na = 3.658 bar L^2/mol\n\nb = 0.04286 L/mol\n\n\nIf you are doing research at this level...have you forgottn about the trusty CRC handbook for Chem/Phys?\n\nThat's where I found my data. Mine is 77th edition page 6-48.\n\nHope that helps.
480675
no. depending on the exact details its either inverse beta decay or electron capture. both release energy in very small amounts tho, in the form of radiation.
87793
44.00995 grams per mole
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)