about 29 decemeter cube or litre.....
Yes. CO is the poisonous gas carbon monoxide.At STP and in all environmental conditions, it is a gas.It is created by the incomplete combustion of some fuels, and can prevent oxygen from entering the bloodstream by binding with the hemoglobin. In moderate to high concentrations in the air, it can be seriously harmful or fatal.
You stated the amount of carbon you had, but how much water (separated into its constituents by electrolysis, otherwise if you dump water on carbon you wind up with nothing more than wet carbon) is available? Let's assume you've got an excess of it and go from there.In a real-life situation you won't get any hydrogen gas. In order of reactivity are hydrogen, oxygen and carbon. One of three things will happen here.The most likely reaction is that the hydrogen and oxygen will recombine into water and, once again, you've got wet carbon...that is, unless the heat from the very exothermic 2H2 + O2 -> 2H2O reaction sets the carbon on fire, in which case it'll scavenge atmospheric O2 and you'll receive some CO2, some CO and a little free carbon, aka "soot."Under different circumstances, you might get some CH4 and some O2. And because you have 1.07 moles C, you'll wind up with 1.07 moles methane - giving you 23.968 liters of methane at STP.If you have a really nice lab that can emulate photosynthesis, you could convert the three elements into glucose. That's not too likely; you will probably wind up with the same block of carbon and glass of water that you started out with.The chemical reaction should be written as shown below : C + H2O -----> CO + H2 The balanced chemical reaction equation indicates that 1.0 mole of H2 gas is produced for each mole of carbon that reacts. Therefore you have : n H2 = (1.07 mole C ) ( 1.0 mole H2 / 1.0 mole C ) = 1.07 moles H2 At STP, there are 22.7 L per mole of ideal gas. Therefore the H2 liters at STP is given by : V H2 at STP = ( 22.7 L at STP / mole ideal gas ) ( 1.07 moles H2 ) V H2 at STP = 24.3 L of H2 at STP
In the relationship CO equals HRSV, SV stands for stroke volume.
CO, or carbon monoxide, is a gas at standard temperature and pressure. It exists in a gaseous state under normal conditions.
no, CO2 does
1.799 kg/m3 at 25 Co .
Using the ideal gas law - the volume of a gas is independent of it composition and is determined solely by the equation PV=nRT. As one mole of CO would produce one mole of CO2 it would take 541 mL of CO to produce 541 mL of CO2.
First convert the number of grams of CO2 into moles, then use the Ideal Gas Law. For how to solve this problem, see the two Related Questions links to the left of this answer.
Not enough information to answer ! You can't make Iron from CO gas.
Yes. CO is the poisonous gas carbon monoxide.At STP and in all environmental conditions, it is a gas.It is created by the incomplete combustion of some fuels, and can prevent oxygen from entering the bloodstream by binding with the hemoglobin. In moderate to high concentrations in the air, it can be seriously harmful or fatal.
There are several different possible reactions of Fe2O3 with CO, depending on temperature and ratio of reactants. The simplest is probably Fe2O3 + CO ==>2FeO + CO21.00 Kg x 1000 g/Kg x 1 mole Fe2O3/160 g = 6.25 moles Fe2O3 moles CO2 produced = 6.25 moles CO2 Volume CO2 at STP = 6.25 moles x 22.4 L/mole = 140 Liters
ok well standard temperature and pressure is stp 22.4 litres per mole of gas. first write a balanced chemical equation 2C + 02 ------> 2CO then write mole ratios 2C + 02 ------> 2CO 2 : 1 : 2 using gay lussacs law and avogadros hypothesis we know that the same volumes of gases in same conditions are mole ratios (only works with gases however) so therefore the volume of CO gas produced will be 2 x that of the volume of oxygen gas (because ratio of O2 gas to CO gas is 1:2) =2x2.24 =4.48 litres of CO gas produced then the moles can be worked out by - number of moles = volume given/volume of 1 mole at STP = 4.48/22.4 =0.2 moles roughly i think. Answer: 1.20 g C is 1/10 mole of C 2.24 liters of O2 is 1/10 mole of O2So when equal volumes molar anounts of C and O2 react: C (1 mole)+ O2(1 mole)-> CO2 (1 mole) (No CO is produced) With the given initial amounts 1/10 mole (2.24 liters) of CO2 would be producedIf oxygen were limited all C could be converted to CO. This would rturn the inital amount of C (1,2 g) and 1.12 liters of O2 into 1/10 mole of CO (2.24 liters)
density = mass/volume = 0.196g/100ml = 0.00196 g/ml = 0.00196 g/cm³
100ml = 1 dm3 0.196g = 196x10-6kg Density = mass/volume Density of CO = 196x10-6 / 1 = 196x10-6 kg/dm3
3,5 g of CO is equivalent to 2,8 L.
You stated the amount of carbon you had, but how much water (separated into its constituents by electrolysis, otherwise if you dump water on carbon you wind up with nothing more than wet carbon) is available? Let's assume you've got an excess of it and go from there.In a real-life situation you won't get any hydrogen gas. In order of reactivity are hydrogen, oxygen and carbon. One of three things will happen here.The most likely reaction is that the hydrogen and oxygen will recombine into water and, once again, you've got wet carbon...that is, unless the heat from the very exothermic 2H2 + O2 -> 2H2O reaction sets the carbon on fire, in which case it'll scavenge atmospheric O2 and you'll receive some CO2, some CO and a little free carbon, aka "soot."Under different circumstances, you might get some CH4 and some O2. And because you have 1.07 moles C, you'll wind up with 1.07 moles methane - giving you 23.968 liters of methane at STP.If you have a really nice lab that can emulate photosynthesis, you could convert the three elements into glucose. That's not too likely; you will probably wind up with the same block of carbon and glass of water that you started out with.The chemical reaction should be written as shown below : C + H2O -----> CO + H2 The balanced chemical reaction equation indicates that 1.0 mole of H2 gas is produced for each mole of carbon that reacts. Therefore you have : n H2 = (1.07 mole C ) ( 1.0 mole H2 / 1.0 mole C ) = 1.07 moles H2 At STP, there are 22.7 L per mole of ideal gas. Therefore the H2 liters at STP is given by : V H2 at STP = ( 22.7 L at STP / mole ideal gas ) ( 1.07 moles H2 ) V H2 at STP = 24.3 L of H2 at STP
In the relationship CO equals HRSV, SV stands for stroke volume.