No, the 88g cartridge is much larger than it's 12g counterpart and will not fit the same applications. Also, the 88g AirSource cartridge has a threaded neck and screws into place, whereas the 12g powerlet is clamped in place.
Molecular weight of CO2 is about 44g (12g Carbon + 32g Oxygen *2 atoms Oxygen). So that means if you are presented with 44g CO2, that means there's 12g Carbon.
usually it is 67% copper and 33% zinc it can range from 55% to 95%
Carbon dioxide emissions vary depending on the fuel. Natural gas (the cleanest fuel) produces about 117 lb CO2/106BTU. Coal, on the oter hand, produces about 225 lb CO2/106BTU. Liquid fuels (gasoline, diesel and furnace fuel) are in the middle of the range.
C = 1 * 12.01 g = 12.01 g O = 2 * 16.00 g = 32.00 g CO2 = 44.01 g (12.01 g/44.01 g) * 100% = 27.289% 0.27 * 88 g = 23.76g Or more accurately.. 0.27289 * 88 g = 24.01 g There is about 24 grams of carbon in 88 grams of carbon dioxide.
the atomic number is 88
Define real? If you mean like a actual gun which shoots bullets, no. If you mean real as in not fiction, yes. There are several CO2 powered rifles, such as the Beretta CX-4 Storm which is powered by a 88 gram CO2 cartridge.
To get a 12 mass% sucrose sol'n dissolve 12.0 g sucrose in 88.0 gram (near 88 ml) of water.
There are 1000 milligrams in a gram, so there are 88000 milligrams in a 88 grams.
The result of 100 minus 12 is 88. This can be calculated by subtracting 12 from 100, which leaves you with 88. Mathematically, this can be written as 100 - 12 = 88.
88% of 12 = 0.88 x 12 =10.56
12 / 88 = 0.136363636
No. 88 is not evenly divisible by 12.
294. To do this quickly, the "trick" is to recognise that 12 + 88 = 100 Then using the laws of commutativity and associativity, 12 + 194 + 88 = 194 + 12 + 88 = 194 + (12 + 88) = 194 + 100 = 294
To find out how many times 12 goes into 88, you can divide 88 by 12. When you do this calculation, 88 ÷ 12 equals approximately 7.33. Therefore, 12 goes into 88 a total of 7 times with a remainder.
100-12=88 200-112=88 240-152=88 1626-1538=88
88 football fields
To find the total mass of CaO that reacts with 88 grams of CO2 to produce 200 grams of CaCO3, we first need to calculate the molar mass of CaCO3 and CO2, then determine the moles of each reactant involved in the reaction. Finally, we use the stoichiometry of the reaction to find the mass of CaO required to react completely with 88 grams of CO2.